8Geometriy 7 1. Introduction In this chapter we will look at the fundamental concepts we need in order to begin our study of geometry. A. POINT, LINE, AND PLANE Definition The word geometry comes from two Greek words, ‘geo’ and ‘metric’, which together mean ‘to measure the earth.’ Geometry is now the branch of mathematics that studies space, shape, area, and volume. geometry Nature displays an infinite array of geometric shapes, from the smallest atom to the biggest galaxy. Snowflakes, the honeycomb of a bees’ nest, the spirals of seashells, spiders’ webs, and the basic shapes of many flowers are just a few of nature’s geometric masterpieces. The Egyptians and Babylonians studied the area and volume of shapes and established general formulas. However, the first real book about geometry was written by a Greek mathematician, Euclid. Euclid’s book, The Elements, was published in about 300 BC. It defined the most basic concepts in geometry and proved some of their properties. Geometry as a science has played a great role in the development of civilization. Throughout history, geometry has been used in many different areas such as architecture, art, house design, and agriculture. After studying this section you will be able to: 1. Understand the fundamental geometric concepts of point, line, and plane. 2. Describe the concepts of line segment, ray, and half line. 3. Understand the concepts of plane and space. 4. Describe the relation between two lines. 5. Describe the relation between a line and a plane. 6. Describe the relation between two planes. Objectives 9 Geometric Concepts The three most basic concepts of geometry are point, line, and plane. Early mathematicians tried to define these terms. In fact, it is not really possible to define them using any other concepts, because there are no simpler concepts for us to build on. Therefore, we need to understand these concepts without a precise definition. Let us look instead at their general meaning. 2. Point When you look at the night sky, you see billions of stars, each represented as a small dot of light in the sky. Each dot of light suggests a point, which is the basic unit of geometry. 3. Line Nature’s Great Book is written in mathemati- cal symbols. Galileo Galilei All geometric figures consist of collections of points, and many terms in geometry are defined using points. We use a dot to represent a point. We name a point with a capital letter such as A, B, C, etc. Concept A point is a position. It has no size, length, width, or thickness, and it is infinitely small. point Concept A line is a straight arrangement of points. It is the second fundamental concept of geometry. There are infinitely many points in a line. A line has no width or thickness, and extends without end in both directions. line 10 Geometriy 7 Let none unversed in geometry enter here. Plato The arrows at each end of a line show that the line extends to infinity in both directions. If any point C is on a line AB or a line d, we write C AB, or C d. C C Î AB A B A B C d A, B, C Î d Property There exists exactly one line passing through any two distinct points. By this property, a line is determined by two distinct points. However, remember that a line consists of more than just two points. There are infinitely many points on a line. 4. Plane A plane is suggested by a flat surface such as a table top, a wall, a floor, or the surface of a lake. We represent a plane with a four-sided figure, like a piece of paper drawn in perspective. Of course, all of these things are only parts of planes, since a plane extends forever in length and in width. We use a capital letter (A, B, C, ...) to name a plane. We write plane P, or (P), to refer to a plane with name P. Concept A plane is the third fundamental concept of geometry. A plane has length and width but no thickness. It is a flat surface that extends without end in all directions. plane A line is usually named by any two of its points, or by a lower-case letter. Look at the diagram. The line that passes through points A and B is written AB. We say it is line AB. The line on the right is simply called line . 11 Geometric Concepts For example, in the diagram above, points A, B, and C lie on the same line d. Therefore A, B, and C are collinear points. However, point P is not on line so M, P, and N are not collinear points. We say that, M, P, and N are noncollinear points. A B C d P M N l 5. Collinear Points Definition Points that lie on the same line are called collinear points. collinear points EXAMPLE 1 Look at the given figure. a. Name the lines. b. Write all the collinear points. c. Give two examples of noncollinear points. Solution a. There are three lines, AC, CN, and SR. b. The points A, B, C, the points S, T, R, and the points C, M, N are on the same line, so they are collinear. c. The points A, N, C and the points M, T, N are not on the same line. They are exam- ples of noncollinear points. Now consider the three noncollinear points in the figure on the right. Since we know that two distinct points determine a straight line, we can draw the lines AB, AC and BC passing through A, B, and C. Therefore, there are three lines that pass through three noncollinear points. A B C P S T N R M C A B d m l We can now understand the meaning of the terms point, line, and plane without a formal definition. We can use these undefined terms to define many new geometric figures and terms. 12 Geometriy 7 When we say, n triwise noncollinear points, we mean that any three of n points are noncollinear. For example, the diagram opposite shows five triwise noncollinear points. Any set of three points in the diagram is noncollinear. Definition If three points are noncollinear then they are also called triwise points. triwise points A E C B D Theorem different lines pass through n triwise points. ( – 1) 2 n n EXAMPLE 2 How many different lines pass through each number of triwise noncollinear points? a. 4 b. 5 c. 9 d. 22 Solution a. b. c. d. = = 22 (22 – 1) 22 21 231 2 2 = = 9 (9 – 1) 9 8 36 lines 2 2 = = 5 (5 – 1) 5 4 10 lines 2 2 = = 4 (4 – 1) 4 3 6 lines 2 2 Check Yourself 1 1. Describe the three undefined terms in geometry. 2. Name the collinear points in the figure. 3. Look at the figure. a. Name the lines. b. Name all the collinear points. c. Give two examples of noncollinear points. A C D B G E F A B C E D 13 Geometric Concepts 4. How many different lines can pass through each number of triwise noncollinear points? a. 8 b. 14 c. 64 d. 120 Answers 1. Point: A point is a position. It has no size, length, width, or thickness, and it is infinitely small. Line: A line a straight arrangement of points. There are infinitely many points in a line. A line has no width or thickness, and extends without end in both directions. Plane: A plane has length and width but no thickness. It is is a flat surface that extends without end in all directions. 2. The points A, B, C and the points D, B, E are collinear points. 3. a. The lines: AC, AB, DG b. The points A, E, B, the points D, B, G are collinear points c. The points A, F, G, and the points D, B, C are non collinear points. 4. a. 28 b. 91 c. 2016 d. 7140 1. Line Segment B. LINE SEGMENT, RAY, AND HALF LINE Definition The line segment AB is the set of points consisting of point A, point B, and all the points between A and B. A and B are called the endpoints of the segment. We write [AB] to refer to the line segment AB. line segment A B line segment AB or [AB] This definition describes one type of line segment: a closed line segment. There are three types of line segment. a. Closed Line Segment A line segment whose endpoints are included in the segment is called a closed line segment. [AB] in the diagram is a closed line segment. b. Open Line Segment A line segment whose endpoints are excluded from the segment is called an open line segment. The line segment AB in the diagram is an open line segment and denoted by ]AB[. We use an empty dot ( ) to show that a point is not included in a line segment. A B A B Physical model of a line segment: a piece of string. 14 Geometriy 7 EXAMPLE 3 Name the closed, open and half-open line seg- ments in the figure on the right. Solution Closed line segment: [AB] Half-open line segments: [AC[, [BC[, and [BD[ Open line segment: ]CD[ A B C D Property If C is a point between A and B, then [AC] + [CB] = [AB]. Using this property, we can conclude that if three points are collinear, then one of them is between the other points. Point B is between the points A and C. A B C A B C [AC] [CB] [AB] c. Half-Open Line Segment A line segment that includes only one of its endpoints is called a half-open line segment. half-open line segment AB [AB[ A B half-open line segment AB ]AB] A B 2. Ray In the diagrams, each ray begins at a point and extends to infinity in one direction. A is the endpoint of [AB, and C is the endpoint of [CD. ray AB, or [AB A B [CD D C Definition The ray AB is the part of the line AB that contains point A and all the points on the line segment that stretches from point A through point B to infinity. The ray AB is denoted by [AB. ray 15 Geometric Concepts A half line extends to infinity in one direction. A half line is like a ray, but it begins at an open endpoint. A B half line AB ]AB 1. Plane We can think of the floor and ceiling of a room as parts of horizontal planes. The walls of a room are parts of vertical planes. A point can be an element of a plane. In the diagram, point A is an element of plane P. We can write A (P). Similarly, B (P), C (P), D (P), and E (P). E D P A C B plane P: (P) vertical planes horizontal planes C. PLANE AND SPACE a. Coplanar Points In the figure, points A, B, and C are all in the plane P. They are coplanar points. Points K, L, and M are also coplanar points. A, K, and M are not coplanar points, because they do not lie in the same plane. Definition Points that are in the same plane are called coplanar points. coplanar points P A C B M L K Q 16 Geometriy 7 b. Coplanar Lines For example, in the figure, the lines m and n are both in the plane P. They are coplanar lines. Definition Lines that are in the same plane are called coplanar lines. coplanar lines P m n 2. Space We have seen that lines and planes are defined by sets of points. According to the definition of space, all lines and planes can be considered as subsets of space. Definition Space is the set of all points. space In the figure, the plane P is determined by the noncollinear points A, B, and C. P A C B Theorem For any three points, there is at least one plane that contains them. For any three non- collinear points, there is exactly one plane that contains them. 1. Intersecting Lines Two lines that intersect each other in a plane are called intersecting lines, or concurrent lines. In the figure on the left, line d and line l intersect each other at point A. They are intersecting lines. P l d A D. RELATION BETWEEN TWO LINES 17 Geometric Concepts 2. Parallel Lines Two lines are parallel if they are in the same plane and do not have a common point. In the figure on the left, line d and line l are parallel lines. We write d l to show that lines d and l are par- allel. 3. Coincident Lines Two lines are coincident if each one contains all the points of the other. In the figure on the left, line d and line l are coincident lines. We write d = l to show that lines d and l are coincident. 4. Skew Lines Two lines are skew if they are non-coplanar and they do not intersect. In the figure on the left, E and F are two non-parallel planes. Hence, lines d and l are in different planes, and since they do not intersect, they are skew lines. l d d l and d Ç l = Æ l d d = l F d E l EXAMPLE 4 In the figure there are three intersecting lines. Decide whether each statement is true or false. a. point A is the intersection of l and d b. point C is the intersection of d and l c. point B is the intersection of l and m Solution a. True, since point A is the common point of l and d. b. False, since point C is not a common point of d and l. c. True, since point B is the common point of l and m. P l d A B C m 18 Geometriy 7 We have seen the different possibilities for the relation between two lines. Let us look at the possible relations between a line and a plane. 1. The Intersection of a Line and a Plane A line can intersect a plane at one point. In the diagram on the left, the line d intersects the plane E at point A. 2. Parallelism of a Line and a Plane A line can be parallel to a plane. In the diagram on the left, there is no common point between line d and plane E. They are parallel. 3. A Line Lies in a Plane If at least two points of a line lie in a plane, then the line lies in the plane. We write d (E) to show that line d lies in plane E. In the diagram, points A and B are in plane E, so the line AB lies in the plane E. E. RELATION BETWEEN A LINE AND A PLANE F. RELATION BETWEEN TWO PLANES E A d Ç (E) = {A} d E d d Ç (E) = Æ d E A B A, B Î d A, B Î (E) d Î (E) 1. Parallel Planes If two planes have no common point, they are called parallel planes. We write (A) (B) to show that two planes are parallel. The opposite walls of a room are an example of parallel planes. P Q (P) (Q) (E) (F) E F 19 Geometric Concepts 2. Intersecting Planes If two planes have only one common line, they are called intersecting planes. 3. Coincident Planes If two planes have three noncollinear points in common, they are called coincident planes. (P) and (Q) in the figure are coincident planes. We write (P) = (Q) to show that planes P and Q are coincident. 4. Half Planes A line in a plane separates the plane into two disjoint regions that are called half planes. (E 1 ) and (E 2 ) in the figure are half planes of (E). E d A B F (E) Ç (F) = d A, B, C Î (P) A, B, C Î (Q) (P) = (Q) P A Q B C E l half plane (E 1 ) half plane (E 2 ) boundary of two half planes (E 1 ) Ç (E 2 ) = Æ (E 1 ) È (E 2 ) È l = (E) EXERCISES 1.1 20 Geometriy 7 1. Explain why the concepts of point, line, and plane cannot be defined in geometry. 2. Draw five points on a piece of paper, and make sure that no three are of them collinear. Draw all the lines passing through these points. How many lines can you draw? 3. Explain the difference between a ray and a half line. 4. At least how many points determine a line? 5. At least how many noncollinear points determine a plane? Why? 6. Give examples from daily life to illustrate the concepts of point, line, and plane. 7. Write words to complete the sentences. a. A point has no __________ and no __________. b. Two points determine a ___________ . c. Three noncollinear points determine a _____ . d. Two lines that lie in different planes and do not intersect are called ___________ lines. 8. Determine whether the statements are true or false for the given figure. a. A, B, and C are collinear points b. points D and E are not in the line l c. B l d. E l e. C, D, and E are noncollinear A B C l E D 13. Write the coplanar points in the given figure. 14. Draw a diagram to show that the intersection of two planes can be a line. 15. Draw a diagram to show that the intersection of three planes can be a point. 12. Describe the intersection of the line and the plane in each figure. 16. Look at the figure. a. How many planes are there? b. Write the intersection of the planes. c. How many lines pass through each point? 11. Write the meaning of the following. a. [CD] b. [PQ[ c. ]AB[ d. [KL e. ]MN f. EF 10. Name all the lines, rays, line segments, and half lines in the given figure. A B C D E F G H L M N P R S K L P Q T S R E D A B C G n A E B l F C m 9. How many different lines can pass through each number of triwise noncollinear points? a. 5 b. 7 c. 21 d. 101 a. b. c. 21 Chapter Review Test 1A CHAPTER REVIEW TEST 1A 1. Which concept is precisely defined in geometry? A) point B) line C) plane D) space E) ____ 2. A plane has no A) thickness. B) length. C) width. D) surface. E) ____ 3. A ray with an open endpoint is called A) a line. B) a half line. C) a line segment. D) an open line segment. E) ____ 4. According to the figure, which statement is true? A) A, B and E are collinear points B) l d = {B} C) C l D) D, B, and E are noncollinear points E) _ l d D C E B A 5. According to the figure, which statement is false? A) l d = {C} B) l m = {A} C) l d m = {A, B, C} D) m d = {B} E) ____ l d A B C m 8. Space is A) the intersection of two planes. B) the set of all points. C) a subset of a plane. D) a very large plane. E) ? 7. How many lines do five points determine if no three of the points are collinear? A) 15 B) 12 C) 10 D) 9 E) ? 6. ABCD is a rectangle in a plane P. E is a point such that E (P). How many planes are there that include point E, with one or more of points A, B, C, and D? A) 7 B) 8 C) 9 D) 10 E) ? 22 Geometriy 7 12. Let [AB] + [BC] = [AC], and [MN] + [NK] = [MK]. Which points are between two other points? A) A and M B) B and M C) C and K D) B and N E) ____ 9. According to the figure, which statement is false? A) C (E) B) l (E) = {A} C) l d = {B} D) l and d are skew lines E) ? E A C d B l 11. According to the figure, which statement is false? A) (P) l = l B) (P) m = m C) (P) n = n D) l m n = {A} E) ? n A P m l 10. Which figure shows ]AB? A) B) C) D) E) ____ A B A B A B A B 24 Geometriy 7 1. Angle One of the basic figures in geometry is the angle. A television antenna is a physical model of an angle. Changing the length of the antenna does not change the angle. However, moving the two antennae closer together or further apart changes the angle. A. REGIONS OF AN ANGLE Definition An angle is the union of two rays that have a common endpoint. The rays are called the sides of the angle. The common endpoint is called the vertex of the angle. angle Look at the diagram. [BA and [BC are the sides of the angle. The vertex is the common endpoint B. The symbol for an angle is Z. We name the angle in the diagram ZABC, or ZCBA, and say ‘angle ABC’, or ‘angle CBA’. We can also name angles with numbers or lower-case letters, or just by their vertex. A B C side vertex s i d e Note In three-letter angle names the letter in the middle must always be the vertex. After studying this section you will be able to: 1. Define the concept of angle and the regions an angle forms. 2. Measure angles. 3. Classify angles with respect to their measures. 4. Classify angles with respect to their positions. 5. Classify angles with respect to the sum of their measures. Objectives 25 Angles EXAMPLE 1 Name the angles in the diagrams. Solution a. ZAOB or ZBOA b. ZA c. Z1 d. Za 1 a A O B A a. b. c. d. EXAMPLE 2 Answer the questions for the angle ABC on the right. a. Which points are in the interior region of the angle? b. Which points lie on the angle? c. Which points are in the exterior region of the angle? Solution a. The points D and E are in the interior region of the angle. b. The points A, B, C, and H lie on the angle. c. The points G and F are in the exterior region of the angle. E D A H C B G F Definition The region that lies between the sides (arms) of an angle is called the interior region of the angle. The region that lies outside an angle is called the exterior region of the angle. interior and exterior region of an angle B. MEASURING ANGLES Angles are measured by an amount of rotation. We measure this rotation in units called degrees. One full circle of rotation is 360 degrees. We write it as 360°. We can show the size of an angle on a diagram using a curved line between the two rays at the vertex, with a number. When we write the size of an angle, we write a lowercase m in front of the angle symbol. For example, mZAOB = 45° means that angle AOB measures 45° degrees. Look at some more examples of angle measures in the diagrams. 360° B A O B 45° Babylonian astronomers chose the number 360 to represent one full rotation of a ray back on to itself. Why this number was chosen? It is because 360 is close to the number of days in a year and it is divisible by 2, 3, 4, 5, 6, 8, 9, 10, 12, and many other numbers. mÐE = 360° mÐC = 90° mÐB = 30° mÐA = 10° 10° A 30° B 90° C 150° D 360° E mÐD = 150° 26 Geometriy 7 Notice that the symbol for a 90° angle is a small square at the vertex. A 90° angle is also called a right angle in geometry. It is important to read angles carefully in geometry problems. For example, an angle in a problem might look like a right angle (90°). However, if it is not labelled as a right angle, it may be a different size. We can only use the given information in a problem. We calculate other information using the theorems in geometry. 0° 10° 20° 30° 40° 50° 60° 70° 80° 90° 100° 120° 130° 140° 150° 160° 170° 0° 10° 20° 30° 40° 50° 60° 80° 90° 100° 110° 120° 130° 140° 150° 160° 170° 180° 0 1 2 3 4 5 6 7 8 9 10 180° 110° 70° To measure angles with a protractor, follow the three steps below. 1. Place the central hole (dot) of the protractor on the vertex of the angle. 2. Place the zero measure on the protractor along one side of the angle. 3. Read the measure of the angle where the other side of the angle crosses the protractor’s scale. 0° 10° 20° 30° 40° 50° 60° 70° 80° 90° 100° 120° 130° 140° 150° 160° 170° 0° 10° 20° 30° 40° 50° 60° 80° 90° 100° 110° 120° 130° 140° 150° 160° 170° 180° 0 1 2 3 4 5 6 7 8 9 10 180° 110° 70° 0° 10° 20° 30° 40° 60° 70° 80° 90° 100° 120° 130° 140° 150° 160° 170° 180° 0° 10° 20° 30° 40° 50° 60° 70° 80° 90° 100° 110° 120° 140° 150° 160° 170° 180° 0 110° 1 2 3 4 5 6 7 8 9 10 50° 130° Notice that there are two semicircular scales of numbers on the protractor. If the angle measure is smaller than 90° then we read the angle using the scale with the smaller number. If the angle measure is greater than 90° then we use the scale with the larger number. Definition The geometric tool we use to measure angles on paper is called a protractor. A protractor has a semi-circular shape and a scale with units from 0 to 180. protractor EXAMPLE 3 Read the protractor to find the measure of each angle. a. mZAOB b. mZAOC c. mZAOD d. mZAOE e. mZAOF f. mZBOC g. mZCOF h. mZDOE F A 0° 10° 20° 30° 40° 50° 60° 70° 80° 90° 100° 120° 130° 140° 150° 160° 170° 0° 10° 20° 30° 40° 50° 60° 80° 90° 100° 110° 120° 130° 140° 150° 160° 170° 180° 0 1 2 3 4 5 6 7 8 9 10 180° 110° 70° E D C B O 27 Angles Solution a. mZAOB = 22° b. mZAOC = 68° c. mZAOD = 90° d. mZAOE = 140° e. mZAOF = 175° f. mZBOC = mZAOC – mZAOB = 68° – 22° = 46° g. mZCOF = mZAOF – mZAOC = 175° – 68° = 107° h. mZDOE = mZAOE – mZAOD = 140° – 90° = 50° For example, let us use a protractor to draw an angle of 56°. 1. Draw a ray. 2. Place the centre point of the protractor on the endpoint (A) of the ray. Align the ray with the base line of the protractor. 4. Remove the protractor and draw [AC. 3. Locate 56° on the protractor scale. Make a dot at that point and label it as C. B A B A C A B C 56° mÐBAC = 56° After learning to how use a protractor we can easily draw and measure angles. Check Yourself 1 1. Name the following angles. 2. Find the following sets of points in the figure. a. ZL · {X} b. int ZL · {X} c. ext ZL · ZL d. int ZL · ext ZL e. int ZL · {S} f. ZL · {T, S, K} g. int ZL · {Z, Y} h. ext ZL · {Z, Y} O B A A 3 b X T M S L K Z Y a. b. c. d. 28 Geometriy 7 3. Measure each angle using a protractor. 4. Draw the angles. a. 45° b. 83° c. 174° d. 180° e. 225° Answers 1. a. ZAOB b. ZA c. Z3 d. Zb 2. a. C b. {x} c. C d. C e. C f. {S, K} g. C h. {Z, Y} a. b. c. d. D C B A We can classify angles according to their measures. 1. Acute Angle An angle that measures less than 90° is called an acute angle. The angles on the left are all examples of acute angles because they measure less than 90°. 2. Right Angle An angle that measures exactly 90° is called a right angle. The angles on the left are all examples of right angles because they measure exactly 90°. We use a special square symbol at the vertex to show a right angle. 3. Obtuse Angle An angle that measures between 90° and 180° is called an obtuse angle. The angles on the left are all obtuse angles. 4. Straight Angle An angle that measures exactly 180° is called a straight angle. In the diagram, ZA is a straight angle. C. TYPES OF ANGLE WITH RESPECT TO THEIR MEASURES C 80° A 25° B 45° M N O O Y 120° 165° X 91° A 180° 29 Angles 5. Complete Angle An angle that measures exactly 360° is called a complete angle. In the diagram, ZE is a complete angle. 360° E EXAMPLE 4 Classify the angles according to their measure. Solution a. 180° is a straight angle. b. 360° is a complete angle. c. 125° is between 90° and 180°, so it is an obtuse angle. d. 90° is a right angle. e. 35° is less than 90°, so it is an acute angle. 35° 125° 360° 180° a. b. c. d. e. D. TYPES OF ANGLE WITH RESPECT TO THEIR POSITION In the diagram, the angles ZAOC and ZBOC have a common vertex and a common side ([OC) with non-intersecting interior regions. Therefore, ZAOC and ZBOC are adjacent angles. EXAMPLE 5 Determine whether the pairs of angles are vertical or not, using the figure. a. Za, Zb b. Za, Zc c. Zd, Za d. Zb, Zd Solution The lines l and k intersect at one point. Therefore, a. Za and Zb are not vertical angles, b. Za and Zc are vertical angles, because they are in opposite directions, c. Zd and Za are not vertical angles, and d. Zb and Zd are vertical angles. c b d a l k 1. Adjacent Angles Definition Adjacent angles are two angles in the same plane that have a common vertex and a common side, but do not have any interior points in common. adjacent angles A B O C Theorem If two angles are vertical then they are also congruent, i.e. they have equal measures. 30 Geometriy 7 E. TYPES OF ANGLE WITH RESPECT TO THE SUM OF THEIR MEASURES Each angle is called the complement of the other angle. For example, in the diagram opposite, ZANB and ZCMD are complementary angles, because the sum of their measures is 90°: mZANB + mZCMD = 30° + 60° = 90°. 1. Complementary Angles Definition If the sum of the measures of two angles is 90°, then the angles are called complementary angles. complementary angles 30° 60° A B C D N M In the diagram, ZXYZ and ZMNO are supplementary angles because the sum of their measures is 180°: mZXYZ + mZMNO = 40° + 140° = 180°. 2. Supplementary Angles Definition If the sum of the measures of two angles is 180°, then the angles are called supplementary angles. Each angle is called the supplement of the other angle. supplementary angles Z X M O 140° 40° N Y EXAMPLE 6 Find x if the given angles are complementary. Solution a. If Zx and Z2x are complementary, then x + 2x = 90°. Therefore, x = 30°. b. 2x + 10° + 3x + 30° = 90° 2x + 3x + 10° + 30° = 90° 5x = 50° x = 10° 2x+10 3x+30 x 2x a. b. 31 Angles EXAMPLE 7 Find x if the given angles are supplementary. Solution a. If Z2x and Z4x are supplemen- tary, then 2x + 4x = 180°. Therefore, x = 30°. b 2x + 60° + 3x + 50° = 180° 2x + 3x + 60° + 50° = 180° 5x + 110° = 180° 5x = 70° x = 14° 4x 2x 3x+50° 2x+60° a. b. Check Yourself 2 1. Find x if the given angles are complementary. 2. Find x if the given angles are supplementary. Answers 1. a. 18° b. 30° c. 9° 2. a. 20° b. 25° c. 30° 6x 3x 4x+40° 2x 10 5x 12° 2 x 1 8 ° a. b. c. 2 x + 3 0 ° 3 x + 1 5 ° x 20° 2x+20° x 4x a. b. c. 32 Geometriy 7 A. CORRESPONDING ANGLES AND ALTERNATE ANGLES Definition Let m and n be two lines in a plane. A third line l that intersects each of m and n at different points is called a transversal of m and n. In the diagram, line AB is a transversal of m and n. supplementary angles Definition In a figure of two parallel lines with a transversal, the angles in the same position at each intersection are called corresponding angles. corresponding angles In the diagram, Z1 and Z5 are corresponding angles. Also, the angle pairs Z2 and Z6, Z3 and Z7, and Z4 and Z8 are corresponding angles. 1. Corresponding Angles l 5 6 7 8 1 2 3 4 m n m//n Let us look at the types of angle formed in the figure of two parallel lines with a transversal. Remember the notation for parallel lines: m n means that m is parallel to n. Property Corresponding angles are congruent. Therefore, in the diagram, mZ1 = mZ5, mZ2 = mZ6, mZ3 = mZ7, and mZ4 = mZ8. l 5 6 7 8 1 2 3 4 m n m//n After studying this section you will be able to: 1. Identify corresponding angles, alternate interior angles, and alternate exterior angles. 2. Identify interior angles on the same side of a transversal. 3. Describe the properties of angles with parallel sides. 4. Define an angle bisector. Objectives 33 Angles In the diagram, the angles Z4 and Z6 are alternate interior angles. Also, Z3 and Z5 are alternate interior angles. 2. Alternate Interior Angles l 5 6 7 8 1 2 3 4 m//n m n Property Alternate exterior angles are congruent. Therefore, in the diagram, mZ1 = mZ7, and mZ2 = mZ8. l 7 8 1 2 m//n m n Definition In a figure of two parallel lines with a transversal, the interior angles between the parallel lines on opposite sides of the transversal are called alternate interior angles. alternate interior angles In the diagram, the angles Z1 and Z7 are alternate exterior angles. Also, Z2 and Z8 are alternate exterior angles. 3. Alternate Exterior Angles l 5 6 7 8 1 2 3 4 m n m//n Definition In a figure of two parallel lines with a transversal, the angles outside the parallel lines on opposite sides of the transversal called alternate exterior angles. alternate exterior angles Property Alternate interior angles are congruent. Therefore, in the diagram, mZ4 = mZ6, and mZ3 = mZ5. l 5 6 3 4 m n m//n 34 Geometriy 7 4. Interior Angles on the Same Side of a Transversal Definition In a figure of two parallel lines intersected by a transversal, interior angles on the same side of the trans- versal are supplementary. Therefore, in the diagram, mZx + mZy = 180°. alternate interior angles l y x m n m//n If [BA [CF then ZABC and Zx are supplementary. mZABC + mZx = 180° 100° + mZx = 180° mZx = 80° ZEDC and Zy are alternate interior angles. mZEDC = mZy mZy = 30° mZBCD = mZx + mZy = 80° + 30° = 110° A B C D E 100° 30° x F A B C D E 100° 30° y F 30° 80° B D C EXAMPLE 8 In the digaram, [BA DE. Find mZBCD. Solution A B C D E 100° 30° 5. Angles with Parallel Sides Theorem The measures of two angles with parallel sides in the same direction are equal. Proof Consider the diagram on the right. 1. ZAOB and ZLTB are corresponding angles. ZAOB = ZLTB 2. ZLTB = ZKLM (corresponding angles) = ZAOB = ZKLM = mZAOB = mZKLM O A B L K M S T [OA [LK [OB [LM 35 Angles Proof A B O L K M R P [OA // [LM [OB // [LK mZCAB = mZFDE 2x + 40° = 6x – 20° 40° + 20° = 6x – 2x 60° = 4x 15° = x So mZCAB = 70°. EXAMPLE 9 In the figure, [AC [DF, [AB [DE, mZCAB = 2x + 40°, and mZFDE = 6x – 20°. Find mZCAB. Solution A C B 6x 20° E F 2x + 40° D mZAOB = mZOAE + mZOBF mZAOB = 40° + 30° = 70° EXAMPLE 10 In the figure, [AE [BF, mZA = 40°, and mZB = 30°. Find mZAOB. Solution E A B F O ? 40° 30° Theorem The measures of two angles with parallel sides in opposite directions are equal. 1. ZKLM = BPL (corresponding angles) 2. ZAOB = ZBPL = ZAOB = ZKLM Property In the figure, if d k and B is the intersection of [BA and [BC, then mZb = mZa + mZc. d k a b c A B C 36 Geometriy 7 Property In the figure, if d k and B is the intersection of [BA and [BC, then mZa + mZb + mZc = 360°. a b c A C B d k Property In a figure such as the figure opposite, the sum of the measures of the angles in one direction is equal to the sum of the measures of the angles in the other direction. mZx + mZy + mZz = mZa + mZb + mZc + mZd A B C D E F H G K L M a b c d z y x mZFHC + mZFHD = 180° (supplementary angles) mZFHC = 180° – 2x mZAEF + mZEFH + mZFHC = 360° 5x + 3x + 180° – 2x = 360° 6x + 180° = 360° 6x = 180° x = 30° EXAMPLE 11 In the figure, AB CD, mZAEF = 5x, mZEFH = 3x, and mZFHD = 2x. Find x. Solution 5x 3x 2x E H F A B C D 35° + 25° = 4x + x 60° = 5x x = 12° Therefore, mZABC = 48°. EXAMPLE 12 In the figure, [AE [DF, mZEAB = 35°, mZBCD = 25°, and mZABC = 4· mZCDF. Find mZABC. Solution A E B C D F 35° 4x 25° x AB CD 37 Angles Property In the diagram, if [OA ± [LK and [OB ± [LB then mZAOB + mZNLB = 180°. K A N L O B EXAMPLE 13 In the figure, [BA ± [FD, [BC ± [FE, [BA [FG, mZABC = 60°, and mZGFE = x. Find x. Solution Let us draw a line [BF parallel to [ED. mZDEB + mZEBF = 180° 118° + mZEBF = 180° mZEBF = 62° mZBAC + mZABF = Z180° mZBAC + 85° = 180° mZBAC = 95° EXAMPLE 14 In the figure, [AC [ED, mZABE = 23°, and mZBED = 118°, Find mZBAC. Solution 23° A B C D E 118° 23° A B C D E 118° 62° F Two lines are called perpendicular lines if they intersect at right angles. We write AB ± CD to show that two lines AB and CD are perpendicular. (interior angles on the same side of a transversal) A D C B O mZGFD = 90° mZGFD + mZGFE + mZDFE = 360° 90° + mZGFE + 120° = 360° 90° + x + 120° = 360° x + 210° = 360° x = 150° mZABC + mZDFE = 180° 60° + mZDFE = 180° mZDFE = 120° 60° x D A G F B E C 38 Geometriy 7 In the figure, [OB is the bisector of ZAOC: mZAOB = mZBOC = ·ZAOC. 1 2 O a a B A C Let us draw a line [BK parallel to [AD. mZDAB+mZABK =180° 112° + mZABK =180° mZABK =68° mZBCE =mZABC + mZABK 120° =mZABC + 68° = mZABC =52° EXAMPLE 15 In the figure, [AD [CE, mZDAB = 112°, and mZBCE = 120°. Find mZABC. Solution 112° 120° ? A B C E D 68° K 112° 120° ? A B C E D 6. Bisector of an Angle Definition A ray that divides an angle into two congruent angles is called the bisector of the angle. angle bisector mZABD + mZCBD = 180° mZEBF = Z Z 90° m ABD m CBD + = 2 2 EXAMPLE 16 In the figure, [BE and [BF are the bisectors of ZABD and ZCBD respectively. Find mZEBF. Solution A B C D E F Definition The bisectors of two adjacent supplementary angles are perpendicular to each other. 39 Angles Let [OH AG. mZHOC = mZGCF = c mZHOB = mZCBE = b mZHOD = mZBAD = a Let mZCOB = mZBOA = x. c + x = b = x = b – c = b + x = a b + b – c = a Therefore, a + c = 2 · b. EXAMPLE 17 In the figure, [OE is the bisector of ZFOD. mZBAD = a, mZEBC = b, and mZFCG = c. Show that a + c = 2· b. Solution E F C G B A O a b c D E F C G B A O a b c D H c EXERCISES 2 40 Geometriy 7 1. Using the given fig- ure, find each set of points. a. ZO · {P} b. ZO · {N} c. ZO · {K, O, M} d. int ZO · {P} e. int ZO · {N} f. int ZO · {K, O, M} g. ext ZO · {N} h. ext ZO · {P} i. ZO · int ZO j. ZO · ext ZO k. int ZO · ext ZO · ZO 4. Draw the angles. a. 20° b. 35° c. 75° d. 120° e. 175° f. 210° g. 240° h. 330° M O K P N 2. Find each set of points for the given figure. a. ZABC · ZACD b. int ZABC · ZCAD c. ZABC · int ZCAD d. ext ZABC · ZCAD e. ZABC · ext ZCAD B C E F D G H A 3. Measure the angles using a protractor. 360° 135° 45° 5. Classify the types of angle. x 4x+30° 2x+25° 3 x 3 5 ° 3x5° x + 1 5 ° 6. Find x in each figure if the angles are complementary. 3x 2x+20° 125° + 2x 45° x 4x+25° 3x+15° 7. Find x in each figure if the angles are supplementary. 8. In the figure, m n, l is a transversal and mZ7 = 115°. Find the measures. a. mZ1 b. mZ2 c. mZ3 d. mZ4 e. mZ5 f. mZ6 g. mZ8 l 6 7 8 5 2 3 4 1 m n a. a. b. d. e. c. a. a. b. c. b. c. b. c. d. e. f. g. h. i. 41 Angles 9. Given [BA [DE, find mZx. x 30° 80° A E B C D 10. In the figure, [BA [ED, [BC [EF, mZABC = 3x – 30°, and mZDEF = 4x – 70°. Find x. B A C D 3x30° E 4x70° K F 13. In the figure, [BA [DE, mZBCD = 40°, and mZCDE = 120°. Find mZABC. A B D E C x 120° 40° 14. In the figure, [BA [FG, mZEFG = 120°, and mZABC = 130°. Find mZx. G A F E D C 120° 130° x B 15. In the figure, [BA [ED] and [CD] [EF. Find the relation between x, y, and z. F A B C D E z y x 16. In the figure, [BA [EF, mZBCD = 100°, mZCDE = 25°, and mZFED = 105°. Find mZABC. x 100° 105° A B C D E F 25° 17. In the figure, d l. Find x. 30° 3x 40° x 2x d l 18. In the figure, [BA [DE, [BC [DF and [BC ± [BD], and mZGDE = 40°. Find mZABC. 40° A B C E D F G 19. In the figure, [BC ± [DF, [BA ± [DG, and [ED] is the angle bisector of mZGDF. Find mZABC. 140° G A B C F D E 20. In the figure, [AB [CD. Find mZAEC. 2x 4x 2x 60° B D A C E 21. In the figure, AB CD. Find mZBFC. 70° A D B F E C 11. In the figure, d l. Find mZx. l d 115° x 12. In the figure, d l. Find mZx. l d 50° 100° x CHAPTER REVIEW TEST 2 42 Geometriy 7 11. In the figure, [BC is the angle bisector of ZABC. Find mZx. A) 65° B) 55° C) 50° D) 45° x 85° 125° A B C 1. The complement of an angle x is 10° more than three times mZx. Find the measure of the bigger angle. A) 50° B) 60° C) 70° D) 80° 2. The sum of the measures of the supplementary and complementary angles of an angle x is 250°. Find mZx. A) 10° B) 20° C) 30° D) 40° 3. What is the measure of the angle between the bisectors of two adjacent supplementary angles? A) 45° B) 60° C) 75° D) 90° 4. In the figure, [OA ± [OB, mZAOC = a, mZCOB = b, and . Find b. A) 30° B) 36° C) 54° D) 60° 2 3 a = b 6. In the figure, l k. Find mZx. A) 110° B) 100° C) 90° D) 80° a b A C B O 5. The ratio of two complementary angles is . Find the measure of the supplementary angle of the smaller angle. A) 170° B) 160° C) 150° D) 110° 2 7 k x 115° 35° l 7. In the figure, m n, mZKAB = 130°, and mZLCD = 40°. Find mZABC. A) 100° B) 90° C) 80° D) 70° n m 130° 40° x B A K L D C 8. In the figure, d l. Find x. A) 40° B) 30° C) 20° D) 10° d l 3x 6x10° 4x20° 9. In the figure, [AB] ± [BE. Find mZx. A) 30° B) 40° C) 50° D) 60° x 70° 60° 130° A B D C E 10. In the figure, [BC ± [DE and [BA ± [DF. What is the relation between mZx and mZy? A) mZx + mZy = 90° B) mZx + mZy = 180° C) mZx = mZy D) mZx – mZy = 30° F A D B E C x y 44 Geometriy 7 A. THE TRIANGLE AND ITS ELEMENTS The roofs of many buildings have a triangular cross-section. A triangle makes a simple musi- cal instrument, and many traffic signs have a triangular shape. These are just some exam- ples of how triangles are used in the world around us. In this section we will consider the main features of triangles and how we can use them to solve numerical problems. 1. Definition The word triangle means ‘three angles’. Every triangle has three angles and three sides. Definition triangle, vertex, side A triangle is a plane figure which is formed by three line segments joining three noncollinear points. Each of the three points is called a vertex of the triangle. The segments are called the sides of the triangle. The plural of vertex is vertices. We name a triangle with the symbol A followed by three capital letters, each corresponding to a vertex of the triangle. We can give the letters in any order, moving clockwise or counterclockwise around the triangle. A B C Challenge! Without lifting your pencil, join the following four points with three segments to form a closed figure. Activity Make a poster to show how triangles are used in everyday life. You can take photographs, make drawings or collect pictures from magazines or newspapers to show buildings, designs, signs and artwork which use triangles. Making a Poster - Triangles After studying this section you will be able to: 1. Define a triangle. 2. Name the elements of a triangle. 3. Describe the types of triangle accordin to sides. 4. Describe the types of triangle according to angles. Objectives 45 Triangles and Construction Definition interior and exterior angles of a triangle In a triangle ABC, the angles BAC, ABC and ACB are called the interior angles of the triangle. They are written as ZA, ZB and ZC, respectively. The adjacent supplementary angles of these interior angles are called the exterior angles of the triangle. They are written as ZA´, ZB´ and ZC´, respectively. We can refer to the sides of a triangle ABC by using the line segments AB, BC and AC, or by using the lower-case form of the vertex opposite each side. For instance, in AABC at the right: a is the side opposite vertex A, b is the side opposite vertex B, and c is the side opposite vertex C. A B C B¢ A¢ C¢ A B C a b c EXAMPLE 1 Look at the figure. a. Name all the triangles in the figure. b. Name all the interior angles of AMNE. c. Name all the vertices of ANEP. d. Name all the sides of AMNP. e. Name all the exterior angles of AENM. Solution a. AMNE, ANEP and AMNP b. ZM (or ZNME), ZMNE and ZMEN. c. points N, E and P d. segment MP, segment PN and segment NM e. ZE´, ZN´ and ZM´ N M E P Notice that a triangle is defined as the union of three line segments. Since an angle lies between two rays (not two line segments), a triangle has no angles by this definition. However, we can talk about the angles of a triangle by assuming the existence of rays: for example, the rays AB and AC form angle A of a triangle ABC. A B C a For example, we can refer to the triangle shown at the right as AABC. We can also call it ABCA, ACAB, AACB, ABAC or ACBA. The vertices of AABC are the points A, B and C. The sides of AABC are the segments AB, BC and CA. 46 Geometriy 7 EXAMPLE 2 In the figure, P(AABC) = P(ADEF). Find x. Solution P(AABC) = P(ADEF) x + 2 + x +10 = 16 + 14 + x (given) 2x + 12 = x + 30 x = 18 A B C x + 2 10 x D E F 16 14 x Check Yourself 1 1. Three distinct points K, M and N lie on a line m, and a fourth point T is not on the line m. Point T is joined to each of the other points. Find how many triangles are formed and name each one. 2. Find and name all the triangles in the figure at the right. 3. Polygon ABCDE is a regular polygon and its diagonals are shown in the figure. Name a. all the triangles whose three vertices lie on the polygon. b. all the triangles which have exactly one vertex on the polygon. c. all the triangles which have two sides on the polygon. d. the sides of all the triangles which do not have a side on the polygon. 4. The side AC of a triangle ABC measures 12.6 cm, and the two non-congruent sides AB and BC are each 1 cm longer or shorter than AC. Find P(AABC). B A C E D K L M N P A E B D F C G L K A regular polygon is a polygon in which all sides have the same length and all angles are equal. Triangles in the world around us For instance, the perimeter of the triangle ABC in the figure is P(AABC) = BC + CA + AB = a + b + c. A B C c b a Challenge! Move exactly three toothpicks in the following arrangement to make five triangles. Definition perimeter of a triangle The sum of the lengths of the three sides of a triangle is called the perimeter of the triangle. We write P(AABC) to mean the perimeter of a triangle ABC. 47 Triangles and Construction 5. Point X is on the side KN of a triangle KMN. Find the length of MX if the perimeters of the triangles KXM, XMN and KMN are 24, 18, and 30, respectively. Answers 1. Three triangles are formed: AKMT, AMNT and ATKN. 2. AAEL, ALEB, ALBC, AAKL, AAGK, AALB, AABC, AAFC, AADF, AAGL, AADC 3. a. AABC, ABCD, ACDE, ADEA, AEAB b. ABKL, ACLM, ADMN, AENP, AAPK c. AABC, ABCD, ACDE, ADEA, AEAB d. sides of ABKL: BK, KL, BL; sides of ACLM: CL, ML, CM; sides of ADMN: DM, MN, DN; sides of AENP: EN, NP, EP; sides of AAPK: AP, KP, AK 4. 37.8 cm 5. 6 2. Regions of a Triangle Any given triangle ABC separates the plane which contains it into three distinct regions: 1. The points on the sides of the triangle form the triangle itself. 2. The set of points which lie inside the triangle form the interior of the triangle, denoted int AABC. 3. The set of points which lie outside the triangle form the exterior of the triangle, denoted ext AABC. The union of a triangle with its interior and exterior region forms a plane. In the figure opposite, the plane is called E. We can write E = int AABC . AABC . ext AABC. E A B C exterior interior EXAMPLE 3 Write whether each statement is true or false according to the figure opposite. a. Point T is in the interior of ADFE. b. M = ext ABDE c. AADF · ABED = C d. ext AFDE · int AFCE = AFCE e. Points T and K are in the exterior of ADFE. Solution a. false b. true c. false d. false e. true P A B C K D E F T M The picture shows the ‘food triangle’ of different types of food. Can you see what the different regions mean? 48 Geometriy 7 We usually use the capital letter V to indicate the length of a median. Accordingly, the lengths of the medians from the vertices of a triangle ABC to each side a, b and c are written as V a , V b and V c , respectively. As we can see, every triangle has three medians. A B C V a D V c V b 3. Auxiliary Elements of a Triangle Three special line segments in a triangle can often help us to solve triangle problems. These segments are the median, the altitude and the bisector of a triangle. Definition median In a triangle, a line segment whose endpoints are a vertex and the midpoint of the side opposite the vertex is called a median of the triangle. In the figure, the median to side BC is the line segment AD. It includes the vertex A and the midpoint of BC. A B C V a D a. Median Auxiliary elements are extra or additional elements. Check Yourself 2 Answer according to the figure. a. Name five points which are on the triangle. b. Name three points which are not on the triangle. c. Name two points which are in the exterior of the triangle. d. What is the intersection of the line ST and the triangle ABC? e. What is the intersection of the segment NS and the exterior of the triangle ABC? Answers a. points A, B, C, T and S b. points J, L and N c. points J and L d. points S and T e. C L J S A B C N T A physical model of a triangle with its interior region 49 Triangles and Construction EXAMPLE 4 Name the median indicated in each triangle and indicate its length. Solution a. median MD, length V m b. median TE, length V t c. median PF, length V p K D L M T S E V P R F N a. b. c. Activity 1. Follow the steps to construct the median of a triangle by paper folding. 2. Cut out three different triangles. Fold the triangles carefully to construct the three medians of each triangle. Do you notice anything about how the medians of a triangle intersect each other? Take a triangular piece of paper and fold one vertex to another vertex. This locates the midpoint of a side. Fold the paper again from the midpoint to the opposite vertex. DM is the median of EF. Paper Folding - Medians Definition centroid of a triangle The medians of a triangle are concurrent. Their common point is called the centroid of the triangle. 50 Geometriy 7 In the figure, AN is the angle bisector which divides ZBAC into two congruent parts. We call this the bisector of angle A because it extends from the vertex A. Since AN is an angle bisector, we can write m(ZBAN) = m(ZNAC). a a A B N C n A We usually use the letter n to indicate the length of an angle bisector in a triangle. Hence the lengths of the angle bisectors of a triangle ABC from vertices A, B and C are written n A , n B and n C , respectively. As we can see, every triangle has three angle bisectors. A B N C n A n B n C The centroid of a triangle is the center of gravity of the triangle. In other words, a triangular model of uniform thickness and density will balance on a support placed at the centroid of the triangle. The two figures below show a triangular model which balances on the tip of a pencil placed at its centroid. G Definition triangle angle bisector An angle bisector of a triangle is a line segment which bisects an angle of the triangle and which has an endpoint on the side opposite the angle. b. Angle bisector Concurrent lines are lines which all pass through a common point. 51 Triangles and Construction Definition incenter of a triangle The angle bisectors in a triangle are concurrent and their intersection point is called the incenter of the triangle. The incenter of a triangle is the center of the inscribed circle of the triangle. As an exercise, try drawing a circle centered at the incenter of each of your triangles from the previous activity. Are your circles inscribed circles? We have seen that n A , n B and n C are the bisectors of the interior angles of a triangle ABC. We can call these bisectors interior angle bisectors. Additionally, the lengths of the bisectors of the exterior angles ZA´, ZB´ and ZC´ are written as n A´ , n B´ and n C´ respectively. These bisectors are called the exterior angle bisectors of the triangle. In the figure at the right, segment KN is the exterior angle bisector of the angle K´ in AKMT and its length is n K´ . O A B C N L K O is the incenter of DABC The inscribed circle of a triangle is a circle which is tangent to all sides of the triangle. n K¢ K N M T Activity Follow the steps to explore angle bisectors in a triangle. 1. Cut out three different triangles. 2. Fold the three angle bisectors of each triangle as shown below. 3. What can you say about the intersection of the angle bisectors in a triangle? Paper Folding - Angle Bisectors Folding the angle bisector of ZA. AN is the angle bisector of ZA. BM is the angle bisector of ZB. 52 Geometriy 7 EXAMPLE 5 Find all the excenters of AKMN in the figure by construction. Solution To find the excenters, we first construct the bisector of each exterior angle using the method we learned in Chapter 1. Then we use a straightedge to extend the bisectors until they intersect each other. The intersection points E 1 , E 2 and E 3 are the excenters of AKMN. K M N E 1 E 2 K E 3 N M An escribed circle of a triangle is a circle which is tangent to one side of the triangle and the extensions of the other two sides. Definition excenter of a triangle The bisectors of any two exterior angles of a triangle are concurrent. Their intersection is called an excenter of the triangle. In the figure, ABC is a triangle and the bisectors of the exterior angles A´ and C´ intersect at the point O. So O is an excenter of AABC. In addition, O is the center of a circle which is tangent to side AC of the triangle and the extensions of sides AB and BC of the triangle. This circle is called an escribed circle of AABC. As we can see, a triangle has three excenters and three corresponding escribed circles. B A C T S V O 53 Triangles and Construction Definition altitude of a triangle An altitude of a triangle is a perpendicular line segment from a vertex of the triangle to the line containing the opposite side of the triangle. In the figure, AH is the altitude to side BC because AH is perpendicular to BC. In a triangle, the length of an altitude is called a height of the triangle. The heights from sides a, b and c of a triangle ABC are usually written as h a , h b and h c , respectively. As we can see, every triangle has three altitudes. A B H C h a c. Altitude A B H C h a h c h b EXAMPLE 6 Name all the drawn altitudes of all the triangles in the figure. A B D C K Solution There are eight triangles in the figure. Let us look at them one by one and name the drawn altitudes in each. 54 Geometriy 7 B A C K altitude BK A B D K altitude AK B D C K altitude CK A B K altitudes AK and BK B C K altitudes BK and CK A D C K altitude DK altitudes AK and DK A D K D C K altitudes CK and DK Activity To fold an altitude, we fold a triangle so that a side matches up with itself and the fold contains the vertex opposite the side. Cut out three different triangles. Fold them carefully to construct the three altitudes of each triangle. What can you say about how the altitudes intersect? Paper Folding - Altitudes 55 Triangles and Construction Definition orthocenter of a triangle The altitudes of a triangle are concurrent. Their common point is called orthocenter of the triangle. Since the position of the altitudes of a triangle depends on the type of triangle, the position of the orthocenter relative to the triangle changes. In the figure opposite, the orthocenter K is in the interior region of the triangle. Later in this chapter we will look at two other possible positions for the orthocenter. Once we know how to draw an altitude of a triangle, we can use it to find the area of the triangle. A B H C h a h c h b K K is the orthocenter of DABC Definition area of a triangle The area of a triangle is half the product of the length of a side (called the base of the triangle) and the height of the altitude drawn to that base. We write A(AABC) to mean the area of AABC. For example, the area of AABC in the figure is Area is usually expressed in terms of a square unit. · · A ( )= = . 2 2 BC AH a h A ABC A B H C h a EXAMPLE 7 Find the area of each triangle. K M N 6 cm 8 cm A B H C 4 cm 3 cm 7 cm D T E F 12 cm 5 cm 2 cm a. b. c. Solution a. (Definition of the area of a triangle) (Substitute) = 20 cm 2 (Simplify) · 10 4 = 2 · A ( )= 2 BC AH A ABC 56 Geometriy 7 Definition circumcenter of a triangle The intersection point of the perpendicular bisectors of a triangle is called the circumcenter of the triangle. The circumcenter of a triangle is the center of the circumscribed circle of the triangle. The circumscribed circle of a triangle is a circle which passes through all the vertices of the triangle. EXAMPLE 8 Find the circumcenter of each triangle by construction. a. b. c. Definition perpendicular bisector of a triangle In a triangle, a line that is perpendicular to a side of the triangle at its midpoint is called a perpendicular bisector of the triangle. In the figure, HN, DN and EN are the perpendicular bisectors of triangle ABC. Perpendicular bisectors in a triangle are always concurrent. A B C H E D N b. (Definition of the area of a triangle) (Substitute) = 35 cm 2 (Simplify) c. (Definition of the area of a triangle) (Substitute) = 24 cm 2 (Simplify) · 6 8 = 2 · A ( )= 2 KM MN A KMN · 5 14 = 2 · A ( )= 2 FT DE A DEF The picture below hangs straight when the hook lies on the perpendicular bisector of the picture’s top edge. 57 Triangles and Construction As an exercise, draw three more triangles on a piece of paper and construct their circumcenters. Check that each circumcenter is the center of the inscribed circle. Solution First we construct the perpendicular bisector of each side of the triangle. Their intersection point is the circumcenter of the triangle. a. b. c. Check Yourself 3 1. Name the auxiliary element shown in each triangle using a letter (n, h or V) and a vertex or side. M S N P X Y W Z K L M J A B C N K M N L M N H P a. b. c. d . e. f. Activity There are three main faculties on a university campus. The university wants to build a library on the campus so that it is the same distance from each faculty. 1. Make a geometric model of the problem. 2. Find the location of the library in the picture opposite. Perpendicular Bisector of a Triangle 58 Geometriy 7 2. In a triangle MNP, the altitude NT of side MP and the median MK of side NP intersect at the point R. a. Name all the triangles in the figure formed. b. Name two altitudes of AMTN. 3. In a triangle DEF, EM is the median of side DF. If DE = 11.4, MF = 4.6 and the perimeter of ADEF is 27, find the length of side EF. 4. In a triangle KLM, LN is the altitude of the side KM. We draw the angle bisectors LE and LF of angles KLN and MLN respectively. If the angles between the angle bisectors and the altitude are 22° and 16° respectively, find m(ZKLM). 5. In the figure, A(AABH) = A(AAHC). Find x. 6. Write one word or letter in each gap to make true statements about the figures. a. Point O is a(n) __________ . b. Segment ________ is a median. c. Point _______ is an excenter. d. Segment ________ is an altitude. e. Point B is a(n) _____________. f. Segment ER is a(n) __________ ___________. g. Point _________ is a circumcenter. h. Line TM is a(n) __________ __________. i. Point ________ is a centroid. Answers 1. a. n B b. h p c. V x d. V l e. h n f. n L 2. a. AMNK, AMKP, AMNT, ANTP, AMRT, AMNR, ARNK, AMNP b. NT, TM 3. 6.4 4. 76° 5. 5 6. a. incenter b. ET c. K d. AB (or BC) e. orthocenter (or vertex) f. angle bisector g. M h. perpendicular bisector i. G D E P F R O A B C Y X Z K G J N S L M T V A B C 8 10 4 x H 59 Triangles and Construction B. TYPES OF TRIANGLE Some triangles are given special names according to the lengths of their sides or the measures of their angles. 1. Types of Triangle According to Sides A triangle can be called scalene, isosceles or equilateral, depending on the lengths of its sides. Definition scalene triangle A triangle is called scalene if all of its sides have different lengths. In other words, a scalene triangle has no congruent sides. a ¹ b ¹ c, so DABC is a scalene triangle A B C b c a Definition isosceles triangle A triangle is called isosceles if it has at least two congruent sides. A B C c b a b = c, so DABC is isosceles Dynamic geometry software is a powerful tool for studying geometric concepts. Geometry programs allow us to change and manipulate figures, so that we can explore and experiment with geometrical concepts instead of just memorizing them. Activity The Euler line of a triangle is the line which passes through the orthocenter, circumcenter and centroid of the triangle. Draw a scalene triangle and find its Euler line using a. a ruler and set square. b. a compass and straightedge. c. dynamic geometry software. Which method was easier? Euler Lines orthocenter circumcenter centroid Euler line 60 Geometriy 7 EXAMPLE 9 Segment EM is a median of an isosceles triangle DEF with base DF. Find the length of EM if the perimeter of AEMF is 65 and the perimeter of ADEF is 100. Solution Let us draw an appropriate figure. In the figure opposite, in ADEM, a + b + x = 65, (1) in ADEF, 2(a + b) = 100. So a + b = 50. (2) Substituting (2) into (1) gives us 50 + x = 65; x = 15. So EM = 15. D E F M b b a a In an isosceles triangle, the congruent sides are called the legs of the triangle. The third side is called the base of the triangle. The two angles between the base and the legs of the triangle are congruent. They are called the base angles of the triangle. The angle opposite the base is called the vertex angle. A B C c b a vertex angle legs base angles base Activity Golden Triangles 36° 72° 72° golden triangle A golden triangle is a triangle in which the ratio of the length of the legs to the length of the base is the golden ratio. The angle between the two legs of a golden triangle is always 36°. A D C B F G E To construct a golden triangle, first draw a square ABCD and mark the midpoint E of AB. Find the point F on the extension of AB by making EF = EC. Then find G by making AF = AG and BA = BG. Finally, draw FG and AG. Then ABGF is a golden triangle. 1. Construct a golden triangle using a straightedge and compass. 2. Repeat the construction using dynamic geometry software. 3. In both constructions, check the measures of the interior angles. The sides of the Great Pyramid of Giza are golden triangles. The head of this knee hammer forms an isosceles triangle. Line segments AB and BC are in the golden ratio if · ¢ ¢ = + = , 1+ 5 = 1.6180339... 2 a b a a b C B A a b 61 Triangles and Construction EXAMPLE 10 In AKMN, ZK = ZN. Given that KN is 4 cm less than MN and MK is 2 cm more than three times KN, find the perimeter of AKMN. Solution We begin by drawing the figure opposite. If MK = x then KN = x – 4. Also, MK = MN because ZK = ZN. Also, we are given MN = 3KN + 2 x = 3(x – 4) + 2 10 = 2x 5 = x. Since P(AKMN) = 3x – 4, P(AKMN) = (3 · 5) – 4 = 11 cm. K N M x x x 4 EXAMPLE 11 In AABC opposite, O is the intersection point of the bisectors of the interior angles of the triangle. Given that OE BC, OD AB, AD = 4 cm, DE = 5 cm and EC = 6 cm, find P(AEOD). Solution Let us join points A and C to O. We know from the question that OA and OC are the bisectors of ZA and ZC, respectively. Since OD AB, m(ZOAB) = m(ZAOD). (Alternate Interior Angles Theorem) So AODA is an isosceles triangle and AD = OD = 4 cm. (1) Similarly, since OE BC, m(ZEOC) = m(ZOCB). (Alternate Interior Angles Theorem) So AEOC is also an isosceles triangle and OE = EC = 6 cm. (2) By (1) and (2), P(AEOD) = OE + OD + DE = 6 + 4 + 5 = 15 cm. A B C E D O 4 5 6 A B C E D O 4 5 6 4 6 62 Geometriy 7 In an equilateral triangle, all of the interior angles are congruent and measure 60°. Notice that an equilateral triangle is also an isosceles triangle, but an isosceles triangle is not always equilateral. A B C 60° 60° 60° EXAMPLE 12 The three sides of a triangle measure 5n + 8, n+12 and 3n+10 with n = N. Which value of n makes this triangle equilateral? Solution If the triangle is equilateral, all the sides must be congruent. So 5n + 8 = n + 12 = 3n + 10. Let us solve the first equality to find n: 5n + 8 = n + 12 4n = 4 n = 1. If we substitute 1 for n, the side lengths become (5 · 1) + 8 = 13, 1 + 12 = 13 and (3 · 1) + 10 = 13. So the triangle is equilateral when n = 1. Definition equilateral triangle A triangle is called equilateral if it has three congruent sides. A B C a a a How many equilateral triangles can you see in the figure below? You will probably ‘see’ two triangles, one on top of the other. This is actually an optical illusion, though, as the white triangle is not actually drawn. Activity Find six toothpicks and try to do each thing below. Some things may not be possible. Can you explain why? 1. Make one equilateral triangle with six toothpicks. 2. Make two equilateral triangles with six toothpicks. 3. Make three equilateral triangles with six toothpicks. 4. Make four equilateral triangles with six toothpicks. Toothpick Triangles 63 Triangles and Construction Check Yourself 4 1. In AABC opposite, DE BC and point O is the incenter of the triangle. If BD = 6 and EC = 4, find DE. 2. The perimeter of an isosceles triangle is 18.4 and its base measures 4 units more than the length of one leg. Find the length of a leg of this triangle. 3. The sides of an isosceles triangle have lengths in the ratio 4 : 5 : 5. Find the length of the base of the triangle if its perimeter is 28. 4. The perimeter of an isosceles triangle is 22.8. An equilateral triangle is drawn such that one side is congruent to the base of the isosceles triangle. If the perimeter of the equilateral triangle is 24.6, find the length of one leg of the isosceles triangle. 5. In an isosceles triangle NTM, MN = NT, MN = 35, TN = 4x +15 and MT = 40 – x 2 . Find MT. 6. In the figure, all triangles are equilateral, AG = 24.12 cm and AC = 3CE = 2EG. Find the perimeter of each triangle. 7. The three sides of a triangle measure 3a, a+10 and 6a – 15. Which value of a makes the triangle equilateral? 8. Construct an isosceles and an equilateral triangle. Answers 1. 10 2. 4.8 3. 8 4. 7.3 5. 15 6. 12.6 cm, 6.3 cm, 4.2 cm 7. 5 B C D F E G A A D E B C O 4 6 2. Types of Triangle According to Angles A triangle can be called acute, right or obtuse, depending on the measures of its angles. Definition acute triangle, right triangle, obtuse triangle A triangle is called an acute triangle if all its angles are acute. A triangle is called a right triangle if it has a right angle. A triangle is called an obtuse triangle if it has an obtuse angle. The picture shows a puzzle called the Three Companions Puzzle. Get your own and try to free one of the triangles from the string. Can you do it? 64 Geometriy 7 In a right triangle, the sides adjacent to the right angle are called the legs of the triangle. The side opposite the right angle is called the hypotenuse of the triangle. A B C hypotenuse legs Triangles Right Scalene Isosceles Obtuse Scalene Isosceles Acute Scalene Isosceles Equilateral Note Notice that a triangle can be only one of obtuse, acute or right. EXAMPLE 14 Classify each triangle according to its side lengths and angle measures. Solution a. isosceles right triangle b. scalene acute triangle c. isosceles acute triangle 50° 80° 50° 70° 80° 30° 45° 45° a. b. c. EXAMPLE 13 Name all the right triangles in the figure. Solution There are four smaller right triangles (AABK, ABKC, ACKD and ADKA) and four larger triangles (AABC, ABCD, ACDA and ADAB). D C A B K Challenge! Try to change the equilateral triangle in the figure so that it points upwards by moving only three balls. Then try to make the triangle in this figure point downwards by using the least number of balls possible. 65 Triangles and Construction EXAMPLE 15 Draw a right triangle and divide it using a. two parallel lines which are perpendicular to one of the legs. b. two parallel lines which are not perpendicular to legs. c. two perpendicular lines to create two more right triangles. d. two intersecting lines which are not perpendicular to each other to create two more right triangles. Activity ‘Tangram’ is a fun puzzle and a good way to exercise your brain. The name comes from tan, which means ‘Chinese’, and gram, which means ‘diagram’ or ‘arrangement’. The puzzle first appeared in China thousands of years ago, and it is now known all over the world. There are seven pieces in a tangram set: five triangles, one square and one parallelogram. The challenge of the puzzle is to use the seven pieces together to make different shapes. You must use all the pieces, and they must all touch but not overlap. All seven tangram pieces are made up of right triangles with this shape: The first tangram challenge is to make a square with all seven pieces. The solution is shown below. Find a tangram set, or copy the figure above to make your own. 1. Make one right triangle using all of the pieces. 2. Can you make an obtuse triangle by using all of the pieces? 3. Can you make an acute triangle by using all of the pieces? 1 2 7 4 6 5 3 Tangram 66 Geometriy 7 Solution a. b. c. d. EXAMPLE 16 Draw each triangle and use a set square to find its orthocenter. Write the orthocenter as an intersection of lines or line segments. a. acute scalene b. right scalene c. obtuse scalene Solution Remember that the orthocenter of a triangle is the intersection point of its altitudes. We draw the altitudes in each triangle by using a set square. a. orthocenter K, K = AF · BD · EC A E B F C D K b. orthocenter A, A = AB · CA · AD B A C D c. orthocenter T, T = AT · BT · TC A B C T Ealier in this chapter we said that the position of the orthocenter of a triangle depends on the type of triangle. One position is in the interior of the triangle. Can you see what the other two possible positions are, after studying the example above? How do they correspond to the types of triangle shown? a 30°-60° set square 30° 60° a 45° set square 45° 45° 67 Triangles and Construction Check Yourself 5 1. Classify each triangle according to its angle measures. 2. Name all the right triangles in the figure. 3. At most how many of each type of angle can one triangle have? a. acute angle b. right angle c. obtuse angle (Hint: Try to draw a suitable figure for each case using a protractor.) 4. Draw a right triangle and divide it using a. two intersecting lines which are perpendicular to each other. b. two intersecting lines which are not perpendicular to each other, to make three more right triangles. 5. Construct a right isosceles triangle. Answers 1. a. right triangle b. acute triangle c. acute triangle d. obtuse triangle e. obtuse triangle 2. ADKB, AKAB, AKBC, AKCD, AKDA 3. a. three b. one c. one 4. a. b. D A B C E A B C 40° 50° 70° 80° M N P 60° 60° S T K X Y Z 20° 60° 120° K L M a. b. c. d. e. This shark’s fin forms a right triangle with the water. How many triangles? 68 Geometriy 7 2. How many triangles can be formed by joining any three points D, E, F and G if no three of the given points are collinear? Name each triangle. 3. In a triangle ABC, AB is of AC, AB = BC and AC = 15 cm. Find P(AABC). 6 5 A. The Triangle and Its Elements 1. Find and name all the triangles in the figure. EXERCISES 3.1 A D B F E C K 4. In a triangle KMN, KM = cm, MN is 75% of KM and KN is 0.1 cm more than KM. Find P(AKMN). 18 5 9. In an isosceles triangle DEF, DF is the base and FT is a median. Given that P(ADEF) = 23 cm and P(AEFT) is 1 cm more than the perimeter of triangle DTF, find DF. 10. Draw three triangles ABC as in the figure and construct each element separately, using a compass and straightedge. a. h a b. V a c. n A 6. Answer according to the figure. a. Name four collinear points on AABC. b. Name a point which is in the interior of AADC. c. What is the intersection of AABC and AADC? d. What is the intersection of AABC and int AMAC? 7. Draw four figures to show how two triangles can intersect to form a four-sided, five-sided, six-sided and three-sided polygon. 8. In a triangle ABC, two points different to A and B on the side AB are joined to the vertex C by line segments. Similarly, three points different to B and C on side BC are joined to the vertex A by line segments. How many regions inside the triangle are formed by the intersection of these segments? A C M F E B D A B C 5. The side AC of a triangle ABC measures 12.8 cm, which is 2.6 cm less than the sum of the lengths of the other sides. Find the perimeter of this triangle. 11. Draw four triangles KMN as in the figure and find each point separately using only a compass and straightedge. a. centroid b. incenter c. orthocenter d. circumcenter 13. Find the excenters of the triangle in question 11 by using a protractor and ruler. 12. Repeat question 11 with a protractor and ruler. 14. The sides AB, BC and AC of a triangle ABC measure 13, 14 and 15 units respectively. Given that the length of the altitude to side BC is 12, find the lengths of the remaining altitudes. N K M 69 Triangles and Construction 21. Write always, sometimes or never to make true statements. a. If a triangle is isosceles then it is ______________ equilateral. b. If a triangle is equilateral then it is ___________ isosceles. c. If a triangle is scalene then it is ______________ isosceles. d. If a triangle is obtuse then it is _______________ isosceles. e. An obtuse triangle is __________________ a right triangle. f. In a triangle DEF, if DE ± EF then DF is ________________ perpendicular to EF. g. A scalene triangle ________________________ has an acute angle. h. If a triangle has two complementary angles then it is ____________________ a right triangle. 20. Complete the table showing the location (in the interior, on the triangle or in the exterior) of the intersection of the segments or lines for each type of triangle. Perpendicular bisectors Angle bisectors Medians Line containing the altitudes Acute triangle a. b. c. d. Right triangle e. f. g. h. Obtuse triangle i. j. k. l. 18. The sides of a triangle measure 2x + 8, 3x – 6, and 12 + x. a. Find the value(s) of x that make(s) the triangle isosceles. b. Which value(s) of x make(s) the triangle equilateral? 19. The sum of the lengths of the legs of an isosceles right triangle is 22 cm. Find the area of this triangle. 22. In each case, draw a triangle with the given property. a. All three angle bisectors are medians. b. No altitude is a median. c. Only one angle bisector is the perpendicular bisector of a side. d. Only one altitude is in the interior region of the triangle. e. The medians, altitudes and angle bisectors coincide. f. Exactly one of the three altitudes is also a median. 23. Divide any right triangle using two lines so that the figure contains a total of a. five right triangles. b. six right triangles. B. Types of Triangle 15. Look at the figure and name a. an isosceles triangle. b. three right triangles. c. an obtuse isosceles triangle. d. an acute triangle. e. an equilateral triangle. 17. State whether each type of triangle is possible or not. a. an isosceles acute triangle b. a right equilateral triangle c. a scalene acute triangle d. an obtuse isosceles triangle e. an obtuse equilateral triangle 16. Find the circumcenter of a right triangle using a ruler and protractor. A B E C F 60° 30° 70 Geometriy 7 A. THE CONCEPT OF CONGRUENCE In the previous section we studied triangles and their features and properties. In this section we will look at possible relations between two or more triangles. If we are given two triangles, how can we compare them? We might notice that they are the same size and shape. This important relation in geometry is called congruence. Let us start our study of congruence with a general definition of congruence in figures and polygons. 1. Congruent Figures The world around us is full of objects of various shapes and sizes. If we tried to compare some of these objects we could put them in three groups: + objects which have a different shape and size, + objects which are the same shape but a different size, and + objects which are the same shape and size. The tools in the picture at the right have different shape and size. The pictures below show tools which have the same shape but different size. In geometry, figures like this are called similar figures. We will study similar figures in Chapter 3. Congruence is a basic geometric relationship. After studying this section you will be able to: 1. Identify congruent triangles 2. Construct a circle 3. Construct congruent segments 4. Find the midpoint of a segment 5. Construct perpendicular lines and parallel lines 6. Construct congruent angles and an angle bisector 7. Construct atriangle tram given information 8. Desctibe and use the properties of isosceles, equilateral and right triangles. 9. Describe and use the triangle Angle Bisector Theorem. Objectives 71 Triangles and Construction The pictures below show objects which are the same size and shape. In this section, we will study figures which have this property. Factories often need to produce many parts with exactly the same size and shape. Definition congruent figures Figures that have the same size and shape are called congruent figures. We say ‘A is congruent to B’ (or ‘B is congruent to A’) if A and B are congruent figures. The pictures at the bottom of the previous page show some examples of congruent objects. The pictures below show two more examples. In these two examples there is only one piece left to fit in the puzzle. Therefore, without checking anything, we can say that each piece and its corresponding place are congruent. Activity Make a poster to show congruent figures in everyday life. You can take photos, draw pictures or collect pictures from magazines or newspapers that show buildings, designs, signs and artwork with congruent parts. Making a Poster - Congruent Figures Congruence in nature: the petals of this flower are congruent. EXAMPLE 17 Which piece is congruent to the empty space? Solution If we compare the vertices and sides, we can easily see that only c. fits into the space. a. b. c. d. 72 Geometriy 7 A car has many congruent parts. Activity When you learned common fractions, you probably learned them by working with figures divided into congruent parts. Often the figures are circles and rectangles, as these are the easiest to divide into any number of congruent parts. Dividing (also called dissecting) a figure into congruent parts can also be a puzzle. As an example, can you see how to dissect the first figure below into two congruent pieces? Answer: Now try the two puzzles below. The answers are at the back of the book. 1. Dissect each figure into four congruent pieces. 2. The polygon below left can be dissected into four congruent polygons, as shown in the figure below right. There is also a way to divide this polygon into five congruent polygons. Can you find it? Congruent Dissections 73 Triangles and Construction Challenge! Remove five toothpicks to make five congruent triangles. In the figure below, AABC and ADEF are congruent because their corresponding parts are congruent. We can write this as follows: ZA = ZD AB = DE ZB = ZE and BC = EF ZC = ZF AC = DF. We can show this symbolically in a figure as follows: DABC A B C DDEF D E F Definition corresponding elements or parts The points, lines and angles which match perfectly when two congruent figures are placed one on top of the other are called corresponding elements or corresponding parts of the congruent figures. We can see that by definition, corresponding parts of congruent figures are congruent. We can write this in a shorter way as CPCFC. You are already familiar with congruent segments (segments that have equal lengths) and congruent angles (angles that have equal measures). In the rest of this section we will look at congruent figures which are made up of segments and angles. These figures are polygons and especially triangles. Sometimes we need to move or modify a figure to see that it is congruent to another figure. The basic changes that we can make to a figure are reflection (flipping), rotation (turning) and translation (sliding). We will study these in Chapter 3. Definition congruent triangles Two triangles are congruent if and only if their corresponding sides and angles are congruent. We write AABC = ADEF to mean that AABC and ADEF are congruent. 2. Congruent Triangles We can think of congruent figures as figures that are exact copies of each other. In other words, we can put congruent figures one on top of the other so that each side, angle and vertex coincides (i.e. matches perfectly). 74 Geometriy 7 EXAMPLE 18 Given that AMNP = ASTK, state the congruent angles and sides in the two triangles without drawing them. Solution The figure at the right shows how the vertices of each triangle correspond to each other. Because AMNP = ASTK and CPCTC (corresponding parts of congruent triangles are congruent), we can write ZM = ZS MN = ST ZN = ZT and NP = TK ZP = ZK PM = KS. As we can see, the order of the vertices in congruent triangles is important when we are considering corresponding elements. Any mistake in the ordering affects the correspondence between the triangles. If two triangles are congruent then we can write this congruence in six different ways. For instance, if AABC is congruent to ADEF, the following statements are all true: AABC = ADEF AACB = ADFE ABAC = AEDF ABCA = AEFD ACAB = AFDE ACBA = AFED. DMNP DSTK @ M corresponds to S N corresponds to T P corresponds to K A short history of the = symbol: Gottfried Wilhelm Leibniz (1640-1716) introduced for congruence in an unpublished manuscript in 1679. In 1777, Johann Friedrich Häseler (1372-1797) used (with the tilde reversed). In 1824, Carl Brandan Mollweide (1774-1825) used the modern symbol = for congruence in Euclid’s Elements. EXAMPLE 19 Complete each statement, given that APRS = AKLM. a. PR = _____ b. _____ = ZK c. _____ = SP d. ZS = _____ e. ML = _____ f. ZL = _____ Solution a. PR = KL b. ZP = ZK c. MK = SP d. ZS = ZM e. ML = SR f. ZL = ZR EXAMPLE 20 Decide whether or not the two triangles in the figure are congruent and give a reason for your answer. 8 4 A B C 30° 8 K M N 60° 4 75 Triangles and Construction EXAMPLE 21 AABC = AEFD is given with AB = 11 cm, BC = 10 cm and EF + ED = 19 cm. Find the perimeter of AEFD. Solution Since AABC = AEFD, AB = EF, BC = FD and AC = ED by the definition of congruence. So by substituting the given values we get 11 = EF, 10 = FD and AC = ED. Since we are given that EF + ED = 19 cm, we have 11 + ED = 19 cm; ED = 8 cm. So P(AEFD) = EF + ED + FD = 11 + 8 + 10 = 29 cm. Check Yourself 6 1. AKLM = AXYZ is given. State the corresponding congruent angles and sides of the triangles. 2. State the congruence AJKM = ASLX in six different ways. 3. Triangles KLM and DEF are congruent. P(AKLM) = 46 cm, the shortest side of AKLM measures 14 cm, and the longest side of the ADEF measures 17 cm. Find the lengths of all the sides of one of the triangles. 4. Triangles DEF and KLM are congruent. If DE = 12.5 cm, EF = 14.4 cm and the perimeter of the triangle KLM is 34.6 cm, find the length of the side DF. Solution Let us calculate the missing angles: m(ZC) = 60° (Triangle Angle-Sum Theorem in AABC) m(ZM) = 30° (Triangle Angle-Sum Theorem in AKMN) Now we can write the congruence of corresponding parts: AB = KM (Given) BC = KN (BC = KN = 4) AC = MN (AC = MN = 8) ZA = ZM (m(ZA) = m(ZM) = 30°) ZB = ZK (m(ZB) = m(ZK) = 90°) ZC = ZN (m(ZC) = m(ZN) = 60°) Therefore, AABC = AMKN by the definition of congruent triangles. A B C 10 11 E F D 10 11 8 What would happen if the blades of this ship’s propellor or these wheels were not congruent? 76 Geometriy 7 In this section we will construct geometric figures using only two instruments, a straightedge and a compass. 1. Basic Constructions We use a straightedge to construct a line, ray, or segment when two points are given. A straightedge is like a ruler without num- bers. We use a compass to construct an arc or a circle, given a point O and a length r (a radius). Result: [CD] [AB]. straightedge r pencil point compass point compass r O A B Set your compass to the length of [AB]. C Use a straightedge to draw a line. Mark a point C on the line. C D Using C as the center, draw an arc intersecting line [CD]. Label the point of intersection D. A B A B B. CONSTRUCTIONS Construction 1 Constructing a congruent segment. Given [AB], construct [CD] such that [CD] [AB]. A B 5. Two line segments KL and AB bisect each other at a point T. If AL = 7 and the lengths of the segments KL and AB are 22 and 18 respectively, find the perimeter of AKTB. Answers 1. 2. 3. 14 cm, 15 cm, 17 cm 4. 7.7 cm 5. 27 APKM = ASLN, AKMP = ALNS, AMPK = ANSL, APMK = ASNL, AKPM = ALSN, AMKP = ANLS ZK = ZX ZL = ZY ZM = ZZ KL = XY LM = YZ KM = XZ 77 Triangles and Construction Result: M is the midpoint of [AB]. Draw an arc with the same radius and center B. Label the points of intersection of the arcs X and Y. A B X Y Draw [XY]. Mark and name the intersection point M. X Y M A B A B Using any radius greater than , draw an arc with center A. |AB| 1 2 Construction 2 Finding the midpoint of a given segment. Given [AB], construct M such that such that [AM] [MB]. A B Result: [MN] l. M l A B Using M as the center and any radius, use a compass to draw arcs intersecting l at A and B. Using centers A and B and a radius greater than |MA|, draw two arcs and find the intersection point N. M l A B N M l A B N Draw [MN]. Construction 3 Constructing a perpendicular to a line at a given point on the line. Given point M on the line l, construct [MN] l. M l 78 Geometriy 7 Result: [MN] l. N A B l M Use A and B as centers to draw arcs with the same radius that intersect at a point M. N A B l Using N as a center, draw an arc that intersects l at two points A and B. Draw [MN]. N A B l M Construction 4 Constructing a perpendicular to a given line through a point outside the given line. Given line l and a point N outside the line, construct [MN] l. Result: AA. Use a straightedge to draw a ray. Name its initial point A¢. Using a compass at center A, draw BïC. Keep the same radius and draw an arc which intersects the ray from A¢ at point B¢. A¢ B¢ A B C Use |BC| as a radius and center B¢ to draw an arc which intersects the first arc at point C¢. A¢ B¢ C¢ Draw [A¢C¢. A¢ B¢ C¢ A B C Construction 5 Constructing a congruent angle. Given A, construct A such that A A. A B C l N 79 Triangles and Construction Result: l t. l M N k t Q P R At N, construct ÐRNQ congruent to ÐNMP. l M N k Draw a line k which intersects line l at point M, and passes through point N. Construction 6 Constructing a parallel to a line through a point outside the line. Given line l with point N which is not on l, construct a line through N which is parallel to l. N l Result: [AD bisects A Draw an arc BïC with center A. A B C A B C D Draw [AD. A B C D With B and C as centers and a radius greater than |BC|, draw arcs intersecting at D. 1 2 Construction 7 Constructing an angle bisector. Given CAB, construct the bisector of CAB. A B C 80 Geometriy 7 c. Look at construction 6. A B A¢ l A¢ B¢ l B A C B A C D Use a straightedge to draw [AB]. Set the compass at the points A and B. A l B C A l B C A l B A l B C Use a straightedge to draw [AB]. Next, open the compass to |AB| and draw two arcs, one with center A and the other with center B. Label the intersection point C. Draw [AC] and [BC]. All the sides have equal length, so ABC is an equilateral triangle. Draw another line l. Choose any point on line l and label it A. Use the radius |AB| and set the compass point at A. Draw an arc intersecting l. Label the point of intersection B. Now [AB] [AB] EXAMPLE 22 a. Construct two congruent line segments. b. Construct an obtuse angle and bisect it. c. Construct two parallel line segments. d. Construct an isosceles triangle. e. Construct an equilateral triangle. Solution a. b. d. e. Draw any obtuse angle ABC. Use B as the center, and draw an arc AïC. Next, draw two arcs, one with center A and the other with center C. Label the point D where the two arcs intersect. Draw [BD. [BD is the angle bisector of ABC. Draw a line segment [AB]. Use any radius greater than and draw two arcs with centers A and B. Name the intersection point C. 1 |AB| 2 Draw the triangle ABC. |AC| = |BC|, so the triangle is isosceles. 81 Triangles and Construction Practice Problems 7 1. Construct a 30° angle. (Hint: construct a 60° angle and bisect it.) 2. Construct a right triangle with legs which are congruent to [AB] and [CD] in the figure. 3. Construct a right triangle whose legs are in the ratio 2:1. 4. Construct a line segment and divide it into four equal parts. Answers 1. By constructing equilateral triangles: 2. 3. 4. A D C E B A D C E B A C B Find the midpoint of [AB]. Find the midpoint of [AC] and the midpoint of [CB]. Label the new points D and E. A B C D A B C A B C Find the midpoint of [AD] (see construction 2) Draw a perpendicular line to [AB] from A. (see construction 3) Open your compass to as [AC] then draw [AD] and [DB]. N B A M B A C D A B Use a straightedge to draw [AB] Open your compass more than [MA] and draw two arcs, one with center M, te other with center N. Draw a line from A to the point of intersection. Open your compass to [CD] and draw an arc with center A. Label the point of intersection D. Draw [AD] and [DB. A B C D 30° A B C D 82 Geometriy 7 2. Constructing Triangles We can construct basic geometric figures using only a straightedge and a compass. However, to construct triangles we need a compass, a ruler and a protractor. We use the ruler to measure the sides of triangle, and the protractor to draw the angles. We have seen that a triangle has six basic elements: three angles and three sides. To construct a triangle, we need to know at least three of these elements, and one of these three elements must be the length of a side. Let us look at the possible cases. a. Constructing a Triangle from Three Known Sides Let us construct ABC, where |AB| = c, |BC| = a, and |AC| = b, given that a < b < c. Note In any triangle, the sum of any two given angles is less than 180° and the sides satisfy the triangle inequality. A B c A C b B C a Construction 1 Draw a line d. d Construction 2 Locate point A on d. A d Construction 3 Open the compass as much as length c and put the sharp point of the compass on A. Then draw an arc. Name the intersection point B. c d A B Construction 4 Again open the compass as much as length b and put the sharp point on A. Then draw an arc on the upper side of d. d A B c 83 Triangles and Construction Construction 5 Finally, open the compass as much as length a and put the sharp point on B. Then draw an arc which intersects the other arc drawn before. Name the intersection point C. d A B c C Note Remember that in a triangle, side a is opposite A, side b is opposite B, and side c is opposite C. When we talk about ‘side b’ we mean the side opposite B, or the length of this side. Construction 5 After determining the point C, draw [AC] and [BC]. The result is the constructed triangle. d A B c C b a EXAMPLE 23 Construct ABC given |AB| = 10 cm, |BC| = 8 cm, and |AC| = 6 cm. Solution C is the intersection point of the two arcs. A B 10 cm C 6 cm 8 cm A B 10 cm Draw a line and locate points A and B. Connect the vertices. A B 10 cm C 6 cm 8 cm b. Constructing a Triangle from Two Known Angles and a Known Side Let us construct the triangle ABC, where A, B, and the side c are given. Construction 1 Draw a line d. d Construction 2 Locate point A on the line. A 84 Geometriy 7 Construction 3 Using a protractor, take the point A as a vertex and draw a ray [AX to construct A. d A X Construction 4 Using a compass, locate the point B on d such that |AB| = c. d A X B c Construction 5 Using a protractor, take the point B as vertex and draw a ray [BY to construct B. Label the intersection point of [AX and [BY as C. The construction is complete. d A Y B X C c EXAMPLE 24 Construct ABC given mB = 40°, mC = 70°, and |BC| = 12 cm. Solution B X 40° Mark the point B and draw ÐB. B A 40° Draw ÐC to find the point A on [BX. 12 cm C Y X 70° B X 40° Using a compass and ruler, find C. 12 cm C 3. Constructing a Triangle from Two Known Sides and a Known Angle Finally, let us construct ABC given |AB| = c, |BC| = a and the known angle B. Construction 1 Draw a line d and use a compass to locate the points B and C such that |BC| = a. d A B c B C a 85 Triangles and Construction Construction 2 Use a protractor to construct B and the ray [BX. B X C a d Construction 3 Use a compass or ruler to locate the point A on [BX such that |AB| = c. B X C a d c A Construction 4 Join the points A and C. The result is the constructed triangle. d A X B c EXAMPLE 25 Construct AABC given |BC| = 5 cm, |AB| = 10 cm, and mB = 70 o . Solution Join A and C. B C 70° 5 cm X A 10 cm Locate A on [BX. B C 70° 5 cm X A 10 cm B C 70° 5 cm Locate the points B and C and draw ÐB. X 86 Geometriy 7 Practice Problems 8 1. State the things you need to know in order to construct a triangle. 2. Draw an equaliteral triangle with sides 6 cm long. 3. Construct ABC given a = 5 cm, b = 4 cm and c = 2 cm. 4. Construct ABC given a = 7 cm, b = 6 cm and c = 8 cm. 5. Construct DEF given d = 6 cm, e = 8 cm and f = 10 cm. 6. Construct ABC given mA = 40 o , mB = 65 o and |AB| = 10 cm. 7. Construct KLM given mM = 45 o , mL = 70 o and |ML| = 7 cm. 8. Construct PQR given mR = 40 o , mQ = 60 o and |RQ| = 4 cm. 9. Construct MNP given mM = 30 o , mN = 65 o and |MN| = 15 cm. 10. Construct ABC given mB = 90 o , |AB| = 5 cm and |BC| = 12 cm. 11. Construct PQR given mQ = 80 o , |PQ| = 7 cm and |QR| = 4 cm. 12. Construct GHK given mH = 50 o , |GH| = 6 cm and |HK| = 9 cm. 13. Construct XYZ given mY = 110 o , |XY| = 3 cm and |YZ| = 5 cm. 14. Can you draw a triangle from only three given angles? C. ISOSCELES, EQUILATERAL AND RIGHT TRIANGLES Isosceles, equilateral and right triangles are useful triangles because they have many special properties. If we can identify one or more of these triangles in a figure then we can often use its properties to solve a geometric problem. In this section we will look at some fundamental theorems about isosceles, equilateral and right triangles, and some useful additional properties. 1. Properties of Isosceles and Equilateral Triangles a. Basic Properties Theorem If two sides of a triangle are congruent then the angles opposite these sides are also congruent. Proof Let us draw an appropriate figure. Given: AB = AC Prove: ZB = ZC A B N C Isosceles Triangle Theorem 87 Triangles and Construction Let AN be the bisector of ZA. Statements Reasons 1. AB = AC 1. Given 2. ZBAN = ZCAN 2. Definition of an angle bisector 3. AN = AN 3. Reflexive property of congruence 4. AABN = AACN 4. SAS Congruence Postulate 5. ZB = ZC 5. CPCTC Theorem If two angles in a triangle are congruent then the sides opposite these angles are also congruent. Proof Let us draw an appropriate figure. Given: ZB = ZC Prove: AB = AC We begin by drawing the bisector AN, and continue with a paragraph proof. + Since AN is the angle bisector, ZBAN = ZCAN. + It is given that ZB = ZC. + By the reflexive property of congruence, AN = AN. + So AABN = AACN by the AAS Congruence Theorem. + Since CPCTC, we have AB = AC. A B N C Converse of the Isosceles Triangle Theorem EXAMPLE 26 In a triangle DEF, T = DF such that DT = DE. Given m(ZEDT) = 40° and m(ZDEF) = 85°, find m(ZTEF). Solution Let us draw an appropriate figure. Since DE = DT, ADET is an isosceles triangle. So by the Isosceles Triangle Theorem, m(ZDET) = m(ZDTE). So by the Triangle Angle-Sum Theorem in ADET, m(ZEDT) + m(ZDET) + m(ZDTE) = 180° 40° + 2m(ZDET) = 180° m(ZDET) = 70°. So m(ZTEF) = m(ZDEF) – m(ZDET) = 85° – 70° = 15°. D 40° E T F 88 Geometriy 7 Corollary If a triangle AABC is equilateral then it is also equiangular. In other words, if a = b = c then m(ZA) = m(ZB) = m(ZC). Corollary of the Isosceles Triangle Theorem Corollary If a triangle AABC is equiangular then it is also equilateral, i.e. if m(ZA) = m(ZB) = m(ZC) then a = b = c. Corollary of the Converse of the Isosceles Triangle Theorem EXAMPLE 27 In the figure, AABC and ADEF are equilateral triangles. If BF = 17 cm and EC = 3 cm, find AB + AH + DH + DF. A B E C H D F Solution In the figure, m(ZHCE) = 60° and m(ZHEC) = 60°. (AABC and ADEF are equilateral) So in AHEC, m(ZH) + m(ZE) + m(ZC) = 180° (Triangle Angle-Sum Theorem) m(ZH) = 60°. So AHEC is equiangular. (m(ZC) = 60°, m(ZE) = 60°, m(ZH) = 60°) Therefore AHEC is equilateral. (By the previous Corollary) So HE = HC = EC = 3 cm. Let a and b be the lengths of the sides of AABC and ADEF, respectively. In AABC, AB = a, BE = a – 3 and AH = a – 3. (EC = 3 cm, given) In ADEF, DF = b, CF = b – 3 and DH = b – 3. (EC = 3 cm, given) So AB + AH + DH + DF = a + a – 3 + b – 3 + b = 2(a + b) – 6. (1) Moreover, BF = a – 3 + 3 + b – 3 (Segment Addition Postulate) 17 = a + b – 3 20 = a + b. (2) Substituting (2) into (1) gives us AB + AH + DH + DF = 34 cm. A B E C H D F 60° 60° 60° 3 60° 60° 60° 60° a a 3 b 3 3 b b 3 a 3 3 89 Triangles and Construction Practice Problems 9 1. In a triangle ABC, the interior angle bisector at the vertex A makes an angle of 92° with the side opposite ZA and has the same length as one of the remaining sides. Find all the angles in AABC. 2. In the figure, CE is the angle bisector of ZC, HD BC and HD = 5 cm. Find the length of AC. 3. In the figure, ADCE is an equilateral triangle and DC = BC. is given. Find m(ZA). Answers 1. 8°, 84° and 88° 2. 10 cm 3. 5° Z Z ( ) 1 = ( 13 m A m B) A D E B C ? A E B C D 5 H b. Further properties Properties 6 1. For any isosceles triangle, the following statements are true. a. The median to the base is also the angle bisector of the vertex and the altitude to the base. b. The altitude to the base is also the angle bisector of the vertex and the median to the base. c. The angle bisector of the vertex is also the altitude and the median to the base. In other words, if AB = AC in any triangle ABC then n A = V a = h a . 2. In an equilateral triangle, the medians, angle bisectors and altitudes from the same vertex are all the same, i.e., h a = n A = V a , h b = n B = V b , and h c = n C = V c . Moreover, all of these lines are the same length: h a = h b = h c = n A = n B = n C = V a = V b = V c . In other words, if AB = BC =AC in AABC then h a = h b = h c = n A = n B = n C = V a = V b = V c . 3. If AB = AC in any triangle ABC then n B = n C , V b = V c and h b = h c . 4. If h a = n A or h a = V a or n A = V a in AABC then AABC is isosceles. 90 Geometriy 7 5. a. If AABC is an isosceles triangle with AB = AC, P = BC, E = AC, D = AB, PE AB and PD AC, then PE + PD = b = c. A B C P D E b. If P, E and D are any three points on the sides of an equilateral triangle ABC such that PE and PD are parallel to two distinct sides of AABC, then PE + PD = AB = BC = AC. 6. a. In any isosceles triangle ABC with AB = AC, the sum of the lengths of two lines drawn from any point on the base perpendicular to the legs is equal to the height of the triangle from the vertex B or C. In other words, if AB = AC, P = BC, H = AC, D = AB, PH ± AC and PD ± AB, then PH + PD = h c = h b . b. In any equilateral triangle ABC, the sum of the lengths of two lines drawn from any point on any side perpendicular to the other sides is equal to the height of the triangle from any vertex. In other words, if AB = BC = AC, P = BC, H = AC, D = AB, PH ± AC and PD ± AB, then PH + PD = h a = h b = h c . 7. In any equilateral triangle ABC, if P = int AABC and points D, E and F lie on the sides of AABC such that PE AB, PD AC and PF BC, then PE + PF + PD = AB. A B C E D F P A B C P D H A B C P D E 91 Triangles and Construction A B C F D E P 8. In any equilateral triangle ABC, if P = int AABC and points D, E and F lie on the sides of AABC such that PE ± AB, PD ± AC and PF ± BC, then PD + PE + PF = AH where AH ± BC. EXAMPLE 28 An isosceles triangle TMS has base TS which measures 8 cm and perimeter 32 cm. The perpendicular bisector of leg TM intersects the legs TM and MS at the points F and K, respectively. Find the perimeter of ATKS. Solution Let us draw an appropriate figure. + Since h k = V k in ATKM, TKM is isosceles by Property 6.4. So TK = KM. (1) + By the Segment Addition Postulate, MK + KS = MS. (2) + Since ATMS is isosceles and P(ATMS) = 32 cm, TM = MS = 12 cm. (3) + So P(ATKS) = TK + KS + ST = KM + KS + ST (By (1)) = MS + ST (By (2)) = 12 + 8 = 20 cm. (By (3)) M F T S K 6 6 8 EXAMPLE 29 In the figure, AB = AC, PD AC and PE AB. Given AB + AC = 32 cm, find PD + PE. Solution Since AB + AC = 32 cm and AB = AC, we have AB = AC = 16 cm. So by Property 6.5b, PD + PE = AB = 16 cm. A D B P C E 92 Geometriy 7 Solution Look at the figure. Given: AT is a median and AB = AC. Prove: AT bisects ZA and is an altitude of AABC. We will write a two-column proof. Proof: EXAMPLE 32 Prove that in any isosceles triangle, the median to the base is also the angle bisector of the vertex and the altitude to the base. EXAMPLE 31 In the equilateral triangle ABC shown opposite, PD BC, PE AB and PF AC. PE = 6 cm, PF = 5 cm and P(AABC) = 42 cm are given. Find the length of PD. Solution Since ABC is equilateral and its perimeter is 42 cm, AB = =14 cm. By Property 6.7, PD + PE + PF = AB. So PD + 6 + 5 = 14 and so PD = 3 cm. 42 3 A D B F C P E 5 6 A B T C Statements Reasons 1. AC = AC 1. Given 2. ZABC = ZACB 2. Isosceles Triangle Theorem 3. BT = TC 3. AT is a median. 4. AABT = AACT 4. By 1, 2, 3 and the SAS Congruence Postulate 5. m(ZTAB) = m(ZTAC) 5. By 4 (CPCTC) 6. AT bisects ZA 6. Definition of angle bisector 7. m(ZATB) = m(ZATC) 7. By 4 (CPCTC) 8. m(ZATB) = m(ZATC) = 90° 8. Linear Pair Postulate 9. AT is also an altitude of AABC 9. Definition of altitude EXAMPLE 30 In the figure opposite, H, A and B are collinear points with CH ± HA, PE ± AC, PD ± AB and AB = AC. PE = 6 cm and HC = 10 cm are given. Find the length of PD. Solution By Property 6.6a, PE + PD = CH. So 6 + PD = 10 and so PD = 4 cm. C P B D A H E 6 10 ? 93 Triangles and Construction 2. In an equilateral triangle ABC, H = BC, N = AC, BN is the interior angle bisector of ZB, and AH is the altitude to the side BC. Given BN = 9 cm, find AH. 3. In the figure, BH = HC, m(ZHAC) = 20° and m(ZBCD) = 30°. Find m(ZBDC). 4. In the figure, AB = AC = 13 cm, PE AC, PF AB and PE = 4 cm. Find the length of PF. 5. In the figure, AB = AC, PE ± HB, PF ± AC and BH ± HC. Given CH = 12 cm, find the value of PE + PF. 6. In the figure, AB = BC, PD = x + 3, PH = 3x – 1, and AE = 14. Find the value of x. Answers 1. 36 2. 9 cm 3. 80° 4. 9 cm 5. 12 cm 6. 3 A D B H C ? 20° E 30° A B P C E F 4 H A E B P F C 12 A B E H C P D Practice Problems 10 1. In the figure, AD and BE are the interior angle bisectors of ZA and ZB, respectively. AC = BC and BE + AD = 12 cm are given. Find the value of BE · AD. A B D C E 94 Geometriy 7 Proof There are many proofs of the Pythagorean Theorem. The proof we will give here uses the dissection of a square. It proves the Pythagorean Theorem for the right triangle ABC shown opposite. Imagine that a large square with side length a + b is dissected into four congruent right triangles and a smaller square, as shown in the figure. The legs of the triangles are a and b, and their hypotenuse is c. So the smaller square has side length c. We can now write two expressions for the area S of the larger square: and S = (a + b) 2 . Since these expressions are equal, we can write This concludes the proof of the Pythagorean Theorem. · ¸ ¸ · ( ¸ , 2 2 2 2 2 2 2 2 4 + =( + ) 2 2 + = +2 + = + . a b c a b ab c a ab b c a b · ¸ ¸ · ( ¸ , 2 =4 + 2 a b S c b a c c c c b b a a a b A C B c b a Theorem In a right triangle ABC with m(ZC) = 90°, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the legs, i.e. c 2 = b 2 + a 2 . Pythagorean Theorem A C B c b a 2. Properties of Right Triangles a. The Pythagorean Theorem The Pythagorean Theorem is one of the most famous theorems in Euclidean geometry, and almost everyone with a high school education can remember it. 95 Triangles and Construction EXAMPLE 33 In the figure, ST ± SQ. Find x and y. Solution First we will use the Pythagorean Theorem in ASKT to find x, then we can use it in ASKQ to find y. + SK 2 + KT 2 = ST 2 (Pythagorean Theorem in ASKT) x 2 + 12 2 = 13 2 (Substitute) x 2 + 144 = 169 x 2 = 25 (Simplify) x = 5 (Positive length) + SK 2 + KQ 2 = SQ 2 (Pythagorean Theorem in ASKQ) 5 2 + y 2 = 6 2 (Substitute) y 2 =36 – 25 (Simplify) y = ò11 S T K Q x y 13 12 6 x = –5 is not an answer because the length of a segment cannot be negative. So the answer is x = 5. From now on we will always consider only positive values for lengths. Activity 1. Cut out a square with side length c, and cut out four identical right triangles with hypotenuse c and legs a and b. Place the four triangles over the square, matching the hypotenuses to the sides and leaving a smaller square uncovered in the centre. Try to obtain the Pythagorean Theorem by considering the area of the original square and the areas of the parts that dissect it. 2. Cut out or construct two squares with sides a and b. Try to dissect these squares and rearrange their pieces to form a new square. Then use the areas of the original squares and the new square to write the Pythagorean Theorem. (Hint: Start by cutting out two right triangles with legs a and b.) b a b c c b c a b c a b c a b c a The Pythagorean Theorem Try the following activity to discover two more proofs of the Pythagorean Theorem. 96 Geometriy 7 Theorem If the square of one side of a triangle equals the sum of the squares of the other two sides, then the angle opposite this side is a right angle. Converse of the Pythagorean Theorem EXAMPLE 34 In the figure, PT = TS = KS, PM = 4 cm and KM = 3 cm. Find ST. Solution Let PT = TS = KS = x. So SM = KM – KS = 3 – x and TM = PM – PT = 4 – x. In ATMS, TS 2 = TM 2 + MS 2 (Pythagorean Theorem) x 2 = (4 – x) 2 + (3 – x) 2 (Substitute) x 2 = 16 – 8x + x 2 + 9 – 6x + x 2 (Simplify) x 2 – 14x + 25 = 0 x 1, 2 = (7 ± ò24) cm (Quadratic formula) Since 7 + ò24 is greater than 3 and 4 which are the lengths of the sides, the answer is x = |ST| = 7 – ò24 cm. K P M T S Quadratic formula The roots x 1 and x 2 of the quadratic equation ax 2 + bx + c = 0 are ± 2 1,2 – – 4 = . 2 b b ac x a EXAMPLE 35 In the figure, m + k = 3 · n. Given A(AKMN) = 30 cm 2 , find the value of n. Solution + m + k = 3 · n (1) (Given) + A(AKMN) = 30 cm 2 (Given) (Definition of the area of a triangle) k · m = 60 (2) + In AKMN, n 2 = k 2 + m 2 (Pythagorean Theorem) n 2 = (k + m) 2 – 2km (Binomial expansion: (k+m) 2 = k 2 + 2km + m 2 ) n 2 = (3n) 2 – 2 · 60 (Substitute (1) and (2)) 8n 2 = 120 (Simplify) n 2 = 15 n = ò15 cm. · =30 2 k m K M N m n k 97 Triangles and Construction Proof We will give a proof by contradiction. Suppose the triangle is not a right triangle, and label the vertices A, B and C. Then there are two possibilities for the measure of angle C: either it is less than 90° (figure 1), or it is greater than 90° (figure 2). Let us draw a segment DC ± CB such that DC = AC. By the Pythagorean Theorem in ABCD, BD 2 = a 2 + b 2 = c 2 , and so BD = c. So AACD is isosceles (since DC = AC) and AABD is also isosceles (AB = BD = c). As a result, ZCDA = ZCAD and ZBDA = ZDAB. However, in figure 3 we have m(ZBDA) < m(ZCDA) and m(ZCAD) < m(ZDAB), which gives m(ZBDA) < m(ZDAB) if ZCDA and ZCAD are congruent. This is a contradiction of ZBDA = ZDAB. Also, in figure 4 we have m(ZDAB) < m(ZCAD) and m(ZCDA) < m(ZBDA), which gives m(ZDAB) < m(ZBDA) if ZCAD and ZCDA are congruent. This is also a contradiction. So our original assumption must be wrong, and so AABC is a right triangle. A B C b c a A C B b c a figure 1 figure 2 A D C B b b c c a A D C b b c c B figure 3 figure 4 EXAMPLE 36 In the triangle ABC opposite, K = AC and AN is the interior angle bisector of ZA. AB = 16 cm, AN = 13 cm, AK = 12 cm and NK = 5 cm are given. Find the area of AABN. Solution Let us draw an altitude NH from the vertex N to the side AB. To find the area of AABN we need to find NH, because and AB is given as 16 cm. 13 2 = 12 2 + 5 2 , so m(ZNKA) = 90°. (Converse of the Pythagorean Theorem) Also, NH = NK = 5 cm. (Angle Bisector Theorem) So (Substitution) · · A 2 5 16 ( )= = =40 cm . 2 2 NH AB A ABN · A ( )= 2 NH AB A ABN B A K C N H 5 13 12 16 98 Geometriy 7 Proof Let us join the point P to the vertices A, B and C and use the Pythagorean Theorem in the six right triangles that are created. + In AATP, AT 2 + TP 2 = AP 2 . (Pythagorean Theorem) + In AANP, AN 2 + NP 2 = AP 2 , (Pythagorean Theorem) AT 2 + TP 2 = AN 2 + NP 2 . (1) (Transitive property of equality) + In ABKP, BK 2 + KP 2 = BP 2 . (Pythagorean Theorem) + In ABTP, BT 2 + TP 2 = BP 2 , (Pythagorean Theorem) BK 2 + KP 2 = BT 2 + TP 2 . (2) (Transitive property of equality) + In ACNP, CN 2 + NP 2 = CP 2 . (Pythagorean Theorem) + In ACKP, CK 2 + KP 2 = CP 2 , (Pythagorean Theorem) CN 2 + NP 2 = CK 2 + KP 2 . (3) (Transitive property of equality) + From (1), (2) and (3) we get AT 2 +TP 2 +BK 2 +KP 2 +CN 2 +NP 2 =AN 2 +NP 2 +BT 2 +TP 2 +CK 2 +KP 2 (Addition property of equality) + AT 2 + BK 2 + CN 2 = AN 2 + BT 2 + CK 2 . (Simplify) Theorem If a triangle ABC with P = int AABC has perpendiculars PK, PN, and PT drawn to the sides BC, AC and AB respectively, then AT 2 + BK 2 + CN 2 = AN 2 + BT 2 + CK 2 . Carnot’s Theorem A T B K C P N A T B K C P N We can use the Pythagorean Theorem to prove other theorems. Here is one example. 99 Triangles and Construction Practice Problems 11 1. Find the length x in each figure. 2. In the figure, TN = NK, ST = 12 cm and SN = ò69 cm. Find the length of TK. 3. In a right triangle ABC, m(ZA) = 90°, AB = x, AC = x – 7 and BC = x + 1. Find AC. Answers 1. a. 15 b. 25 c. 9 d. 5ñ3 e. 20 f. 10 2. 10 cm 3. 5 cm S T N K 12 ò69 D E F 8 17 x A B C 7 24 x P M S N 17 x 6 10 K L M N 8 6 x 5 C A B 7 15 x E 9 M N K J x 12 ò19 15 a. b. c. d. e. f. Solution AT 2 + BK 2 + CN 2 = AN 2 + BT 2 + CK 2 (Carnot’s Theorem) x 2 + 4 2 + (2x) 2 = 2 2 + 4 2 + 6 2 (Substitute) 5x 2 = 40 (Simplify) x 2 = 8 x = 2ñ2 EXAMPLE 37 Find the length x in the figure. A T B K C N x 4 4 6 P 2 2x A Pythagorean triple is a set of three integers a, b and c which satisfy the Pythagorean Theorem. The smallest and best-known Pythagorean triple is (a, b, c) = (3, 4, 5). 100 Geometriy 7 b. Further properties Properties 7 1. The length of the median to the hypotenuse of a right triangle is equal to half of the length of the hypotenuse. 2. a. In any isosceles right triangle, the length of the hypotenuse is ñ2 times the length of a leg. (This property is also called the 45°-45°-90° Triangle Theorem.) b. In any right triangle, if the hypotenuse is ñ2 times any of the legs then the triangle is a 45°-45°-90° triangle. (This property is also called the Converse of the 45°-45°-90° Triangle Theorem). 3. In any 30°-60°-90° right triangle, a. the length of the hypotenuse is twice the length of the leg opposite the 30° angle. b. the length of the leg opposite the 60° angle is ñ3 times the length of the leg opposite the 30° angle. (These properties are also called the 30°-60°-90° Triangle Theorem.) 4. In any right triangle, a. if one of the legs is half the length of the hypotenuse then the angle opposite this leg is 30°. b. if one of the legs is ñ3 times the length of the other leg then the angle opposite this first leg is 60°. (These properties are also called the Converse of the 30°-60°-90° Triangle Theorem.) 5. The center of the circumscribed circle of any right triangle is the midpoint of the hypotenuse of the triangle. B A C r r r O Activity + Fold a corner of a sheet of paper, and cut along the fold to get a right triangle. + Label the vertices A, B and C so that m(ZB) = 90°. + Fold each of the other two vertices to match point B. + Label the point T on the hypotenuse where the folds intersect. What can you say about the lengths TA, TB and TC? Repeat the steps with a different right triangle, and see if your conclusions are the same. Paper Folding - Median to the Hypotenuse 101 Triangles and Construction EXAMPLE 39 EXAMPLE 38 In the figure at the right, find m(ZADC) if m(ZBAC) = 90°, m(ZBAD) = 2x, m(ZACB) = 3x and BD = DC. Solution Since AD is a median, by Property 7.1 we have So AD = BD = DC. Hence ADCA and ABDA are isosceles triangles. Since ADCA is isosceles, m(ZDAC) = m(ZACD) = 3x. Additionally, m(ZBAC) = m(ZBAD) + m(ZDAC) by the Angle Addition Postulate. So 2x + 3x = 90° and x = 18°. By the Triangle Angle-Sum Theorem in ADCA, m(ZADC) + 3x + 3x = 180°. So m(ZADC) = 180° – (6 · 18)°, i.e. m(ZADC) = 72°. · 1 = . 2 AD BC A B D C 2x 3x In the figure, m(ZBAC) = 90°, m(ZC) = 60° and BD = DC. Find BC if AD = 2x + 3 and AC = 6x – 1. Solution + Since AD is a median and the length of the median to the hypotenuse of a right triangle is equal to half the length of the hypotenuse, + By the Triangle Angle-Sum Theorem in AABC, m(ZB) = 30°. + By the 30°-60°-90° Triangle Theorem, because m(ZB) = 30° and BC is the hypotenuse. + So by the transitive property of equality, AC = AD, i.e. 6x – 1 = 2x + 3 and so x = 1. + Finally, BC = 2 · AC = 2 · AD = 2 · (2x + 3) = 10. AC BC 1 = 2 · · 1 = . 2 AD BC C A B D 2x + 3 6x 1 60° 102 Geometriy 7 EXAMPLE 40 One of the acute angles in a right triangle measures 16°. Find the angle between the bisector of the right angle and the median drawn from the same vertex. Solution Let us draw an appropriate figure. We need to find m(ZNAT). According to the figure, + AN is the angle bisector, AT is the median, and m(ZBAC) = 90°. + m(ZACB) = 16° by Property 5.3. + Since AT is median to hypotenuse, AT = CT = BT. + So AATC is isosceles. + Therefore, by the Isosceles Triangle Theorem, m(ZTAC) = m(ZACT) = 16°. + Since AN is an angle bisector and m(ZBAC) = 90°, m(ZNAC) = 45°. + So m(ZNAT) = m(ZNAC) – m(ZTAC)= 45° – 16° = 29°. 45° A B N T C 16° Solution In AABC, since m(ZC) = 30°, m(ZB) = 60°. In AABH, since m(ZB) = 60°, m(ZBAH) = 30°. In AABH, by Property 7.3, AB = 2 · BH = 2 · 2 = 4 cm. In AABC, again by Property 7.3, BC = 2 · AB = 2 · 4 = 8 cm. So HC = BC – BH = 8 – 2 = 6 cm. A B H C 60° 30° 60° 30° 8 4 2 EXAMPLE 41 In the figure, AB ± AC and AH ± BC. Given m(ZC) = 30° and BH = 2 cm, find the length of HC. A B H C 2 ? 30° This set square is in the form of a 30°-60°-90° triangle. Property 5.3: In any triangle ABC, if m(ZB) > m(ZC) or m(ZB) < m(ZC) then h a < n a < V a . 103 Triangles and Construction Solution We will construct a right triangle in which one leg is half of the hypotenuse. Then by the Converse of the 30°-60°-90° Triangle Theorem, the angle opposite this leg will measure 30°. m m B n m B n C K k m B n C K A r r 2 r 2 m B n C K A r 30° r 2 r 2 m B n C K r 2 r 2 r 2 r 2 A m n B C r 1. 2. 3. 5. 6. 7. 4. EXAMPLE 43 Construct a 30° angle by using the Converse of the 30°-60°-90° Triangle Theorem. EXAMPLE 42 Find the value of x in the figure. Solution Let us draw an altitude from C to AB. + In ABHC, BC = ñ2 · BH (45°-45°-90° Triangle Theorem) 6ñ2 = ñ2 · BH (Substitute) BH = 6. (Simplify) + AB = AH + HB (Segment Addition Postulate) 10 = AH + 6 (Substitute) AH = 4. (Simplify) + In AAHC, AC = 2 · AH (30°-60°-90° Triangle Theorem) AC = 2 · 4 (Substitute) AC = x = 8. (Simplify) A B C 45° 60° 10 x 6ñ2 A B C 45° 60° 6ñ2 H This set square is in the form of 45°-45°-90° right triangle. 104 Geometriy 7 EXAMPLE 44 Construct the circumscribed circle of a given right triangle. Solution 1. Label AABC with m(ZA) = 90° and m(ZB) > m(ZC). 2. Construct the perpendicular bisector of each side and label their point of intersection M. 3. Draw a circle with center M and radius MB. This is the circumscribed circle. C A B M C A B M C A B 1. 2. 3. 1. Draw a line m. 2. Construct a perpendicular to the line at any point. Name the line n and label the intersection point of these lines B. 3. Draw an arc with center B and any radius r. Label the intersection point C of this arc and the line m. 4. Construct the midpoint K of the segment BC and draw the line k perpendicular to m through this point. 5. Draw an arc with center K and radius BC. Label the intersection point A of the arc and line n. 6. Join A and K to make AK = 2 · KB. 7. By the Converse of the 30°-60°-90° Triangle Theorem, m(ZBAK) = 30°. Activity The Kobon triangle problem is an unsolved problem in geometry which was first stated by Kobon Fujimura. The problem asks: What is the largest number of non-overlapping triangles that can be produced by n straight line segments? This might seem like a simple question, but nobody has yet found a general solution to the problem. The mathematician Saburo Tamura proved that the largest integer below n(n – 2) / 3 is an upper bound for the number of non-overlapping triangles which can be produced by n lines. If the number of triangles is exactly equal to this upper bound, the solution is called a perfect solution. Perfect solutions are known for n = 3, 4, 5, 7, 9, 13 and 15, but for other n-values perfect Kobon triangle solutions have not been found. + The perfect Kobon solution for five lines creates 5 · (5 – 2) / 3 = 5 triangles. Can you find this solution? + Make Kobon patterns with seven lines. Can you find the perfect solution for seven lines? Unsolved Problem - Kobon Triangles A perfect Kobon solution with 15 lines and 65 triangles (T. Suzuki, Oct. 2, 2005) 105 Triangles and Construction Check Yourself 12 1. In an isosceles right triangle, the sum of the lengths of the hypotenuse and the altitude drawn to the hypotenuse is 27.3. Find the length of the hypotenuse. 2. In the figure, AABC is a right triangle with m(ZABC) = 90° and CF = FE, and CE is the angle bisector of ZC. If m(ZADB) = 102°, find the measure of ZCAB. 3. One of the acute angles in a right triangle measures 48°. Find the angle between the median and the altitude which are drawn from the vertex at the right angle. 4. In a triangle ABC, m(ZB) = 135°, AC = 17 cm and BC = 8ñ2 cm. Find the length of AB. 5. In a right triangle, the sum of the lengths of the hypotenuse and the shorter leg is 2.4. Find the length of the hypotenuse if the biggest acute angle measures 60°. 6. In the figure, m(ZC) = 60°, HC = 4 cm and DH = 2ñ3 cm. Find the length AC = x. 7. AABC in the figure is an equilateral triangle with DH ± BC, BH = 5 cm and HC = 3 cm. Find the length AD = x. 8. The distance from a point to a line k is 10 cm. Two segments non-perpendicular to k are drawn from this point. Their lengths have the ratio 2 : 3. Find the length of the longer segment if the shorter segment makes a 30° angle with k. 9. ACAB is a right triangle with m(ZA) = 90° and m(ZC) = 60°, and D is the midpoint of hypotenuse. Find the length of the hypotenuse if AD = 3x + 1 and AC = 5x – 3. 10.The hypotenuse of an isosceles right triangle measures 18 cm. Find the distance from the vertex at the right angle to the hypotenuse. Answers 1. 18.2 2. 22° 3. 6° 4. 7 cm 5. 1.6 6. 5 cm 7. 2 cm 8. 30 cm 9. 14 10. 9 cm A B H C D x 5 3 A D H B C 4 x 2ñ3 60° C D A E B F 102° 106 Geometriy 7 D. THE TRIANGLE ANGLE BISECTOR THEOREM Theorem 1. The bisector of an interior angle of a triangle divides the opposite side in the same ratio as the sides adjacent to the angle. In other words, for a triangle ABC and angle bisector AN, 2. In any triangle ABC, if the bisector AN of the exterior angle A´ cuts the extension of side BC at a point N, then · . AB BN AC CN = . AB BN AC CN A B N C A B C N Triangle Angle Bisector Theorem Proof of 1 We begin by drawing two perpendiculars NK and NL from N to the sides AB and AC respectively, then we draw the altitude AH ± BC. + (1) (Definition of the area of a triangle and simplify) Now let us find the same ratio by using the sides AB and AC and the altitudes NK and NL. Since N is the point on the angle bisector, by the Angle Bisector Theorem we have NK = NL. A A ( ) = ( ) AH A ABN A ANC · 2 BN AH · 2 NC = BN CN + (2) (Definition of the area of a triangle and simplify) + (By (1), (2) and the transitive property of equality) = AB BN AC CN A A ( ) = ( ) NK A ABN A ANC · 2 AB NL · 2 AC = AB AC A K B H N C L 107 Triangles and Construction Proof of 2 We begin by drawing two perpendiculars NK and NL from N to the extension of the sides AB and AC respectively, then we draw the altitude AH ± BN. + (1) (Definition of the area of a triangle and simplify) Now let us find the same ratio by using the sides AB and AC, and the altitudes NK and NL. Since N is the point on the angle bisector, by the Angle Bisector Theorem we have NK = NL. + (2) (Definition of the area of a triangle and simplify) + (By (1), (2) and the transitive property of equality) = AB BN AC CN A A ( ) = ( ) NK A ABN A ACN · 2 AB NL · 2 AC = AB AC A A ( ) = ( ) AH A ABN A ACN · 2 BN AH · 2 CN = BN CN A B C N H L K EXAMPLE 45 Find the length x in the figure. Solution (Triangle Angle Bisector Theorem) 12 6 = 8 3 6 = 2 =4 x x x A B D C 12 8 6 x 108 Geometriy 7 1. In any triangle ABC, if AN is an interior angle bisector then AN 2 = (AB · AC) – (BN · NC). A B N C 2. In any triangle ABC, if AN is an exterior angle bisector then AN 2 = (BN · CN) – (AB · AC). A B C N EXAMPLE 47 Find the length x in the figure. Solution By the Triangle Angle Bisector Theorem, By Property 8.1, x 2 = 6 · 4 – 3 · 2 = 18 x = 3ñ2. 6 3 = 2 =4. y y A B C 6 x 3 D 2 y EXAMPLE 46 Find the length x in the figure. Solution (Triangle Angle Bisector Theorem) 3 1 = ; 3 =12+ ; =6. 12+ x x x x x 12 4 = 12+x x A D B C 4 12 x x Properties 8 109 Triangles and Construction EXAMPLE 48 In the figure, m(ZCAB) = 2 · m(ZABC). Given that AC = 4 cm and AB = 5 cm, find the length of BC. Solution Let AD be the bisector of angle A. Then m(ZB) = m(ZDAB) = m(ZDAC), since m(ZCAB) = 2 · m(ZABC). So ADAB is an isosceles triangle. Let AD = DB = x. If BC = a then CD = a – x. By the Triangle Angle Bisector Theorem in ABAC, 5(a – x) = 4x x = (1) Now we can use Property 8.1: x 2 = 5 · 4 – x(a – x) x 2 = 20 – ax + x 2 ax = 20. (2) Substituting (1) into (2) gives a · = 20 ; a 2 = 36 ; a = 6 cm. 5 9 a 5 . 9 a 5 = 4 – x a x A B C 4 5 A B C 4 5 x D a x x a Check Yourself 13 1. In a triangle ABC, N is on side BC and AN is the angle bisector of ZA. If AB = 8 cm, AC = 12 cm and BC = 10 cm, find AN. 2. In a triangle KMN, points M, Z, N and T are collinear, KZ is the angle bisector of the interior angle K, and KT is the angle bisector of the exterior angle K´. If MZ = 5 cm and ZN = 3 cm, find NT. 3. MP is the angle bisector of the interior angle M of a triangle KMN. Find MN if KP : PN = 3 : 4 and KM = 15 cm. 4. In a triangle ABC, point D is on side BC and AD is the bisector of the angle A. Find BD if BA : AC = 5 : 3 and BD + DC = 8 cm. 5. ET is an angle bisector in a triangle DEF. Find the length of ET if DE = 14, EF = 21 and DF = 15. Answers 1. 6ñ2 cm 2. 12 cm 3. 20 cm 4. 5 cm 5. 4ò15 110 Geometriy 7 A. The Concept of Congruence 1. Find two pairs of congruent figures in each picture. Draw each pair. a. b. EXERCISES 3.2 2. In the figure, polygon ABCDE is congruent to polygon KLMNP. Find each value, using the information given. a. x b. y c. n d. a e. b L K P 80° M M 7 (n+10°) a+2 10y+7 A E D C B 70° (x+10°) 8 10 b 20y 3 4. A triangle ABC is congruent to a second triangle KMN. Find the unknown value in each statement, using the properties of congruence. a. AC = 5m – 25, KN = 11 – m b. m(ZBCA) = 45° – v, m(ZMNK) = 2v – 15°, c. m(ZB) = 18°, m(ZM) = u – 4°, d. BA = 22x – 30, MK = 3 – 2x 5. Two triangles ABC and CMN are congruent to each other with AB = 8 cm, CN = 3 cm, and CM = AC = 6 cm. Find BC and MN. B. Constructions 6. Construct each angle. a. 15° b. 105° c. 75° d. 37.5° e. 97.5° 3. AADE = AKLN is given. List the congruent corresponding segments and angles of these triangles. 7. a. Construct an isosceles triangle with base 5 cm and perimeter 19 cm. b. Construct an equilateral triangle with perimeter 18 cm. c. Try to construct a triangle with sides of length 2 cm, 3 cm and 6 cm. What do you notice? Can you explain why this is so? d. Construct a triangle ABC with side lengths a = 5 cm and c = 7 cm, and m(ZB) = 165°. e. Construct a right triangle ABC in which m(ZC) = 90°, b = 5 cm and the hypotenuse measures 7 cm. f. Construct AABC with angles m(ZC) = 120° and m(ZB) = 45°, and side b = 5 cm. g. Construct an isosceles triangle PQR with vertex angle m(ZQ) = 40° and side QP = 3 cm. In each case, construct a triangle using only the three elements given. a. a, b, V c b. a, b, h c c. a, b, n C d. h a , h b , h c e. ZA, a, h a f. V a , V b , V c 8. 111 Triangles and Construction 12. In a triangle KLM, m(ZLKM) = 30°, m(ZLMK) = 70° and m(ZKLM) = 80°. O = int AKLM and KO = LO = MO are given. Find m(ZOKM), m(ZOML) and m(ZOLK). 10. In the figure, AB = BC, AD = DB and BE = EC. If DC = 3x + 1 and AE = 4x – 1, find the length x. A D B E C 11. In a triangle DEF, m(ZE) = 65° and m(ZF) = 15°, and DK is an angle bisector. Prove that ADEK is isosceles. C. Isosceles, Equilateral and Right Triangles 9. AKMN is an isosceles triangle with base KN. The perpendicular bisector of the side MK intersects the sides MK and MN at the points T and F, respectively. Find the length of side KN if P(AKFN) = 36 cm and KM = 26 cm. 13. In the figure, ABC is an equilateral triangle. PE BC, PD AC, PE = 3 and PD = 5 are given. Find the length of one side of the equilateral triangle. A P B D C E 3 5 14. In the figure, AABC is an equilateral triangle, PE AC, PD BC, and PF AB. Given P(AABC) = 45, find the value of PD + PE + PF. A E B F C D P 15. In the figure, BH = HC, AB = DC and m(ZB) = 54°. Find m(ZBAC). A B H C D 54° ? 16. In the figure, BE = EC and AD bisects the interior angle A. Given AB = 12 and AC = 7, find the length of ED. A B E C 12 7 D 17. In the figure, O is the center of the inscribed circle of AABC, OB = CD, AB = AC and m(ZECD) = 20°. Find the measure of ZBEC. A B C D O 20° E ? 18. In a right triangle ETF, the perpendicular bisector of the leg ET intersects the hypotenuse at the point M. Find m(ZMTF) if m(ZE) = 52°. 19. In triangle DEF, DE = EF, m(ZDEF) = 90° and the distance from E to DF is 4.8 cm. Find DF. 112 Geometriy 7 29. In the figure, m(ZBAC) = 90°, m(ZAHC) = 90°, m(ZB) = 30° and HC = 1 cm. Find the length of BH. A B H C 30° 1 ? 30. In the figure, m(ZDBA) = 30°, m(ZABC) = 60° and AD = 4. Find the length of BC. A D B C 30° 60° 4 ? 26. In a triangle ABC, BC = 7ñ2, m(ZBAC) = 30° and m(ZBCA) = 45°. Find the length of the side AB. 27. The larger acute angle A in an obtuse triangle ABC measures 45°. The altitude drawn from the obtuse angle B divides the opposite side into two segments of length 9 and 12. Find the length of the side BC. 28. In the figure, m(ZM) = 150°, LM = 2 and MN = 3ñ3. Find the length of KL. K L M N 2 ? 3ñ3 150° 22. In the figure, BC = 12, m(ZBAC) = 90°, m(ZADC) = 90° and m(ZABC) = 60°. If AC is the angle bisector of ZC, find the length of AD. D A B C 12 ? 60° 24. In the figure, m(ZA) = m(ZB) = 60°, AD = 7 and BC = 4. Find DC. D A B C 60° 60° 4 7 ? 25. In the figure, AB = AC, AD = DB and CD = 8 cm. Find the length of DB. C A D B ? 8 23. In the figure, m(ZA) = 30° and AB = AC = 16 cm. Find the value of PH + PD. A H B P C D 30° In the figure, APMN is a right triangle, MH ± PN and m(ZN) = 15°. Find (Hint: Draw the median to the hypotenuse.) MH PN . 21. 15° P M H N 20. TF is the hypotenuse of a right triangle TMF, and the perpendicular bisector of the hypotenuse intersects the leg MF at the point K. Find m(ZKTF) given m(ZMTF) = 70°. 113 Triangles and Construction 34. In the figure, m(ZA) = 90°, m(ZADC) = 60°, m(ZB) = 30° and BD = 4 cm. Find the length of AD. A D B C 60° 30° 4 ? 31. AABC in the figure is an equilateral triangle. BH = 5 cm and HC = 3 cm are given. Find the length AD = x. 33. In the figure, AABC is an equilateral triangle, BD = 4 cm and AE = 6 cm. Find the length of EC. A B D C E 6 4 ? 32. In the figure, m(ZC) = 90°, m(ZBAD) = m(ZDAC), BD = DA and DC = 3 cm. Find the length of BD. A B D C 3 ? A B H C D x 5 3 38. In the figure, NK is the bisector of the interior angle N and NL = LK. m(ZNMP) = 90° and m(ZP) = 24° are given. Find m(ZPSM). M N S P K 24° L ? 39. In the figure, m(ZBAE) = m(ZDAE) = 60°, CB = 6 cm and CE = 3 cm. Find the length of CD. 60° 60° A B C E 3 D 6 35. In the figure, AABC is an equilateral triangle, PE AC and PD AB. PD = 5 cm, PE = 3 cm and P(AABC) = 48 cm are given. Find the length of PH. P 3 5 A B D C H E ? 37. In the figure, AABC is an isosceles triangle, AB = AC, m(ZBAC) = 30° and 2PE = PD = 4. Find AC. A E B P C D 2 4 30° 36. In a right triangle KLM, m(ZKLM) = 90° and the perpendicular bisector of the leg LM cuts the hypotenuse at the point T. If m(ZLMK) = 20°, find m(ZTLK). 40. In the figure, AB = BC m(ZADB) = 30° and AC = 6. Find the length of CD. A B D C 30° 6 ? 114 Geometriy 7 52. In the figure, AT = DB = DC and m(ZA) = 36°. Find m(ZDBT). A T D B C E ? 36° 48. In the figure, ED = AC, BD = DC and m(ZC) = 63°. Find m(ZEDC). A E B D C 63° ? 49. In the figure, AB = AC = 18 cm, PH = 5 cm and PD = 4 cm. Find m(ZHPD). A H B P C D 5 4 ? 50. In the figure, AABC is an equilateral triangle, AH = ED, AE = EC and CD = 4 cm. Find the length of AB. A B H C D E 4 ? 51. In the figure, AB = AC, AH = HB, m(ZA) = 120° and PB = 8 cm. Find the length of CP. C A B P 8 H 120° ? 42. A rectangular opening in a wall has dimensions 21 cm by 20 cm. Can an empty circular service tray with a diameter of 28 cm pass through the opening? 43. Two ships are in the same port. One starts to travel due west at 40 km/h at 3 p.m. Two hours later the second ship leaves port, traveling due south at 60 km/h. How far apart will the ships be at 7 p.m.? 44. The median drawn from the vertex at the right angle of a right triangle divides the right angle in the ratio 13 : 5. Find the smallest angle in this triangle. 45. In a triangle DEF, m(ZDEF) = 120°, DE = 2ñ5 and DF = 8. Find the length of the side EF. 46. In the figure, BD = DC and BC = 12. Find the length of AD. A B D C ? 47. In the figure, AD = EB, CD = DB and m(ZDEB) = 80°. Find the measure of ZACB. C A E B D 80° ? 41. The base and a leg of an isosceles triangle measure 14 cm and 25 cm respectively. Find the height of the altitude drawn to the base. 115 Triangles and Construction 53. A line m divides a segment KN with the ratio 2 : 3 at the point M. Find the distances from the points K and N to m if KN = 40 and the angle between m and the segment KN is 30°. 54. In a triangle KMN, T is the intersection point of the three angle bisectors and the distance from T to the side MN is 4. Find the distance from T to the vertex K if m(ZK) = 60°. 55. Prove that in a right triangle ABC with m(ZA) = 90° and acute angles 15° and 75°, the altitude to the hypotenuse measures of the length of the hypotenuse. 1 4 In the figure, ABC is an equilateral triangle, m(ZBCE) = 15°, EF = FC, DF ± EC and AD = 2 cm are given. Find AE. 58. A E B C F D 15° 2 ? 57. Prove that if AH is an altitude and AD is a median of a triangle ABC then |b 2 – c 2 | = 2a · HD. A B H D C In the figure, AC BD, CE = 2AB and m(ZC) = 15°. Find m(ZCAF). 56. C A B D E F 15° ? 60. In the figure, BE = EC, BD ± AC, m(ZA) = 45°, m(ZD) = 30° and DC = 6. Find the length x. A B C D E 45° 30° x 6 59. In the figure, NL = LK and NK is the bisector of ZN. If m(ZNMP) = 90° and m(ZMSP) = 120º, what is m(ZP)? M N S P K L ? 120° 63. In a triangle ABC, D lies on the side AC and BD is the interior angle bisector of ZB. If AD = 5 cm, DC = 6 cm and the sum of the lengths of the sides a and c is 22 cm, find the value of a. D. The Triangle Angle Bisector Theorem 61. In a triangle ABC, D = AB and CD is the interior angle bisector of ZC. Given AC = 9 cm, BD = 8 cm, AD = m and BC = n, find the value of m · n. 64. In a triangle ABC, points C, A and D are collinear and C, B and E are also collinear. BD is the angle bisector of ZEBA. If AC = AD and AB = 6 cm, find the length of BC. 62. In a triangle ABC, D lies on the side BC and AD is the interior angle bisector of ZA. If AC = BD, AB = 9 cm and DC = 4 cm, find the length of BD. 116 Geometriy 7 67. In a triangle ABC, D lies on the side BC and AD is the interior angle bisector of ZA. If AB = AD = 12 cm and AC = 16 cm, find the length of BD. 72. In a triangle ABC, points D, B and C are collinear and AD is the angle bisector of ZA´. If AB = 2m, AC = 2m + 6, BC = 2m + 2 and DB = 28, find the value of m. 70. In a triangle ABC, points B, C and D are collinear and AD is the angle bisector of ZA´. If CD = 3 · BC and AB = 12 cm, find the length of AC. 71. In a triangle ABC, D = AB, E = BC and AE and CD are the angle bisectors of ZA and ZC respectively, intersecting at the point K. If AD = 4, DB = 6 and AC = 8, find the value of . AK KE 73. In the figure, AN and BE are the angle bisectors of ZA and ZB respectively. If AB = 4, AC = 6 and BC = 5, find the length of BE. A B C E 4 6 5 N ? 74. In the triangle ABC at the right, AD is the angle bisector of ZA´ and BN is the angle bisector of ZB. DB = m, BC = 3, AB = 4 and NC = 2 are given. Find the value of m. 75. In a triangle ABC, points D, B and C are collinear and AD is the angle bisector of ZA´. If AB = 12 and BD = 4 · BC, find the length of AC. A N C B D 4 3 2 m 69. In the figure, AD and BE are the angle bisectors of ZA and ZB, respectively, and AC = 12 cm. Find the length of segment DC. AE ED 3 = 2 A B D C E 12 ? 68. In a triangle DEF, E, F and K are collinear and DK is the exterior angle bisector of the angle D. If DE = 4 cm, DF = 3 cm and EF = 2 cm, find DK. 66. In a triangle ABC, D lies on the side BC and AD is the interior angle bisector of ZA. If BD = 3 cm, DC = 4 cm and AB + AC = 14 cm, find the length of AD. 65. In a triangle EFM, points F, N and M are collinear and EN is the interior angle bisector of ZE. If EN = 6, NM = 3 and FN = 4, find the length of EF. 117 Triangles and Construction A. RELATIONS BETWEEN ANGLES In this section we will look at some basic rules related to the basic and auxiliary elements of a triangle. Activity Do the following activities and then try to find a common conjecture. 1. Cut out three identical triangles and label their vertices K, M and N. Draw a straight line and place the triangles along the line as in the diagram at the far right. What can you say about the sum of ZK, ZM and ZN? 2. Cut out an acute triangle, a right triangle and an obtuse triangle. Number the angles of each triangle 1, 2 and 3 and tear them off. Finally, put the three angles of each triangle adjacent to each other to form one angle as in the figures at the far right. What can you say about the sum of the angles in each triangle? 3. Cut out a triangle ABC and label its longest side BC. Fold the triangle so that A lies on the fold line and C lies on BC. Label the intersection T of BC and the fold line. Unfold. Now fold the paper so that A, B and C coincide with T. How does this activity support the results of activities 1 and 2 above? Angles in a Triangle K N M K N M K N M N K MNK K M M N ® 1 2 3 1 2 3 ® 2 1 3 ® 2 3 1 1 2 2 1 ® ‘It is indeed wonderful that so simple a figure as the triangle is so inexhaustible in properties. How many as yet unknown properties of other figures may there not be?’ August Crelle (1780-1856), civil engineer and mathematician After studying this section you will able to 1. Describe and use relations between angles 2. Describe and use relations between angles Objectives 118 Geometriy 7 EXAMPLE 49 In the figure, points A, B, F and E, B, C are respectively collinear. Given that ZC and ZF are right angles, m(ZE) = 40° and m(ZA) = o, find the value of o. Solution In AEFB by the Triangle Angle-Sum Theorem, m(ZE) + m(ZF) + m(ZB) = 180° 40° + 90° + m(ZB) = 180° m(ZB) = 50°. By the Vertical Angles Theorem, m(ZEBF) = m(ZABC) 50° = m(ZABC). In AABC by the Triangle Angle-Sum Theorem, m(ZA) + m(ZB) + m(ZC) = 180° o + 50° + 90° = 180° o = 40°. B 40° a A C F E B a A C F E 40° 50° Theorem The sum of the measures of the interior angles of a triangle is 180°. Proof Given: AABC Prove: m(Z1) + m(Z2) + m(Z3) = 180° We begin by drawing an auxiliary line DE through A, parallel to BC. Then we continue with a two-column proof. Statements Reasons 1. Z1 = Z4; Z3 = Z5 1. Alternate Interior Angles Theorem 2. m(ZDAE)=m(Z4)+m(Z2)+m(Z5)=180° 2. Angle Addition Postulate, by the definition of straight angle 3. m(Z1) + m(Z2) + m(Z3) = 180° 3. Substitute 1 into 2 4 2 5 1 3 D A E B C Triangle Angle-Sum Theorem An auxiliary line is a line which we add to a figure to help in a proof. 119 Triangles and Construction Theorem The measure of an exterior angle in a triangle is equal to the sum of the measures of its two nonadjacent interior angles. Triangle Exterior Angle Theorem Proof Given: AABC Prove: m(Z1) = m(Z3) + m(Z4) Statements Reasons 1. m(Z1) + m(Z2) = 180° m(Z2) = 180° – m(Z1) 1. Linear Pair Postulate 2. m(Z2) + m(Z3) + m(Z4) = 180° 2. Triangle Angle-Sum Theorem 3. 180° – m(Z1) + m(Z3) + m(Z4) = 180° 3. Substitute 1 into 2. 4. m(Z1) = m(Z3) + m(Z4) 4. Simplify. 1 2 3 4 Solution m(ZA) + m(ZB) + m(ZC) = 180° (Triangle Angle-Sum Theorem) 3x – 10° + 2x + 20° + 5x = 180° 10x + 10° = 180° x = 17° Activity Complete the table for the figure at the right, using the Triangle Angle-Sum Theorem and the Linear Pair Postulate. What do you notice about the values in the last two columns of the table? m(Z4) m(Z3) m(Z2) m(Z1) m(Z3) + m(Z4) 75° 55° 63° 135° 77° 46° 39° 85° 1 2 3 4 EXAMPLE 50 In a triangle ABC, m(ZA) = 3x – 10°, m(ZB) = 2x + 20° and m(ZC) = 5x. Find the value of x. The two interior angles which are not adjacent to an exterior angle in a triangle are sometimes called remote angles. A B C exterior angle remote angles 120 Geometriy 7 EXAMPLE 53 In a triangle KMN, D lies on side KM. Decide whether each statement about the figure is possible or impossible. If it is possible, sketch an example. a. Triangles KDN and MDN are both acute triangles. b. Triangles KDN and MDN are both right triangles. c. Triangles KDN and MDN are both obtuse triangles. d. Triangle KDN is obtuse and triangle KNM is acute. Solution a. impossible b. possible c. possible d. possible K D N M 50° 80° 55° 30° 45° N K D M K N D M 10° 120° 50° 20° 100° EXAMPLE 52 In the figure, AB = BD, AD = DC and m(ZDAC) = 35°. Find m(ZB). Solution m(ZDCA) = m(ZDAC) (Base angles of an isosceles triangle) = 35° m(ZBDA) = m(ZDAC) + m(ZDCA) (Triangle Exterior Angle Theorem) = 35° + 35° = 70° m(ZBAD) = m(ZBDA) (Base angles of an isosceles triangle) = 70° m(ZB) + m(ZBAD) + m(ZBDA) = 180° (Triangle Angle-Sum Theorem) m(ZB) + 70° + 70° = 180° m(ZB) = 40° A B D C 35° EXAMPLE 51 In a triangle ABC, AB ± AC and m(ZB´) = 136°. Find m(ZC). Solution m(ZB´) = m(ZA) + m(ZC) 136° = 90° + m(ZC) 46° = m(ZC) 136° B C A 121 Triangles and Construction Theorem The sum of the measures of the exterior angles of a triangle is equal to 360°. Proof Given: AABC Prove: m(ZA´) + m(ZB´) + m(ZC´) = 360° We will give a flow-chart proof. m(ÐB¢) = m(ÐA) + m(ÐC) Triangle Exterior Angle Theorem m(ÐA¢) = m(ÐB) + m(ÐC) Triangle Exterior Angle Theorem m(ÐC¢) = m(ÐA) + m(ÐB) Triangle Exterior Angle Theorem Addition Property of Equality m(ÐA¢) + m(ÐB¢) + m(ÐC¢) = = 2(m(ÐA) + m(ÐB) + m(ÐC)) Triangle Angle-Sum Theorem m(ÐA¢) + m(ÐB¢) + m(ÐC¢) = =2 × 180° = 360° Triangle Exterior Angle-Sum theorem EXAMPLE 54 In the figure, m(ZTCA) = 120°, m(ZKAB) = 5x and m(ZPBC) = 7x. a. Find the value of x. b. Find m(ZBAC). Solution a. m(ZA´) + m(ZB´) + m(ZC´) = 360° (Triangle Exterior Angle-Sum Theorem) 5x + 7x + 120° = 360° 12x = 240° x = 20° b. m(ZKAB) + m(ZBAC) = 180° (Linear Pair Postulate) (5 · 20°) + m(ZBAC) = 180° m(ZBAC) = 80° A B C A¢ B¢ C¢ 5x 120° 7x P B A K T C 122 Geometriy 7 Check Yourself 14 1. The two acute angles in a right triangle measure 0.2x + 6.3° and 3.8x – 2.7. Find x. 2. The measures of the interior angles of a triangle are in the ratio 4 : 6 : 8. Find the degree measures of these angles. 3. The vertex angle of an isosceles triangle measures 42°. An altitude is drawn from a base angle to one of the legs. Find the angle between this altitude and the base of the triangle. 4. In an isosceles triangle, the angle between the altitude drawn to the base of the triangle and one leg of the triangle measures 16° less than one of the base angles of the triangle. Find the measure of the vertex angle of this triangle. 5. Two points E and F are drawn on the extension of the side MN of a triangle MNP such that point M is between the points E and N and point N is between points M and F. State which angle is the smallest angle in AEPF if EM = MP, NF = NP, m(ZPMN) = 30° and m(ZPNM) = 40°. Solution m(ZA´) + m(ZA) = 180° (Linear Pair Postulate) m(ZA´) = 180° – m 180° – m + n + k = 360° (Triangle Exterior Angle-Sum Theorem) n + k = 180° + m (1) Also, m + n + k = 280° (Given) m + 180° + m = 280° (Substitute (1)) 2m = 100° m = 50°. m 180° m n k A B C P T EXAMPLE 55 In the figure, m(ZA) = m, m(ZB´) = n and m(ZC´) = k. Find the value of m if m + n + k = 280°. m n k A B C P T 123 Triangles and Construction 6. In a triangle DEF, point M lies on the side DF such that ZMDE and ZDEM are acute angles. Decide whether each statement about the figure is possible or impossible. If it is possible, sketch an example. a. AFME is an acute triangle. b. AFME is a right isosceles triangle. c. AFME and ADME are both acute triangles. d. ADME is equilateral and AEMF is isosceles. e. ADME is isosceles and ADEF is isosceles. 7. In the figure, AKLN is an isosceles triangle in a plane, m(ZKLN) = 120°, and L is the midpoint of the segment KM. A point P is taken in the same plane such that MP = KL. Find the measure of ZLPM when a. the distance between N and P is at its maximum. b. the distance between N and P is at its minimum. 8. One of the exterior angles of an isosceles triangle measures 85°. Find the measure of the vertex angle of this triangle. 9. State whether each triangle is a possible or impossible figure. If it is possible, sketch an example. If it is impossible, give a reason why. a. A triangle with two obtuse exterior angles. b. A triangle with one acute exterior angle. c. A triangle with two right exterior angles. d. A triangle with two acute exterior angles. 10.Find the value of x in each figure, using the information given. 70° 115° A B C x 105° B A 80° x 72° M N P x 3x 40° 60° C A E D B x a. b. c. d. N K L M 120° 124 Geometriy 7 Answers 1. 21.6° 2. 40°, 60° and 80° 3. 21° 4. 74° 5. ZPEF 6. a. possible b. possible c. not possible d. possible e. possible 7. a. 30° b. 60° 8. 95° 9. a. possible b. possible c. impossible because the third exterior angle would be 180° d. impossible because the third exterior angle would have to be more than 180° 10. a. 45° b. 25° c. 27° d. 80° 15° 50° 30° 100° 130° 150° E D M F 40° 20° 65° 60° 55° 50° 40° 45° 45° E D M F 30° 30° 30° E D M F 60° 60° 30° 30° E D M F 60° Properties 3 1. For any triangle, the following statements are true: a. The measure of the angle formed by the bisectors of two interior angles of the triangle is 90° more than half of the third angle, i.e. in the figure, m(ZBKC) = b. The measure of the angle formed by the bisectors of two exterior angles of a triangle is 90° minus half of the third angle, i.e. in the figure, m(ZBTC) = 90° – c. The measure of the angle which is formed by the bisector of one interior angle and the bisector of a second exterior angle is the half the measure of the third interior angle. We can refer to properties 1a, 1b and 1c as the Angle Bisector Relations Theorem. Z ( ) . 2 m BAC Z ( ) 90° + . 2 m BAC a 2 a 2 90° + a 2 90° A S C T B K a So far we have looked at some basic properties of angles. Now we will study some other useful and important properties. 125 Triangles and Construction 2. In any triangle, the measure of the angle formed by the altitude and the angle bisector which both extend from the same vertex is equal to the half the absolute value of the difference of the other two angles of the triangle. 3. In any triangle KLM, if N is any point in the interior of AKLM then a. m(ZLNM) = m(ZLKM) + m(ZKLN) + m(ZKMN). b. m(ZKNM) = m(ZKLM) + m(ZLKN) + m(ZLMN). c. m(ZKNL) = m(ZKML) + m(ZMLN) + m(ZMKN). A B C H N x x= |m(ÐB) m(ÐC)| 2 L M N K Solution Since the incenter is the intersection of the angle bisectors, both AO and CO are bisectors. By Property 3.1a, m(ZAOC)= Z ° ( ) 80° 90° + =90° + =130 . 2 2 m B 80° A B C O EXAMPLE 56 The triangle ABC at the right has incenter O. Find m(ZAOC). 80° A B C O EXAMPLE 57 In the figure, K is the intersection point of the bisectors of the exterior angles at vertices A and B with m(ZA) = 120° and m(ZB) = 40°. Find m(ZBKA). A K B C 120° 40° 126 Geometriy 7 Solution m(ZA) + m(ZB) + m(ZC) = 180° (Triangle Angle-Sum Theorem in AABC) 120° + 40° + m(ZC) = 180° m(ZC) = 20° (1) m(ZBKA)= (Property 3.1b) = (Substitute (1)) = 80° 20° 90° – 2 Z ( ) 90° – 2 m C EXAMPLE 58 Find the value of x in the figure. A B C K x E 3x Solution m(ZC) + m(ZC´) = 180° (Linear Pair Postulate) m(ZC) = 180° – 3x (1) m(ZAEB)= (Property 3.1c) x = (Substitute (1)) 5x = 180° x = 36° 180° – 3 2 x Z ( ) 2 m C EXAMPLE 59 In the figure at the right, AN is an angle bisector, m(ZANC) = 100° and m(ZB) = 2m(ZC). Find the value of x. A B N C x 100° 127 Triangles and Construction Solution m(ZC) = x is given, so m(ZB) = 2x. Let us draw the altitude AH ± BC. Since ZANC is an exterior angle of AAHN, m(ZHAN) + m(ZAHN) = m(ZANC) m(ZHAN) + 90° = 100° m(ZHAN) = 100° – 90° = 10° m(ZHAN) = (Property 3.2) 10° = 10° = x = m(ZC) = 20°. 2 x - |2 | 2 x x Z Z | ( ) – ( )| 2 m B m C A B H N C 10° 2x 100° x EXAMPLE 60 One of the acute angles in a right triangle measures 20°. Find the angle between the altitude and the angle bisector which are drawn from the vertex of the right angle of the triangle. Solution Let us draw an appropriate figure. In the figure at the right, ZA is the right angle, AN is the angle bisector and m(ZNAB) = m(ZNAC) = 45°. Let m(ZC) = 20°, then m(ZB) = 70° and m(ZHAB) = 20°. Therefore, m(ZHAN) = m(ZNAB) – m(ZHAB) = 45° – 20° = 25°. This is the required angle measure. Note that we can also solve this example by using Property 3.2. This is left as an exercise for you. A B H N C 70° 20° 25° 45° 20° EXAMPLE 61 In a triangle KLM, prove that if N is a point in the interior of AKLM then a. m(ZLNM) = m(ZLKM) + m(ZKLN) + m(ZKMN), b. m(ZKNM) = m(ZKLM) + m(ZLKN) + m(ZLMN) and c. m(ZKNL) = m(ZKML) + m(ZMLN) + m(ZMKN). 128 Geometriy 7 Solution Let us draw an appropriate figure. Given: N is an interior point of AKLM Prove: m(Z1) = m(Z2) + m(Z3) + m(Z4) Proof: Let us extend segment MN through N and label the intersection point T of ray MN and segment KL. m(ZLTN) = m(Z3) + m(Z4) (Triangle Exterior Angle Theorem) m(Z1) = m(Z2) + m(Z3) + m(Z4) (Triangle Exterior Angle Theorem) This means m(ZLNM) = m(ZLKM) + m(ZKLN) + m(ZKMN). The proofs of b. and c. are similar. They are left as an exercise for you. 2 3+4 3 1 4 K T L M N EXAMPLE 62 AABC is an equilateral triangle and a point D = int AABC such that AD ± DC and m(ZDCA) = 42°. Find m(ZBAD). Solution Let us draw an appropriate figure. m(ZB) = 60° (AABC is equilateral) m(ZBCD) = 60° – 42° = 18° (m(ZBCA) = 60°) m(ZADC) = m(ZB) + m(ZBAD) + m(ZBCD) (Property 3.3) 90° = 60° + m(ZBAD) + 18° m(ZBAD) = 12° A B C D 60° 42° 129 Triangles and Construction Check Yourself 15 1. Each figure shows a triangle with two or more angle bisectors. Find the indicated angle measures in each case. 2. In the triangle ABC at the right, AN is an angle bisector and AH is an altitude. Given m(ZC) – m(ZB) = 36°, find m(ZHAN). 3. A student draws the altitude and the angle bisector at the vertex of the right angle of a right triangle. The angle between them is 18°. Find the measure of the larger acute angle in the right triangle. 4. Find the value of x in the figure. Answers 1. a. 110° b. 80° c. 35° d. 40° e. p f. 80° 2. 18° 3. 63° 4. 15° 4x 2x x 105° A B N H C P Q R 40° x S x = ? S K T x 50° M x = ? x 70° P Q R M x = ? E Q P R M p y x x y = ? (in terms of p) S T K M 70° x x = ? S M K T y = ? 3x+70° 2x y a. b. c. d. e. f. 130 Geometriy 7 B. RELATIONS BETWEEN ANGLES AND SIDES Theorem If one side of a triangle is longer than another side of the triangle then the measure of the angle opposite the longer side is greater than the measure of the angle opposite the shorter side. In other words, if two sides of a triangle have unequal lengths then the measures of the angles opposite them are also unequal and the larger angle is opposite the longer side. Proof Given: AABC with AB > AC Prove: m(ZC) > m(ZB) A K B C 3 2 1 longer side opposite larger angle We begin by locating K on AB such that AK = AC. We then draw CK and continue with a two- column proof. Statements Reasons 1. AB > AC 1. Given 2. AAKC is isosceles 2. Definition of isosceles triangle (AK = AC) 3. Z3 = Z2 3. Base angles in an isosceles triangle are congruent. 4. m(ZACB) = m(Z2) + m(Z1) 4. Angle Addition Postulate 5. m(ZACB) > m(Z2) 5. Definition of inequality 6. m(ZACB) > m(Z3) 6. Substitution property 7. m(Z3) > m(ZB) 7. Triangle Exterior Angle Theorem 8. m(ZACB) > m(ZB) 8. Transitive property of inequality EXAMPLE 63 Write the angles in each triangle in order of their measures. Solution a. Since 7 > 5 > 3, m(ZA) > m(ZB) > m(ZC). b. Since 5 = 5 > 4, m(ZE) = m(ZF) > m(ZD). 5 5 4 D E F A B C 5 3 7 a. b. 131 Triangles and Construction Theorem If two angles in a triangle have unequal measures then the sides opposite them have unequal lengths and the longer side is opposite the larger angle. Proof Given: AABC with m(ZB) > m(ZC) Prove: AC > AB We will give a proof by contradiction in paragraph form. According to the trichotomy property, exactly one of three cases holds: AC < AB, AC = AB or AC > AB. + Let us assume that either AC = AB or AC < AB and look for a contradiction. + If AC < AB then m(ZB) < m(ZC) by the previous theorem. Also, if AC = AB then m(ZB) = m(ZC) by the definition of an isosceles triangle. + In both cases we have a contradiction of the fact that m(ZB) > m(ZC). That means that our assumption AC s AB must be false. By the trichotomy property, it follows that AC > AB. A B C larger angle opposite longer side Trichotomy property For any two real numbers a and b, exactly one of the following is true: a < b, a = b, a > b. EXAMPLE 64 Order the sides of triangle in the figure according to their length. Solution m(ZA) + m(ZB) + m(ZC) = 180° 2x + 40° + 20° + 3x – 10° = 180° 5x = 130° x = 26° So m(ZA) = (2 · 26°) + 40° = 92° and m(ZC) = (3 · 26°) – 10° = 68°. Since m(ZA) > m(ZC) > m(ZB), by the previous theorem we have a > c > b. A B C 20° (2x + 40°) (3x 10°) c a b EXAMPLE 65 In the figure at the right, KN = KM. Prove that KT > KM. K T N M 1 2 4 3 132 Geometriy 7 Solution Given: KN = KM Prove: KT > KM Proof: Statements Reasons 1. KN = KM 1. Given 2. Z2 = Z3 2. Base angles of isosceles triangle KNM 3. m(Z2) = m(Z1) + m(Z4) 3. Triangle Exterior Angle Theorem 4. m(Z2) > m(Z1) 4. By 3 5. m(Z3) > m(Z1) 5. Substitute 2 into 4. 6. KT > KM 6. By the previous theorem EXAMPLE 66 Prove that in any triangle ABC, a + b + c > h a + h b + h c , where h a , h b and h c are the altitudes to the sides a, b and c, respectively. Solution Given: AABC with altitudes h a , h b and h c Prove: (a + b + c) > (h a + h b + h c ) Proof: Look at the figure. By the previous theorem, in right triangle BCD, BC > BD, i.e. a > h b ; (1) in right triangle AEC, AC > CE, i.e. b > h c ; (2) in right triangle ABH, AB > AH, i.e. c > h a . (3) Adding inequalities (1), (2) and (3) gives (a + b + c) > (h a + h b + h c ). A E B H C D EXAMPLE 67 Find the longest line segment in the figure using the given angle measures. A B C D 62° 60° 59° 61° 55° 63° Solution In AABC, m(ZB) > m(ZA) > m(ZC) so AC > BC > AB. (1) (By the previous theorem) In AADC, m(ZC) > m(ZA) > m(ZD) so AD > CD > AC. (2) (By the previous theorem) Combining (1) and (2) gives us AD > DC > AC > BC > AB. So AD is the longest segment in the figure. 133 Triangles and Construction Check Yourself 16 1. Write the measures of the angles in each triangle in increasing order. 2. Write the lengths of the sides of each triangle in increasing order. 3. Find the longest line segment in each figure using the given angle measures. Answers 1. a. m(ZB) < m(ZA) < m(ZC) b. m(ZM) < m(ZP) < m(ZN) c. m(ZN) < m(ZK) < m(ZM) 2. a. c < b < a b. n = m < k c. k < s = t 3. a. CD b. PK c. BC 65° 60° 62° 60° A B C D P M N K 60° 61° 60° E A B C D 61° 59° 66° 60° 67° 57° a. b. c. 80° 50° 50° K N M S K T 70° 40° A B C 80° 60° 40° a. b. c. A B C 8 4 7 P M N 10 5 9 K M N 6 8 a. b. c. Activity For this activity you will need a piece of string and a ruler. + Cut the string into eight pieces of different lengths. Measure the lengths and label or mark each piece with its length. + Take any three pieces of string and try to form a triangle with them. + Make a table to note the lengths of the pieces of string and whether or not they formed a triangle. + Repeat the activity until you have two successes and two failures at making a triangle. + Look at your table. Which lengths of string together made a triangle? Which lengths didn’t make a triangle? What conjecture can you make about the sides of a triangle? Triangle Inequality 134 Geometriy 7 Properties 4 Triangle Inequality Theorem In any triangle ABC with sides a, b and c, the following inequalities are true: |b – c| < a < (b + c), |a – c| < b < (a + c), |a – b| < c < (a + b). The converse is also true. This property is also called the Triangle Inequality Theorem. Solution In AABC, |10 – 5| < x < (10 + 5) (Triangle Inequality Theorem) 5 < x < 15. (1) In ADBC, |7 – 4| < x < (7 + 4) (Triangle Inequality Theorem) 3 < x < 11. (2) The possible values of x are the elements of the common solution of inequalities (1) and (2), i.e. 5 < x < 11. So x = {6, 7, 8, 9, 10}. EXAMPLE 69 Find all the possible integer values of x in the figure. A B D C 10 5 4 7 x EXAMPLE 68 Is it possible for a triangle to have sides with the lengths indicated? a. 7, 8, 9 b. 0.8, 0.3, 1 c. 1 1 , , 1 2 3 Solution We can check each case by using the Triangle Inequality Theorem. a. |9 – 8| < 7 < (8 + 9) |8 – 9| < 8 < (7 + 9) |7 – 8| < 9 < (7 + 8). This is true, so by the Triangle Inequality Theorem this is a possible triangle. b. |0.8 – 0.3| < 1 < (0.8 + 0.3) |1 – 0.3| < 0.8 < (1 + 0.3) |1 – 0.8| < 0.3 < (1 + 0.8). This is true, so by the Triangle Inequality Theorem this is a possible triangle. c. This is impossible, since 1< 1 1 . 2 3 + 135 Triangles and Construction EXAMPLE 70 Find the greatest possible integer value of m in the figure, then find the smallest possible integer value of n for this case. Solution In AABD, |9 – 6| < m < (9 + 6) (Triangle Inequality Theorem) 3 < m < 15. So the greatest possible integer value of m is 14. In AADC, |8 – m| < n < (m + 8) (Triangle Inequality Theorem) |8 – 14| < n < (14 + 8) (m = 14) 6 < n < 22. So when m = 14, the smallest possible integer value of n is 7. A B D C m 9 6 n 8 Properties 5 1. In any triangle ABC, a. if m(ZA) = 90° then b. if m(ZA) < 90° then |b – c| < a < c. if m(ZA) > 90° then < a < (b + c). 2. In any triangle ABC, if P = int AABC then (BP + PC) < (BA + AC). 3. In any triangle ABC, if m(ZB) > m(ZC) or m(ZB) < m(ZC) then h a < n A < V a . A B H N D C h a n A V a A B H N D C h a n A V a A B C P 2 2 + b c 2 2 + . b c 2 2 + = <( + ). b c a b c 136 Geometriy 7 EXAMPLE 71 In a triangle ABC, m(ZA) > 90°, c = 6 and b = 8. Find all the possible integer lengths of a. Solution Since m(ZA) > 90°, < a < (b + c) by Property 5.1. Substituting the values in the question gives < a < (8 + 6), i.e. 10 < a < 14. So a = {11, 12, 13}. 2 2 8 +6 2 2 + b c EXAMPLE 72 In the triangle ABC shown opposite, P = int AABC, AB = 10, AC = 8 and BC = 9. Find the sum of all the possible integer val- ues of PB + PC. Solution + In APBC, BC < (BP + PC) by the Triangle Inequality Theorem. So 9 < BP + PC. (1) + In AABC, (PB + PC) < (AB + AC) by Property 5.2. So PB + PC < 10 + 8. (2) + Combining (1) and (2) gives 9 < (PB + PC) < 18. + So the possible integer values for PB + PC are 10, 11, 12, 13, 14, 15, 16 and 17. + The required sum is therefore 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 = 108. A P B C 10 9 8 EXAMPLE 73 Prove that the sum of the lengths of the medians of a triangle is greater than half of the perimeter and less than the perimeter. Solution Let us draw an appropriate figure. Given: AABC with centroid G Prove: Proof: We need to prove two inequalities. + + <( + + )<( + + ). 2 a b c a b c V V V a b c A B D E C F G Remember! The centroid of a triangle is the point of intersection of its medians. 137 Triangles and Construction Proof that We will use the Triangle Inequality Theorem three times. + In ACEB, (CE + EB) > BC, i.e. (V c + ) > a. (Triangle Inequality Theorem) + In AADC, (AD + DC) > AC, i.e. (V a + ) > b. (Triangle Inequality Theorem) + In AABF, (BF + FA) > AB, i.e. (V b + ) > c. (Triangle Inequality Theorem) So (V c + V a + V b + + + ) > (a + b + c). (Addition Property of Inequality) So (1) (Subtraction Property of Inequality) Proof that V a + V b + V c < a + b + c: For the second part of the inequality, let us draw another figure as shown at the right and extend the median AD through D to a point K such that AD = DK. (2) Then join K and B. Now, BD = DC (AD is a median) m(ZBDK) = m(ZADC) (Vertical angles) AD = DK. (By (2)) So by the SAS Congruence Postulate, ADBK = ADCA and so |BK| = |CA| = b. Then, in AABK, 2V a < b + c. (3) (Triangle Inequality Theorem) By considering the other medians in a similar way we get 2V b < (a + c) and 2V c < (a + b). (4) Adding the inequalities from (3) and (4) side by side gives us 2(V a + V b + V c ) < 2(a + b + c). So (V a + V b + V c ) < (a + b + c). (5) Finally, by (1) and (5), as required. + + <( + + )<( + + ) 2 a b c a b c V V V a b c A B D C a 2 a 2 V a b c A B D C K a 2 a 2 V a V a b c b + + ( + + . 2 a b c a b c V V V ) > 2 b 2 a 2 c 2 b 2 a 2 c : 2 a b c a+b+c <V +V +V 138 Geometriy 7 EXAMPLE 75 Prove that for any triangle ABC, if P = int AABC and x, y and z are as shown in the figure then (x + y + z) < (a + b + c) < 2(x + y + z), i.e. + + <( + + )<( + + ). 2 a b c x y z a b c A P B C c a b y z x Solution Let M and N be as in figure 1, and let be a line representing the river. Then we can use the following method to locate A: 1. Draw a ray from N perpendicular to , intersecting at point B. 2. Locate point C on the extension of NB such that NB = BC. 3. Draw KC. 4. Locate A at the intersection of KC and , as shown in figure 2. Now we need to show that A is really the location which makes AN + AK as small as possible. Figure 3 shows an alternative location X on l. Notice that in AKXC, (KX + XC) > KC by the Triangle Inequality Theorem. So (KX + XC) > (KA + AC) (1) by the Segment Addition Postulate. Since AB ± NC and NB = BC, ANXC is isosceles with XC = NX (2). By the same reasoning, ANAC is isosceles with NA = AC (3). Substituting (2) and (3) into (1) gives us (KX + XN) > (KA + AN). So A is the best location for the station. The result we have just proved does not mean that for a given triangle, the sum of the medians can be anything between the half perimeter and full perimeter of the triangle. This is because the lengths of the medians are directly related to the lengths of the sides. As we will see in the next chapter, once we know the lengths of the three sides of a triangle then we can calculate the lengths of its medians. Their sum is a fixed number. Remark EXAMPLE 74 Two towns K and N are on the same side of the river Nile. The residents of the two towns want to construct a water pumping station at a point A on the river. To minimize the cost of constructing pipelines from A to K and N, they wish to locate A along the Nile so that the distance AN + AK is as small as possible. Find the corresponding location for A and show that any other location requires a path which is longer than the path through A. X C A B K N l K N figure 1 figure 3 K N A B figure 2 C l l 139 Triangles and Construction Check Yourself 17 1. Two sides of a triangle measure 24 cm and 11 cm respectively. Find the perimeter of the triangle if its third side is equal to one of other two sides. 2. Determine whether each ratio could be the ratio of the lengths of the sides of a triangle. a. 3 : 4 : 5 b. 4 : 3 : 1 c. 10 : 11 : 15 d. 0.2 : 0.3 : 0.6 3. The lengths of the sides DE and EF of a triangle DEF are 4.5 and 7.8. What is the greatest possible integer length of DF? 4. The base of an isosceles triangle measures 10 cm and the perimeter of the triangle is an integer length. What is the smallest possible length of the leg of this triangle? 5. In an isosceles triangle KLM, KL = LM = 7 and m(ZK) < 60°. If the perimeter of the triangle is an integer, how many possible triangle(s) KLM exist? 6. In a triangle ABC, AB = 9 and BC = 12. If m(ZB) < 90°, find all the possible integer lengths of AC. Answers 1. 59 cm 2. a. yes b. no c. yes d. no 3. 12 4. 5.5 cm 5. six triangles 6. AC = {4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14} The example that we have just seen shows an application of triangle inequality. But the result we obtained does not mean that the value of x + y + z can be any number less than a + b + c. In other words, the maximum value of x + y + z may be a lot less than a + b + c. In fact, the maximum value of x + y + z is always less than the sum of the lengths of the two longer sides of the triangle, because as the interior point moves towards one of the vertices, two distances increase but the third distance decreases. When this interior point reaches the vertex point, the distance to this point becomes zero and the sum of the distances becomes the sum of the two sides which include this vertex. So the maximum value of x + y + z will always be less than the sum of the length of the two longer sides. Remark Solution + In AABP, c < (x + z). (Triangle Inequality Theorem) + In AAPC, b < (y + z). (Triangle Inequality Theorem) + In ABPC, a < (x + z). (Triangle Inequality Theorem) So (a + b + c) < 2(x + y + z). (1) (Addition property of inequality) + Also, (x + y) < (c + b), (Property 5.2) (y + z) < (a + c) and (Property 5.2) (x + z) < (b + a). (Property 5.2) So (x + y + z) < (a + b + c) (2) (Addition property of inequality) As a result, (x +y +z) <(a +b +c) <2(x +y +z), (By (1) and (2)) or equivalently, + + <( + + )<( + + ). 2 a b c x y z a b c 140 Geometriy 7 8. In the triangle ABC at the right, AB = AD = BE, m(ZA) = 114° and m(ZB) = 60°. Find m(ZEDC). A B E D C 60° 114° ? A. Relations Between Angles 1. Each figure below shows triangles with two or more congruent sides. Find the value of x in each figure, using the information given. 2. An angle in a triangle measures 20° less than the measure of the biggest angle in the triangle. The measure of the third angle is half the measure of the biggest angle. Find the measures of all three angles. 4. In a triangle ABC, the angle bisector of the interior angle ZC makes an angle of 40° with the side AB. Find the angle between the bisector of the exterior angle ZC´ and the extension of the side AB. 6. In the triangle MNP opposite, MS = MP, ST = TP, m(ZM) = 94° and m(ZN) = 26°. Find m(ZMST). 5. In a triangle KMN, the altitudes to sides KM and MN intersect each other at a point P. Find m(ZKPN) if m(ZKNM) = 72° and m(ZNKM) = 64°. 3. The two acute angles in a right triangle measure and respectively. Find x and the measures of these angles. ° ( – 25) 4 x ° 2 ( +5) 3 x EXERCISES 3.3 15° x M N R P 48° A B C E D m(ÐBAC) = x x Q T S M R P A B D C x T x 9° M N S TM = TN 30° S T M J x M N S Q x 2x M K P N J K 52° x a. b. c. d. e. f. g. h. M 94° 26° S N T P ? 7. In the triangle ABC at the right, BD = BE = BC and segment EB bisects ZB. If m(ZACD) = 18°, find m(ZABC). A 18° E F D B C 141 Triangles and Construction B. Relations Between Angles and Sides 17. Each figure shows the lengths of two sides of a triangle. Write an interval for the possible length of the third side. A B C 8 6 a P R S 12 8 p m 3 4 K L M a. b. c. 11. The bisectors of the interior angles ZD and ZF in a triangle DEF intersect at the point T. Find the measure of ZDEF if its measure is one-third of m(ZDTF). 12. a, b and b are the measures of the interior angles of an isosceles triangle such that a and b are integers and 24° < b < 38°. Find the smallest possible value of a. 13. In a triangle DEF, point M is on the side DF and ZMDE and ZDEM are acute angles. Draw an appropriate figure for each of the following, if it is possible. a. AFME is obtuse b. AFME is equilateral c. ADME is equilateral and ADEF is isosceles d. ADME is isosceles and AEMF is isosceles e. ADME is isosceles and ADEF is equilateral 14. x, y and z are the exterior angles of a triangle. Determine whether each ratio is a possible ratio of x : y : z. a. 2 : 3 : 5 b. 1 : 2 : 3 c. 6 : 11 : 19 d. 12 : 15 : 21 9. In the figure, PQ = PS = PR and m(ZSPR) = 24°. Find m(ZSQR). P 24° Q R S E ? 10. In the figure opposite, m(ZNMT) = 16°, m(ZTMP) = 44°, m(ZP) = 38° and m(ZSNT) = 22°. Find m(ZTSN). 22° 16° 44° M N P S 38° T 15. Find the value of x in the figure. x 140° 145° 130° 16. In an isosceles triangle KMN, the bisectors of the base angles ZK and ZM intersect each other at a point T. Prove that m(ZKTM) = m(K´). 18. For each figure, state the interval of possible values for the length x. K M N P x 12 4 7 8 A B D C x 8 6 4 5 A B D C 6 10 7 5 x a. b. c. 142 Geometriy 7 28. For each figure, order the numbered angles according to their size. 27. A triangle has side lengths 2x + y, 2y + 3x and 2x. Which one is bigger: x or y? 18 20 20 1 2 3 1 3 2 n 1 n n + 1 1 2 3 ò17 4.20 82 20 16 2 15 1 3 a. b. c. d. 19. A triangle ABC has sides a, b and c with integer lengths. How many triangles can be formed such that b = c and a · b = 18? 20. In the figure, a, b and c are integers. Calculate the smallest possible value of a + b + c, using the information given. D A F C E B c b a 5 9 6 7 5 6 22. In the figure, m(ZA) > 90° and m(ZC) > 90°. If AB = 6 cm, AD = 10 cm, BC = 12 cm and CD = 5 cm, find the sum of the all the possible integer lengths of the side BD. A B D C 10 6 12 5 23. In the triangle ABC at the right, AD = 9 cm, BD = 6 cm, DC = 8 cm, AC = x cm and AB = y cm. Find the sum of the smallest and largest possible integer values of x + y. A B D C x 9 y 6 8 24. In the figure, AB = 8 cm, AC = 10 cm, BD = 3 cm, CD = 7 cm and BC = 2x + 1 cm. Find the sum of all the possible integer values of x. A B D C 10 8 3 7 2x + 1 25. In the figure, AB = 8 cm, BC = 12 cm, CD = 6 cm and DA = 4 cm. Find the number of possible integer lengths of AC. 6 12 8 4 C D A B 21. In the figure, AC = 9 cm, BC = a, AB = c and m(ZBAC) > 90°. Find the smallest possible value of a + c if a, c = . 9 c a A B C 26. In each case determine whether it is possible for a triangle to have sides with the lengths given. a. 13, 9, 5 b. 5, 5, 14 c. 8, 8, 16.1 d. 17, 11, 6 e. 0.5, 0.6, 1 f. 18, 18, 0.09 143 Triangles and Construction 31. A student has five sticks, each with an integer length. He finds that he cannot form a triangle using any three of these sticks. What is the shortest possible length of the longest stick, if a. the lengths of the sticks can be the same? b. all the sticks have different lengths? (Hint: Use the Triangle Inequality Theorem.) 32. How many distinct isosceles triangles have integer side lengths and perimeter 200 cm? 30. State the longest line segment in each figure. C B A (x + 10)° 110° (x 8)° A B C D 70° 70° 10° A C D B 10° 20° A B D C 20° 65° 60° 80° a. b. c. d. 29. Determine whether each statement is true or false. a. In a triangle ABC, if the measure of ZA is 57° and the measure of ZB is 64° then the shortest side of AABC is AB. b. In a triangle KMN, if the measure of ZK is 43 and the measure of ZM is 47 then the shortest side of AKMN is KM. c. In a triangle ABC, if ZB is an obtuse angle and AH ± BC then HA < AB. d. If an isosceles triangle KTA with base KA has TA < KA then the measure of ZT is always less than 90°. e. An angle bisector in an equilateral triangle is shorter than any of the sides. f. All obtuse triangles are isosceles. g. Some right triangles are equilateral. 33. How many triangles can be drawn if the length of the longest side must be 11 units and all side lengths must be integer values? 35. In a triangle ABC, AB = 8 cm, BC = x and AC = y. If m(ZA) > 90° and x, y = , find the smallest possible value of x + y. 34. In the figure, AD = 5 cm, AB = 12 cm, BC = 9 cm and DC = 8 cm. If m(ZA) > 90° and m(ZC) < 90°, find all possible integer values of BD. A B C D 5 8 9 12 ? 144 Geometriy 7 40. In a triangle DTF, m(ZD) = 90° and m(ZF) < 45°. a. 2 · m(ZT), m(ZD) b. FT, 2DF c. FD, DT d. m(ZF´), 2m(ZT) 41. In a triangle DEF, m(ZD) > m(ZE) = m(ZF). a. m(ZD), 60° b. m(ZE), 60° 42. In a triangle DEF, m(ZE) = 120° and EF > DE. a. 120°, 3 · m(ZD) b. 2 · m(ZE), 3 · m(ZD) 39. a. (x – 10)°, (y + 20)° b. MB + MC, AB + AC A B C 30° y x M 36. How many distinct triangles have integer side lengths and perimeter 11? 37. Prove each theorem. a. Hinge Theorem: If two sides of one triangle are congruent to two sides of another triangle, and if the included angle of the first triangle is larger than the included angle of the second, then the third side of the first triangle is longer than the third side of the second. b. Converse of the Hinge Theorem: If two sides a and b of one triangle are congruent to two sides d and e of another triangle, and if the third side of the first triangle is longer than the third side of the second, then the angle between a and b is larger than the angle between d and e. 38. a. 130°, x b. y, 90° c. y, x d. KM, MN x 130° y K M N Instructions for questions 38 to 42 Each question gives two quantities to be compared, separated by a comma. In each case, use the figure or extra information to compare the quantities. Write A if the first quantity is greater than the second, B if the first quantity is smaller than the second, C if the quantities are equal, or D if the extra information is not enough for you to be able to compare the quantities. All variables represent real numbers. Figures are generally not drawn to scale. 145 Triangles and Construction A. DISTANCE FROM A POINT TO A LINE Let A(x 1 , y 1 ) be a point and d: ax + by + c = 0 be a line, then the distance from A to the line d is . 1 1 2 2 | + + | = + ax by c l a b Theorem Proof Let the distance of A(x 1 , y 1 ) to the line d: ax + by + c = 0 be l = AH. Take C(x 2 , y 2 ) = AD · d. x 2 = x 1 and y 2 = CD C is a point on the line ax + by + c = 0, so ax 1 + b · CD + c = 0 b · CD = –a · x 1 – c So we have the coordinates of C, Now, o is the inclination of d and o = m(ZCBD) = m(ZCAH) (angles with perpendicular sides). In the right triangle ACH, and AH = AC · cos o ...(1) Now, let’s find the equivalent expressions for AC and coso. AC = AD – CD = We know sec 2 o = 1 + tan 2 o so , and so o ± ± - - 2 2 1 1 cos = = ...(3) 1+( ) 1 ( ) a a b b tan = a m b o · - o ± o 2 1 cos = 1+tan - - - - - 1 1 1 1 ( ), so = ...(2) a c a c y x AC y x b b b b o cos = AH AC 1 1 ( , – – ). a c C x x b b · - - 1 . a c CD x b b O y A(x 1 , y 1 ) C(x 1 , |CD|) a a H B D x d distance from a point to a line After studying this section you will be able to: 1. Find the distance from a point to a line. 2. Find the distance between two parallel lines. Objectives 146 Geometriy 7 Substituting (2) and (3) in (1), l = AH = AC · cos o, and since l is the distance, 1 1 1 1 1 1 2 2 2 2 2 2 + + + + 1 = ( ) = = . 1 + + 1+ a c y x ax by c a c b b l y x b b a a b a b b b - - O(0, 0) = O(x 1 , y 1 ). Using the formula, - - 1 1 2 2 2 2 | + + | |0 0+4| 4 4 2 = = = = =2 2. 2 2 + 1 +( 1) ax by c l a b Solution EXAMPLE 76 Find the distance from the point O(0, 0) to the line x – y + 4 = 0. A(5, 2) · A(x 1 , y 1 ). Using the formula, - 1 1 2 2 2 2 | + + | |3 5 4 2+5| 12 = = = =2.4. 5 + 3 +4 ax by c l a b Solution EXAMPLE 77 Find the distance from A(5, 2) to the line 3x – 4y + 5 = 0. ± ± 1 1 2 2 | + + | |5 12 – 12 5+5 | = = =10 25+144 + |5 | = =10, so |5 |=130, i.e. 5 = 130, and so = 26. 13 ax by c k l a b k l k k k Solution EXAMPLE 78 The distance from A(12, 5) to the line 5x – 12y + 5k = 0 is ten units. Find the possible values of k. Check Yourself 18 1. Find the distance from the point P(–2, 3) to the line 3x + 4y + 9 = 0. 2. Find the distance from the point A(1, 4) to the line y = 3x – 4. 3. The distance between the point P(k, 3) and the line 4x – 3y + 5 is 4 units. Find k. Answers 1. 3 2. 3. k {–4, 6} 10 2 147 Triangles and Construction Let d 1 : a 1 x + b 1 y + c 1 = 0 d 2 : a 2 x + b 2 y + c 2 = 0 be two parallel lines. Since d 1 d 2 , we can write , so a 1 = k · a 2 and b 1 = k · b 2 . Now, let’s substitute these values into d 1 : k · a 2 x + k · b 2 y + c 1 = 0 k(a 2 x + b 2 y + ) = 0. k ÷ 0, so we get d 1 : a 2 x + b 2 y + = 0. When we compare d 1 with d 2 , we see that their difference is a constant number. In general, we can write two parallel lines d 1 and d 2 as: d 1 : ax + by + c 1 = 0 d 2 : ax + by + c 2 = 0. 1 c k 1 c k 1 1 2 2 a b k a b · · Let d 1 : ax+ by + c 1 = 0 and d 2 : ax + by + c 2 = 0 be two parallel lines. Then the distance between d 1 and d 2 is . - 2 1 2 2 | | = + c c l a b Theorem B. DISTANCE BETWEEN TWO PARALLEL LINES distance between two parallel lines Solution EXAMPLE 79 Find the distance between the parallel lines x – 2y + 5 = 0 and 3x – 6y + 9 = 0. Proof The distance of any point A(x, y) on line d 1 to the line d 2 is In the equation of d 1 : ax + by + c = 0 ax + by = –c 1 , and so 2 1 2 2 | | = . + c c l a b - 2 2 2 | + + | . + ax by c a b l A(x, y) d 2 d 1 It is important to notice that to find the distance between two parallel lines, first of all we need to equalize the coefficients of x and y. Remark - - ¦ ¦ ¦ ¦ = ` ` - - ¦ ¦ ' ' ¦ - ¦ = ` - ¦ ' 1 1 2 1 1 2 2 2 2 : 2 +5=0 3 6 +15=0 was multiplied by 3. : 3 6 +9=0 3 6 +9=0 Now, we have =15 | | 15 – 9 6 6 2 5 = = = = = . 5 9+36 45 3 5 =9 d x y x y d d x y x y c c c l a b c 148 Geometriy 7 There is also another way to solve the problem: The distance between d 1 and d 2 is the same as the distance of any point on d 1 or d 2 to the other line. For example, A(0, –4) is one point on d 2 , and the distance of A to d 1 is The solution is the same. 1 1 2 2 |3 2 +5| |3 0 2(– 4)+5| 13 = = = 13. 13 13 3 +(–2) x y l - - · - - ¦ ¦ ¦ ¦ = ` ` - - - ¦ ¦ ' ' ¦ - ¦ = ` - ¦ ' 1 1 2 2 1 1 2 2 2 2 2 2 : 3 2 +5=0 : 3 2 +5=0 : 3 +2 +8 =0 : 3 2 8 =0 So =5 | | |5+8| 13 = = = = 13. 13 3 +(–2) =–8 d x y d x y d x y d x y c c c l a b c Solution EXAMPLE 80 Find the distance between the parallel lines 3x – 2y + 5 = 0 and –3x + 2y + 8 = 0. Check Yourself 19 1. Find the distance between the lines 4x – 3y – 5 = 0 and –12x + 9y + 4 = 0. 2. The lines x + 2y + 1 = 0 and 3x + 6y + k = 0 are parallel and the distance between them is ñ5. Find k. 3. Find the area of the square whose two sides are on the parallel lines 2x + y – 2 = 0 and 4x + 2y + 6 = 0. Answers 1. 2. k {–12, 18} 3. 5 11 15 149 Triangles and Construction A. Distance from a Point to a Line 1. Find the distance from the point A(–2, 3) to the line 8x + 6y – 15 = 0. 4. The distance between P(1, –2) and the line 7x – y + k = 0 is 4ñ2 units. Find k. 5. The points A(1, 3), B(–2, 1) and C(3, –1) are the vertices of the triangle ABC. Find the length of the altitude of BC. 6. The distance from P( , k) to the line 12x + 9y – 1 = 0 is 2 units. Find k. 1 2 8. The distance between the parallel lines 12x + 9y – 2 = 0 and ax + 3y + c = 0 is three units. Find the ratio > , if 0. a c c B. Distance Between Two Parallel Lines 7. Find the distance between each pair of parallel lines. a. –2x + 3y – 4 = 0 and –2x + 3y – 17 = 0 b. x – y – 4 = 0 and –2x + 2y – 7 = 0 c. y = 2x + 1 and 2y = 4x – 3 3. The distance from a line with equation y – 4 = m(x + 2) to the origin is 2. Find m. 2. The distance between B(2, 3) and the line 12y – 5x = k is . Find k. 5 13 Write the equations of the lines which are four units away from the line 3x + 4y + 10 = 0. 9. 10. The distance between the parallel lines 3x + 4y – 6 = 0 and 4x – ky + 4 = 0 is p. Find k + p. EXERCISES 3.4 Angles Chapter 3 Review Test A 1. In the triangle ABC in the figure, m(ZA) = 4x, m(ZB) = x and m(ZC) = 30°. Find the value of x. A) 10° B) 15° C) 20° D) 25° E) 30° A B C 4x x 30° 6. Which is the longest side in the figure, according to the given angle measures? A) BC B) AB C) BD D) CD E) BE 7. In a triangle DEF, DE = EF and DF > EF. Which statement is true? A) DE < (DF – EF) B) m(ZE) > m(ZF) C) m(ZE) < m(ZD) D) m(ZE) = 60° E) m(ZE) = m(ZD) A B E D C 62° 61° 60° 60° 8. What is the sum of the smallest and largest possible integer values of x in the figure? A) 26 B) 24 C) 22 D) 20 E) 17 A B D C 8 10 14 10 x 4. In the figure, m(ZP) = 45°, m(ZN) = 36° and m(ZR) = 25°. Find the value of x. A) 260° B) 256° C) 254° D) 248° E) 244° P R N S 25° 36° x 45° 2. In a triangle MNP, the interior angle bisectors of ZM and ZP intersect at the point S. Given that ZN measures 40°, find m(ZPSM). A) 95° B) 100° C) 105° D) 110° E) 120° 5. Two sides of a triangle have lengths 8 and 12. What is the sum of the minimum and maximum possible integer values of the length of the third side? A) 24 B) 22 C) 19 D) 18 E) 16 3. In the triangle STK opposite, N = TK and SN is the interior angle bisector of ZS. If m(ZT) – m(ZK) = 40°, find m(ZSNK). A) 110° B) 105° C) 100° D) 95° E) 90° S T N K ? CHAPTER 3 REVIEW TEST A Chapter Review Test 1A 151 9. In a triangle ABC, D = BC and AD bisects ZA. If AB = 6 cm, BD = 3 cm and DC = 2 cm, find the length of AD. A) 5ñ2 cm B) 4ñ3 cm C) 3ñ2 cm D) 2ñ3 cm E) ñ3 cm 10. In the figure, BD = DC, AD = AE and m(ZC) = 20°. Find m(ZEDC). A) 70° B) 65° C) 60° D) 45° E) 30° 11. If AMNP = ASTK, which of the following statements is false? A) MN = ST B) MP = TK C) NP = TK D) ZPNM = ZSTK E) KT = PN A B D C E ? 20° 12. In the figure, BD bisects ZB, BD = BE and DE = EC. If m(ZA) = 80° and, m(ZACD) = 20°, what is m(ZBDC)? A) 100° B) 110° C) 120° D) 140° E) 150° A B E C D 20° 80° 14. In the figure, m(ZBAC) = 90°, m(ZC) = 60° and BD = DC. Find BC if AD = 2x + 3 and AC = 6x – 1. A) 6 B) 8 C) 10 D) 12 E) 14 C A B D 2x + 3 6 x 1 60° 15. In the figure, m(ZBAC) = 90°, m(ZBAD) = 12°, BC = 8 cm and AD = 4 cm. What is m(ZABC)? A) 52° B) 54° C) 58° D) 60° E) 64° A B C D 8 4 12° ? 13. a, b and c are the lengths of the sides of a triangle ABC. Given that a, b and c are integers and a 2 – b 2 =17, what is the sum of the minimum and maximum possible values of c? A) 7 B) 13 C) 17 D) 18 E) 23 16. In the figure, ND = DP and . What is ? A) B) C) D) E) 3 6 3 5 3 4 3 3 3 2 MH NP 3 = 3 HD MH M N H D P Angles Chapter 3 Review Test B CHAPTER 3 REVIEW TEST B 1. In the figure, AB = AD, AC = BC and m(ZDAC) = 15°. Find m(ZC). A) 40° B) 45° C) 50° D) 60° E) 65° A B C D 15° ? 2. In the triangle MNP in the figure, MS = NS and KN = KP. If m(ZMRP) = 117°, what is m(ZMNP)? A) 39° B) 41° C) 43° D) 45° E) 47° 3. In a triangle ABC, Dis a point on the side AB and CD is the interior angle bisector of ZC. If AB = 15 cm and 3 · AC = 2 · BC, find the length of DB. A) 2 cm B) 3 cm C) 4 cm D) 6 cm E) 9 cm 5. In a triangle ABC, points D and E are the midpoints of the sides AB and AC respectively. DE = (x + 5)/4 and BC = 8x – 5 are given. What is the value of x? A) 1 B) 2 C) 3 D) 4 E) 5 6. AABC is a right triangle with m(ZA) = 90°, and AH is the altitude to the hypotenuse. If m(ZC) = 30° and BH = 2 cm, find HC. A) 4 cm B) 5 cm C) 6 cm D) 7 cm E) 8 cm M K N S P 117° R ? 8. In the figure, MS = SN and MP = PN. If m(ZP) = 20°, m(ZKMP) = 40° and m(ZKNP) = 30°, what is m(ZSKN)? A) 50° B) 45° C) 40° D) 35° E) 30° 30° 40° M 20° K N P S ? 4. In the figure, MN = MP and ML = MK. If m(ZPLK) = 12°, what is m(ZLMN)? A) 18° B) 20° C) 24° D) 30° E) 36° P K M N L 12° ? 7. In the figure, AB = AD. What is ? A) B) 1 C) D) E) 4 3 3 2 2 3 1 2 DE EC A B C E D Chapter Review Test 1A 153 9. In the figure, HK = KN, m(ZDAC) = 40° and m(ZHKB) = 20°. Find m(ZBKD). A) 20° B) 30° C) 40° D) 60° E) 70° A B H K N C D 20° 40° ? 13. In the triangle MNP shown opposite, point O is the center of the inscribed circle of AMNP. If KS NP, KN = 6 and SP = 8, what is the length of KS? A) 10 B) 12 C) 14 D) 16 E) 18 M K N P S O 6 8 ? 15. In the figure, ACDE is a square, m(ZABC) = 60° and BD = 2 cm. Find the length of one side of the square. A) (3 – ñ3) cm B) (ñ3 – 1) cm C) (ñ3 + 1) cm D) (4 – 2ñ3) cm E) (2ñ5 – 3) cm A B C D E 2 60° 11. In the figure, MK = KL, MN = m, KN = 2m and NL = 3m. Find m(ZKNL). A) 45° B) 50° C) 60° D) 70° E) 75° M N K L m 2m 3m ? 10. In the figure, BD bisects angle B. Given m(ZADB) = 90°, DE BC, AB = 12 and BC = 16, find the length of DE. A) 1 B) C) 2 D) E) 3 5 2 3 2 A B D E C 16 12 ? 12. The lengths of the sides of a triangle ABC are integers a, b and c such that b = c and (a + b + c) · (a + b – c ) = 15. Find the value of a. A) 1 B) 2 C) 3 D) 5 E) 7 16. In the figure, AE = BD = DC and AB = AC. What is m(ZFDC)? A) 45° B) 50° C) 60° D) 62.5° E) 67.5° E A B D C F ? 14. In the figure, CH = HB, AD = 3 and DB = 8. What is the sum of the all possible integer values of the length AC? A) 30 B) 34 C) 40 D) 42 E) 51 C A D B H 3 8 ? Angles Chapter 3 Review Test C CHAPTER 3 REVIEW TEST C 1. In the figure, m(ZKBC) = m(ZKCA). and m(ZLKB) = 80°. What is the measure of ZACB? A) 40° B) 60° C) 70° D) 75° E) 80° A B C K L 80° 5. In the figure, AABC, ACDE, and AFEG are equilateral triangles. If BG = 16, what is the sum of the perimeters of the three triangles? A) 32 B) 36 C) 42 D) 46 E) 48 A B C D E G F 6. In the triangle ABC in the figure, CD is the bisector of ZC, AE is the median to BC and DE AC. If m(ZB) = 50°, what is m(ZBAC)? A) 30° B) 35° C) 40° D) 45° E) 50° A D B C E F 50° 7. Which of the following is a possible sum of the lengths of AB and BC in the figure? A) 11 B) 12 C) 13 D) 14 E) 37 A E B D C 10 7 6 13 12 2. Which is the largest angle in the figure, according to the given lengths? A) ZM B) ZN C) ZS D) ZSPK E) ZK 3. In an isosceles triangle XYZ, m(ZY) = m(ZZ) and m(ZX) < m(ZY). What is the largest possible integer measure of the angle Y? A) 59° B) 60° C) 89° D) 90° E) 110° 4. In a triangle ABC, points B, C and D are collinear and AD is the angle bisector of the exterior angle A´. If AC = BC, DB = 12 and AB = 4, find the length of BC. A) 2 B) 3 C) 4 D) 5 E) 8 S N P K M 12 9 8 6 2ò22 8 8. In the figure, MP = PS = SN = PT and ST = TN. What is m(ZNMP)? A) 36° B) 60° C) 72° D) 84° E) 108° M S N T P ? Chapter Review Test 1A 155 9. In the figure, AABC is an equilateral triangle. If BD = AE, what is the measure of ZEFC? A) 45° B) 60° C) 75° D) 90° E) 120° A D B C E F ? 11. In the figure, O is the incenter of AABC, AB OT and AC OV. If BT = 6 cm, TV = 7 cm and VC = 5 cm, what is the perimeter of the triangle OTV? A) 12 cm B) 15 cm C) 16 cm D) 18 cm E) 20 cm A B T V C O 6 7 5 13. In the figure, AF = FB and AE = EC. If EH + FH = 12, what is AB + AC? A) 16 B) 18 C) 22 D) 24 E) 36 A F B H C E 14. In the triangle ABC in the figure, BH is the exterior angle bisector of ZB and AE = EC. If m(ZBHC) = 90°, BC = 8 and EH = 7, what is the length of AB? A) 6 B) 7 C) 8 D) 10 E) 12 A B H C E ? 16. In the figure, m(ZM) = 90°, m(ZMST) = 150° and PM = MS = ST. What is m(ZN)? A) 5° B) 10° C) 15° D) 22.5° E) 30° P M S T N 150° ? 10. In the figure, AC = BC and AB = AD. If m(ZCAD) = 18° and m(ZEBD) = 12°, what is m(ZAEB)? A) 82° B) 80° C) 78° D) 72° E) 42° A B D E C 18° 12° ? 12. In the figure, AABC is an equilateral triangle. If PB = 16 and PN = 10, what is the length of AH? A) 2ñ3 B) 3ñ3 C) 4ñ3 D) 5ñ3 E) 6ñ3 N A C B P H K 16 10 ? 15. In the figure, MK = NK = PK. What is x + y + z? A) 270° B) 180° C) 90° D) 60° E) 45° M N P K x y z Angles Chapter 3 Review Test D CHAPTER 3 REVIEW TEST D 1. In the triangle ABC in the figure, BN is the bisector of ZABC and H is the intersection point of the altitudes of AABC. If m(ZAHC) = 110° and m(ZHBN) = 20°, what is m(ZBAC)? A) 50° B) 55° C) 65° D) 75° E) 80° A H B C N 110° 20° 2. In the figure, MN = MP, KP = KT, m(ZNMP) = m, m(ZPKT) = k, and points N, P and T are collinear. If m + k = 130°, what is m(ZMPK)? A) 50° B) 55° C) 60° D) 65° E) 70° M N P T K m k ? 5. In the triangle MNP opposite, MK = TK, NS = TS and m(ZKTS) = 50°. What is m(ZMPN)? A) 70° B) 65° C) 60° D) 55° E) 50° M T N S P K 50° ? 6. According to the figure, what is the value of ? A) 5 B) C) D) 1 E) 1 2 1 2 2 1 2 2 (2 ) m n m + D B E A A m n 8 12 7. The measure of one angle in a triangle is equal to the sum of the measures of the other two angles. Which statement about this triangle is always true? A) The triangle is equilateral. B) The triangle is acute. C) The triangle is a right triangle. D) The triangle is obtuse. E) The triangle is isosceles. 8. In the triangle XYZ in the figure, m(ZYZX) = 90°, XZ = PK and XP = PY. What is m(ZPKZ)? A) 120° B) 135° C) 140° D) 150° E) 160° Y K Z X P ? 3. In the figure, AB = AC = b, BC = a, and a < b. What is the largest possible integer value of m(ZA)? A) 59° B) 60° C) 44° D) 30° E) 29° A B C b b a 4. In the figure, AD = BD, m(ZDAC) = x and m(ZBCE) = 2x. If m(ZEAB) = 110°, what is the value of x? A) 30° B) 35° C) 40° D) 45° E) 50° 110° x 2x A B D C E Chapter Review Test 1A 157 9. In the figure, PM is the angle bisector of ZNPK, MN = MP, NS = SP and m(ZMKP) = 90°. What is m(ZSTP)? A) 90° B) 85° C) 80° D) 75° E) 60° K M N S P T ? 13. In the figure, m(ZDAC) = m(ZB) and m(ZEAB) = m(ZC). If m(ZAEC) = 130°, what is m(ZADE)? A) 50° B) 55° C) 60° D) 70° E) 80° A B D E C 130° ? 14. In the figure, KS = KN, m(ZM) = 70°, m(ZP) = x and m(ZMKS) = 2x. What is the value of x? A) 55° B) 60° C) 65° D) 70° E) 75° M K N S P 2x 70° x 15. In the figure, AD and CB bisect angles A and C, respectively. If m(ZAEC) = 75° and m(ZB) = 30°, what is m(ZADC)? A) 5° B) 10° C) 15° D) 20° E) 25° A B D C 75° 30° E ? 10. In the figure, ABC is an equilateral triangle and BD = CE. If AD = 6ñ3, what is the length of DE? A) 6 B) 8 C) 4ñ3 D) 13 E) 6ñ3 A B D C E ? 6ñ3 12. In the equilateral triangle ABC in the figure, AF = FC and AH = BD. What is the measure of ZEDC? A) 5° B) 10° C) 15° D) 20° E) 30° A E D B H C F ? 11. In the figure, AB = BC, DE = BE m(ZABD) = 36° and m(ZEDC) = 48°. What is m(ZACB)? A) 76° B) 72° C) 68° D) 58° E) 52° A B E C D 36° 48° ? 16. In the figure, KL = LM and LH = MH. If NH = 5 and m(ZK) = 30°, what is KM? A) 15 B) 20 C) 25 D) 30 E) 40 M N K L H 30° 5 Angles Chapter 3 Review Test E CHAPTER 3 REVIEW TEST E 1. In the figure, DE = DC and DB = BF. If m(ZA) = 45°, what is m(ZABC)? A) 30° B) 45° C) 50° D) 60° E) 75° A F D B C E 45° ? 5. In the figure, points K, S, T, M, N and P are the midpoints of the sides on which they lie. If AB = 12, AC = 8 and BC = 16, what is P(AMNP)? A) 6 B) 8 C) 9 D) 10 E) 12 A K B S C T M N P 6. In the figure, KL RS and KM bisects ZRKL. If KL = 6, KR = 4 and MS = 8, what is the length of PK? A) 4 B) 5 C) 6 D) 7 E) 8 P K R M S L 4 6 8 ? 7. In the figure, m(ZBAC) = 90°, m(ZC) = 15° and BC = 24. What is the length of AH? A) 4 B) 5 C) 6 D) 8 E) 12 A B H C 15° ? 24 8. According to the figure, what is the smallest possible value of a + b + c if a, b and c are integers? A) 7 B) 8 C) 9 D) 10 E) 11 D A F C E B 4 8 5 4 6 5 c b a 2. In the triangle ABC at the right, AB = AC, m(ZA) = x + 13° and m(ZB) = y – 38°. What is the sum of the minimum integer value of y and the maximum integer value of x? A) 248° B) 243° C) 240° D) 233° E) 204° x+13° A B C y38° 3. According to the figure, what is the value of x? A) 10° B) 15° C) 20° D) 25° E) 30° x 105° 130° 140° 4. In the figure, AABC is an equilateral triangle and AD = EC = CF. If BC = 12, what is the length of CF? A) 2 B) 3 C) 4 D) 5 E) 6 A D B C F E 12 ? Chapter Review Test 1A 159 9. In the figure, AH = BH = HC. If AC = 1, what is HD? A) B) C) D) E) 3 5 1 5 1 4 1 3 1 2 A D B H C ? 1 10. In the triangle ABC in the figure, CD ± AB and BE ± AC. If m(ZBFC) = 140°, what is m(ZA)? A) 20° B) 30° C) 40° D) 45° E) 50° 140° A D B C E F ? 13. In the figure, BD = DC, CE = 3AE and 2AB = AC. If m(ZA) = 90°, what is m(ZDEC)? A) 30° B) 40° C) 45° D) 50° E) 60° C E A B D ? 14. In the figure, m(ZA) = 90°, m(ZB) = 15° and AB = 6 + 3ñ3. What is the length of AC? A) 1 B) 2 C) 3 D) 4 E) 5 15° 6+3ñ3 A B C ? 15. In the figure, MK = KP, m(ZM) = 90°, NS = 9 cm and SP = 3 cm. Find the length of KS. A) 2 cm B) 3 cm C) 4 cm D) 5 cm E) 6 cm M N S P K 9 3 ? 16.In the figure, CD is the bisector of ZC. If m(ZBAE) = 15°, m(ZEAC) = 60° and m(ZB) = 45°, what is m(ZDEA)? A) 10° B) 15° C) 20° D) 22.5° E) 30° A B E C D 45° 15° 60° ? 11. In the figure, AABC is an equilateral triangle and DEFH is a square. Find the measure of ZAKD. A) 65° B) 67.5° C) 70° D) 75° E) 80° A D B E F C H K ? 12. In the figure, ABCD is a square and AABF and ABEC are equilateral triangles. What is m(ZFEC)? A) 5° B) 10° C) 15° D) 20° E) 22.5° D C E B A F ? Angles 160 162 Geometriy 7 1. Definition You can see many circular or ring-shaped geometric figures all around you. For example, wheels, gears, compact discs, clocks, and windmills are all basic examples of circles in the world around us. All radii of a circle are congruent. A circle is named by its center. For example, the circle on the left is named circle O. We write a circle with center O and with radius r as O or C(O, r). In this book, the point O in a circle is always the center of the circle. wheels compact disc gears O center radius circle It is easy to recognize a circle, but how can we define it as a shape? Let us look at a geometric definition. A. BASIC CONCEPTS Note The word ‘circle’ is derived from the latin word circus, which means ‘ring’ or ‘racecourse’. Definition circle A circle is the set of all the points in a plane that are at the same distance from a fixed point in the plane. The distance is called the radius of the circle (plural radii), and the fixed point is called the center of the circle. After studying this section you will be able to: 1. Define the concept of a circle and its basic elements. 2. Describe and use the properties of chords. 3. Describe and use the properties of tangents. 4. Describe the possible relative positions of two circles in the same plane. Objectives 163 Circles 2. Regions Separated by a Circle in a Plane A circle divides a plane into three separate regions. The set of points whose dis- tance from the center of a circle is less than the radius of the circle is called the interior of the circle. For example, if R is a point in the plane and |OR| < r, then R is in the interior of the circle. The set of points whose distance from the center is greater than the radius of the circle is called the exterior of the circle. P interior R P Q O exterior circle exterior region interior region circle For example, if Q is a point in the plane and |OQ| > r, then the point Q is in the exterior of the circle. The set of points whose distance from the center is equal to the radius is called the circle itself, and the points are on the circle. For example, if P is a point in the plane and |OP| = r, the point P is on the circle. To construct a circle, fix a pin on a piece of paper, connect a string of any length to the pin, tie the other end of the string to your pencil, and turn your pencil on the paper around the pin for one complete revolution, keeping the string taut. You will get a circle. You can also use a compass to draw a circle. Mark a point O as the center and set your compass to the length of the radius. Turn your compass around the center for one complete revolution. You will get a circle. Note The union of a circle and its interior is called a circular closed region or a disc. r O EXAMPLE 1 Name the points in the figure which are a. in the interior of the circle. b. on the circle. c. in the exterior of the circle. Solution a. Since |OA| < r and |OB| < r, points A and B are in the interior of the circle. b. Since |OC| = |OD| = r, points C and D are on the circle. c. Since |OE| > r, |OF| > r and |OG| > r, points E, F, and G are in the exterior of the circle. A O B E C F D G 164 Geometriy 7 In the figure, chord [CD] passes through the center of the circle, so [CD] is a diameter. We can see that the length of every diameter in a given circle is the same. For this reason, we usually talk about ‘the diameter’ of a circle to mean the length of any diameter in the circle. The length of the diameter of a circle is twice the radius. For example, if r is the radius of a circle and d is the diameter, then d = 2 · r, or . The diameter of a circle is the longest chord in the circle. 2 d r = Definition diameter A chord which passes through the center of a circle is called a diameter of the circle. 3. Auxiliary Elements of a Circle For example, [AB] and [CD] in the figure are chords. Definition chord A line segment which joins two different points on a circle is called a chord. A B C D chord diameter O EXAMPLE 2 1. Find the length of the diameter for each given radius. a. b. 3x cm c. 2x + 5 cm d. 7x – 12 cm 2. The length of the diameter of a circle is 20 cm and the radius is 2x – 4. Find x. 1 3 cm 2 Solution 1. a. d = 2 · r = d = = 7 cm b. d = 2 · (3x) = 6x cm c. d = 2 · (2x + 5) = 4x + 10 cm d. d = 2 · (7x – 12) = 14x – 24 cm 2. d = 2 · r 20 = 2 · (2x – 4) 2x – 4 = 10 2x = 14 x = 7 cm 1 2 3 2 · 165 Circles 4. Relative Position of a Line and a Circle in the Same Plane A line and a circle in the same plane can have one of three different positions relative to each other. If the distance from the center of the circle to the line is greater than the radius of the circle, then the line does not intersect the circle. In the figure, [OH] J l and |OH| > r, and l ì C(O, r) = ,. If the distance from the center of the circle to the line is equal to the radius, then we say that the line is tangent to the circle. In the figure, |OH| J l, |OH| = r, and l ì C(O, r) = {H}. H is the only point of intersection of the line and the circle. r l H O tangent point of tangency H O l If the distance from the center of the circle to the line is less than the radius, then the line intersects the circle at two points. In the figure, [OH] J and |OH| < r, and ì C(O, r) = {A, B}. For example, line is a secant in the figure on the left. Definition tangent A line which intersects a circle at exactly one point is called a tangent of the circle. The inter- section point is called the point of tangency. Definition secant A line which intersects a circle at two different points is called a secant of the circle. l H O A B EXAMPLE 3 Name all the radii, diameters, chords, secants, and tangents of the circle in the figure. Solution [OF], [OC], and [OB] are radii. [FC] is a diameter. l is a secant line. [EF], [ED], and [FC] are chords. GH is a tangent, and A is a point of tangency. l A H G F E O D C B 166 Geometriy 7 Check Yourself 1 1. Define the terms center, radius, chord, diameter, tangent, and secant. Show them in a figure. 2. How many regions does a circle divide the plane into? 3. Sketch all the possible relative positions of a circle and a line in the same plane. 4. Look at the figure on the right. a. Name the tangents. b. Name the secants. c. Name the chords. d. Name the radii. e. Name the diameters. Answers 1. center: a point inside the circle that is equidistant from all the points on the circle. radius: a distance from the center to a point on the circle. chord: a line segment joining two different points of a circle. diameter: a chord passing through the center of a circle tangent: a line intersecting a circle at exactly one point. secant: a line intersecting a circle at two different points. 2. three parts: the interior of the circle, the circle, and the exterior of the circle. 3. 4. a. EF, EB b. BC, DB c. [AB], [DB], [BC] d. [OD], [OA], [OB], [OC] e. [BD] n l m B C A secant tangent diameter chord r a d i u s A O D B C E F 167 Circles Remember that a chord is a line segment which joins two different points on a circle. In this section we will look at the properties of chords. B. CHORDS Property A radius that is perpendicular to a chord bisects the chord. For example, in the figure, if [OH] J [AB] then |AH| = |HB|. A O B H EXAMPLE 4 A chord of length 10 cm is 12 cm away from the center of a circle. Find the length of the radius. Solution Look at the figure. In AAHO, r 2 = 5 2 + 12 2 r 2 = 25 + 144 r 2 = 169 r = 13 cm. 2 10 5 cm 2 |AB| A O B H 12 r 5 Property In the same circle or in congruent circles, two chords which are equidistant from the center are congruent. For example, in the figure, if |OM| = |ON|, then |AB| = |CD|. The converse of this property is also true: if |AB| = |CD|, then |OM| = |ON|. A B C M N D O 168 Geometriy 7 Property In the same circle or in congruent circles, if two chords have different lengths, then the longer chord is nearer to the center of the circle. For example, in the figure, if |CD| > |AB|, then |OF| < |OE|. The converse of this property is also true: if |OF| < |OE|, then |CD| > |AB|. O A B C D E F EXAMPLE 6 In the circle in the figure, |OM| < |ON| and r = 9 cm. |AB| = 3x + 2 cm and |CD| = 5x – 2 cm are given. Find the possible integer values of x. B A N M D C O EXAMPLE 5 In the figure, |AB| = 8 cm, |CN| = 4 cm, and |OM| = 3 cm. Find |OC| = x. Solution |CD| = 8 cm, since |CN| = 4 cm. So |AB| = |CD|, and by the property, |OM| = |ON| = 3 cm. Let us use the Pythagorean Theorem to find the length of [OC]: |OC| 2 = |ON| 2 + |NC| 2 x 2 = 3 2 + 4 2 x = 5 cm. O A B C D M N x 3 4 Solution If |OM| < |ON|, then |CD| > |AB|. 5x – 2 > 3x + 2 2x > 4 x > 2 (1) Since the longest chord is the diameter, the greatest possible value of |CD| is the diameter. d = 2r, d = 2 · 9 = 18 cm |CD| ± 18 5x – 2 ± 18 5x ± 20 x ± 4 (2) From (1) and (2), the possible integer values of x are 3 and 4. 169 Circles Check Yourself 2 1. In the figure, the radius of the circle is 5 cm and |AB| = |CD| = 8 cm. Find|OE|. 2. In the figure, |AB| = |CD|, [OM] ± [AB], [ON] ± [CD], and |ON| = |OM| = 4 cm. Given |AB| =5x + 1 cm and |CD| = 4x + 2 cm, find the radius of the circle. 3. In the figure, |AP| = 12 cm, |PB| = 4 cm, and |OP| = 11 cm. Find the radius of the circle. 4. In the figure, |AB| = 12 cm, |DC| = 2 cm, [OD] ± [AB]. Find the radius of the circle. Answers 1. 3 cm 2. 5 cm 3. 13 cm 4. 10 cm O A B D C 12 4 A C B D O P B A M N C D O C D E F A B O Remember that a tangent is a line in the plane which intersects a circle at exactly one point. The point is called the point of tangency. In this section we will look at the properties of tangents. T H C. TANGENTS Property If a line is tangent to a circle, then the line is perpendicular to the radius drawn to the point of tangency. For example, in the figure, if l is tangent to the circle C at point H, then [OH] J l. O l H 170 Geometriy 7 Property If two segments from the same exterior point are tangent to a circle, then they are congruent. For example, in the figure, if [PA and [PB are tangent to the circle at points A and B respectively, then |PA| = |PB|. Property Two tangent line segments from the same external point determine an angle that is bisected by the ray from the external point through the center of the circle. For example, in the figure, if [PA and [PB are tangent to the circle then [PO is the angle bisector of /APB, i.e. m/APO = m/BPO. O A B P O A B P EXAMPLE 7 The circle in the figure is inscribed in the triangle ABC. |AK| = x + 5 cm, |BM| = 2x + 3 cm, |CL| = 2x + 5 cm, and the perimeter of triangle ABC is 46 cm. Find |MC|. Solution |AK| = |AL|, |BK| = |BM|, and |CM| = |CL|. P(AABC) =|AB|+|BC|+|AC| =|AK|+|KB|+|BM|+|MC|+|CL|+|LA| = 2 ·|AK| + 2 ·|BM| + 2 ·|CL| = 2 · (x + 5) + 2 · (2x + 3) + 2 · (2x + 5) = 2x + 10 + 4x + 6 + 4x + 10 = 10x + 26 P(AABC) = 10x + 26 = 46 = x = 2 cm So |MC| = 2x + 5 = 9 cm. O K L M A B C 171 Circles a r 2 O 2 O 1 r 1 C 1 Ç C 2 = Æ r 1 r 2 > a O 1 r 1 a r 2 O 2 C 1 Ç C 2 = Æ r 1 + r 2 < a O C 1 Ç C 2 = Æ a = 0 nonintersecting circles D. RELATIVE POSITION OF TWO CIRCLES IN THE SAME PLANE Tangent circles can be externally tangent or internally tangent, as shown in the figure. Definition nonintersecting circles Two circles which have no common point are called nonintersecting circles. Definition tangent circles Two circles which have only one common point are called tangent circles. If two or more circles share the same center, then they are called concentric circles. concentric circles l O 1 O 2 a r 1 r 2 A C Ç C 2 = {A} and r 1 + r 2 = a externally tangent circles: l is the common tangent l O 1 a r 1 A r 2 O 2 C 1 Ç C 2 = {A} and r 1 r 2 = a internally tangent circles: l is the common tangent 172 Geometriy 7 r 1 +r 2 +r 3 =21 16+r 3 =21 r 3 =5 cm. r 1 + r 2 +r 3 =21 r 1 +12=21 r 1 =9 cm. r 1 +r 2 =16 9+r 2 = 16 r 2 =7 cm. |AB| = r 1 + r 2 = 16 |BC| = r 2 + r 3 = 12 |CA| = r 1 + r 3 = 14 2 · (r 1 + r 2 + r 3 ) = 42 r 1 + r 2 + r 3 = 21 + EXAMPLE 8 The circles in the figure with centers A, B, and C are externally tangent to each other. |AB| = 16 cm, |BC| = 12 cm, and |AC| = 14 cm are given. Find the radii of the circles. Solution Let the radii of circles A, B, and C be r 1 , r 2 and r 3 respectively. Then we can write, B A C O 1 O 2 r 1 r 2 a B A C 1 Ç C 2 = {A, B} and r 1 + r 2 > a O 1 O 2 B A [AB] is the common chord [O 1 O 2 ] ^ [AB] and |AH| = |HB| H intersecting circles Definition intersecting circles Two circles which have two common points are called intersecting circles. 173 Circles 1. Describe each line and line segment in the figure as an element of the circle. 2. The points in the figure are in the same plane as the circle. State the position of each point with respect to the circle. E D F C B O G A H 5. In the figure, X, Y, and Z are points of tangency. |AX| = 6 cm, |CZ| = 4 cm, and |BY| = 2 cm. Find the perimeter of AABC. O A B C E F G D 6. In the figure, [OA] J [BC], |AK| = 2 cm, and |KC| = 4 cm. Find |OK|. O 2 4 K B C A 10. In the figure, |O 1 O 2 | = 3 cm and r 1 + r 2 = 11 cm. Find r 1 and r 2 . O 1 O 2 A B r 1 r 2 11. In the figure, |AB| = 3x + 4, |CD| = 2x + 9, and |OM| > |ON|. Find the greatest possible integer value of x. A B C D N M O 7. In the figure, |AC| = 6 cm and |AB| = 3 cm. Find |OB| = r. O B A C r 8. In the figure, |BC| = 12 cm and |AD| = 8 cm. Find the radius of the circle. O A B C D E 8 12 9. In the figure, |AP| = 6ñ3 cm and m/APB = 60°. Find the radius of the circle. C O A B P 3. In the figure, the radius of circle O is 15 cm, |CD| = 24 cm, and |OH| = 12 cm. a. Find |OI|. b. Find |AB|. H O I A B D C 4. Complete each state- ment about the figure with a suitable symbol. a. If |OE| = |OF|, then |AB|...|CD| b. If |OE| > |OF|, then |AB|...|CD|. E O F A B D C O X Z Y A B C EXERCISES 4.1 174 Geometriy 7 We use the ï sign over two or more points to denote the arc which includes the points. For example, in the fig- ure, we write AïB to denote the arc between A and B, and AùCB to denote the arc ACB. Notice that any two points of a circle divide the circle into two arcs. If the arcs are unequal, the smaller arc is called the minor arc and the larger arc is called the major arc. In the figure on the right, AïB is the minor arc and AùCB is the major arc. minor arc A B O C A B O After studying this section you will be able to: 1. Describe the concepts of arc and central angle. 2. Name inscribed angles and calcu- late their measure. 3. Use the properties of arcs, central angles, and inscribed angles to solve problems. Objectives A. ARCS AND CENTRAL ANGLES Definition arc of a circle An arc of a circle consists of two points on the circle and the unbroken part of the circle between these two points. We use the ï sign over two or more points to denote the arc which includes the points. For example, in the figure, we write AïB to denote the arc between A and B, and AùCB to denote the arc ACB. Notice that any two points of a circle divide the circle into two arcs. If the arcs are unequal, the smaller arc is called the minor arc and the larger arc is called the major arc. In the figure on the right, AïB is the minor arc and AùCB is the major arc. O A B C Definition central angle of a circle An angle whose vertex is at the center of a circle is called a central angle of the circle. 175 Circles EXAMPLE 9 Find the measure of the indicated central angle of each circle. Solution Remember that the measure of a minor arc is equal to the measure of its central angle. a. m/AOB = mAïB = 50° b. m/COD = mCïD = 120° c. m/AOB = mAïB = 180° A B O 50° 120° O C D O 180° A B B. INSCRIBED ANGLES Property In the same circle or in congruent circles, if two chords are congruent, then their corresponding arcs and cen- tral angles are also congruent. For example, in the figure, if [AB] = [CD] then AïB = CïD and m/AOB = m/COD. O A B C D r r r r Property If a line through the center of a circle is perpendicular to a chord, it bisects the arcs defined by the endpoints of that chord. For example, in the figure, if [PK] J [AB] then [AH] = [HB] AïP = PïB AïK = KïB. O r r K A B P H For example, angle /ABC in the figure is an inscribed angle. [AB] and [BC] are both chords of the circle. The arc AïC in the figure is called the intercepted arc of the inscribed angle /ABC. Definition inscribed angle of a circle An angle whose vertex is on a circle and whose sides contain chords of the circle is called an inscribed angle. a. b. c. O A C B intercepted arc 176 Geometriy 7 m/OBC = m/OCB = a° m/OAB = m/OBA = b° m/ABC = m/OBA + m/OBC = a° + b° m/COE = m/CBO + m/OCB = 2a° m/AOE = m/OAB + m/OBA = 2b° m/AOC = m/AOE + m/EOC = 2 · (a° + b°) So m/ABC = 2 Z m AOC . Property The measure of an inscribed angle is half of the measure of the central angle which intercepts the same arc. Proof In the figure, let m/BCO = a° and m/BAO = b°. Since the triangles ABOC and AAOB are isosceles triangles, we can write a° a° b° b° 2b° 2a° B C O A E Now remember that the measure of a minor arc is the same as the measure of its central angle. So we can write the property in a slightly different way: Property The measure of an inscribed angle is equal to the half the measure of its intercepted arc. For example, in the figure, m/ABC = . 2 ï mAC B C O A a 2a 2a EXAMPLE 10 Find the measure of x in each figure. x O A C B 120° 50° O A C B x x O A C B a. b. c. 177 Circles Solution a. m/ABC = 50 = mAïC = 100° m/x = 100° 2 ï mAC 2 ï mAC b. m/ABC = m/x = m/x = 60° 120 2 2 ï mAC c. m/ABC = = m/x = 45° 90 2 2 Z m AOC Solution a. m/CAD = m/CBD = x° = y° = x = y = 20 40 2 2 mCDï b. m/BAC = x° = x = 25 and y = 50 ° 50 = 2 2 y 2 2 Z m BOC mBC = ï Property If two inscribed angles intercept the same arc of a circle, then the angles are congruent. For example, in the figure, /ABC = /ADC, because they both intercept AïC. C A D B EXAMPLE 11 Find the value of x and y in each figure. x° y° A B C D 40° x° y° O A B C 50° A B C D y° x° 20° O a. b. c. 178 Geometriy 7 Property An angle inscribed in a semicircle is a right angle. For example, in the figure, if mAùLB = mAùMB = mAùNB = 180°, then m/ALB = m/AMB = m/ANB = 90° or m/L = m/M = m/N = 90°. O A B M N L c. m/BAC = 20 = y = 40 ° 2 y 2 Z m BOC m/BDC = x° = x = 20 40° 2 2 Z m BOC EXAMPLE 12 Find the value of x in each figure. Solution a. Since AC is the diameter, the arc AùBC is a semicircle. So ZABC is inscribed in a semicircle, and therefore m/ABC = x° = 90°, x = 90. A B C O x° A D B O 20° x° y° 60° C A D B O x° C 10° a. b. c. b. m/DAC = = = 30° 60° 2 mDC 2 ï y = 2 · m/BAC = 2 · 20 = 40 mAïD + mDïC + mCïB = 180 x° + 60° + 40° = 180° x = 80 179 Circles Property The measure of the angle formed by a tangent and a chord is equal to the half of the meas- ure of its intercepted arc. For example, in the figure, mZCAB = mZAOB. 1 2 Proof Let us draw the diameter [AD] and the chord [BD]. [AC] J [AD] (a radius is perpendicular to a tangent at the point of tangency) [AB] J [BD] (definition of a semicircle) m/DAB + m/BAC = 90° m/DAB + m/ADB = 90° (in triangle AABD) m/ADB = m/BAC m/ADB = (inscribed angle rule) m/BAC = (inscribed angle rule) 2 ï mAB 2 ï mAB A C D B O A C B O Rule Let [AB] and [CD] be two chords of a circle. If [AB] [CD], then m/ABC = m/BCD (alternate interior angles). So mAïC = mBïD. A B C D c. Let us draw the chord [BD]. m/ADB = 90° m/CAB = m/CDB = 10° m/ADC + m/CDB = 90° x + 10 = 90 x = 80 A D B O x° C 10° 180 Geometriy 7 EXAMPLE 13 Find the value of x and y in each figure. Solution C A O B 30° x° y° O A C D B E 15° x° [AB] || [CD] O E 30° x° A F C D B a. Let us draw the radius [OA]. Then [AC] J [AO]. AAOB is isosceles triangle. m/OAB = 30° and m/OAB + m/BAC = 90° 30° + x° = 90° x = 60 y = 2 · 60 = 120 b. [AB] [CD] and m/BAD = mBïD = 2 ·15° = 30° mAïC + mCïD + mDïB = 180° 30 + mCïD + 30 = 180° mCïD = 120° m/DCE = x° = = = 60 120° 2 2 mCD ï mBD 2 ï C A O B 30° x° y° 30° r r O A C D B E 15° x° 30° 30° a. b. c. [AD] J [BC] and [AF] J [AD]. So [AF] [BC]. Therefore, m/ABC = m/BAE and m/EAB = x° = 30°, x = 30. c. m/ADC = mAïC = 2 · 30° = 60° m/ABC = m/ABC = = 30° 60° 2 ï mAC 2 ï mAC 2 O E 30° x° A F C D B 181 Circles mÐP = mCïD mAïB 2 mÐP = mCïB mAïB 2 O P A B D C O P A C B O P B C A angle formed by two secants: mÐP = mAùCB mAïB 2 mÐP + mAïB = 180° angle formed by a secant and a tangent: angle formed by two tangents: Rule The measure of an angle formed by two secants, a secant and a tangent, or two tangents drawn from a point in the exterior of a circle is equal to half of the difference of the measures of the intercepted arcs. b mÐBAC + mÐBOC = 180° mÐBAC + mBïC = 180° a + b = 180° O A B C a b Rule The measure of an angle formed by two chords that intersect in the interior of a circle is equal to half the sum of the measures of the intercepted arcs. E A B C D a° x° y b a b For example, in the figure, m/AED = m/BEC = and m/AEB = m/CED = and 2 o J = x+y a+b = . 2 2 ï ï mAB +mCD 2 mBC + mAD ï ï 182 Geometriy 7 EXAMPLE 14 Find the value of x in each figure. Solution a. m/BED = 70° = mAïC = 140 – 60 = 80 m/ADC = b. m/CAE = m/CAE = x° = = 35°, x = 35 c. m/QPR + mQïR = 180° 60 + mQïR = 180 mQïR = 120° mQïR + mQùTR = 360° mQùTR = 240° m/QSR = x° = = 120°, x = 120 240° = 2 2 mQTR ù 70° 2 100 – 30 2 2 Z mCE – mBD = (mBD =2×m BCD) ï ï ï 80 = =40°, =40 2 2 ï mAC x +60 ( =2 ) 2 · Z ï ï mAC mBD m BAD 2 ï ï mAC + mBD A B C D E 70° 30° x° 60° O P Q T R x° S E 15° x° O A B C D 100° a. b. c. 183 Circles 4. In the figure, m/BOC = 100° and m/ACO = 20°. Find m/AOC. O 20° 100° x° A B C 9. In the figure, m/AOC = m/ABC = 3x°. Find the value of x. O 3x° 3x° B A C 5. In the figure, [AB] is a diameter and m/OCB = 40°. Find m/OAC. O A B C 40° x° 10. In the figure, m/AOC = 100° and m/OAB = 70°. Find m/OCB. O A B C 100° 70° x° 12. In the figure, m/DPA = 50°. Find m/BCA. O 50° D C B P A x° 11. In the figure, m/APD = 30°, m/DKA = 60°, m/BAC = a°, and m/DCA = b°. Find a and b. A D C P B a° K 60° b° 30° 7. In the figure, mAïD = and m/DPC = 75°. Find m/BAC. 2 mBC ï x° 75° P A B C D 8. In the figure, [AE is tangent to the circle at the point B, and m/EBC = 75°. Find m/A. O A D B E x° 75° C 6. In the figure, m/CDB = 10° and m/ABD = 50°. Find m/P. O 50° 10° A B P C D K x° 1. In the figure, m/AOC = 120°. Find m/ABC. O 120° A B C x° 3. In the figure, m/CBD = 120°. Find /AOC. O 120° x° A B D C 2. In the figure, m/BAC = 30° and m/BKC = 70°. Find m/OCA. O K 70° 30° A B C EXERCISES 4.2 184 Geometriy 7 1. Circumference of a Circle Remember from chapter 4 that the distance around a polygon is called the perimeter of the polygon. C = 2¬r . A. CIRCUMFERENCE AND ARC LENGTH If you measure the circumference and diameter of a circle and divide the circumference by the diameter, you always get the same constant. This constant is approximately equal to 3.14, and denoted by ¬. Definition circumference The distance around a circle is called the circumference of the circle. Note Pi (¬, pronounced like the English word ‘pie’) is a Greek letter. It is the first letter of a Greek word that means ‘measure around.’ d r r C Property For all circles, the ratio of the circumference to the diameter is always the same number. The number is called r (pronounced like‘pie’). This is the formula for the measure of the circumference of a circle. So if the circumference of a circle with a diameter d is C, then we can write or C = ¬·d or r C = d After studying this section you will be able to: 1. Describe the concepts of circumference and arc length. 2. Find the area of a circle, an annulus, a sector, and a segment. Objectives 185 Circles 1. Find five different circular objects. Use a piece of string to measure their circumference (C), and use a ruler to measure their diameter (d). Write the values in a table. 2. For each circular object calculate the ratio and then calculate the average of all the ratios. 3. How do the number ¬ and the formula C = ¬ · d relate to this activity? C d 2. Arc Length Remember that an arc is a part of a circle. The measure of an arc is equal to the measure of its central angle. arc length of = , circumference of the circle 360° arc length of so = . 2ð r 360° o · AB mAB AB ï ï ï arc length of AïB = 2¬ · r · o 360° In the above formula the measure of the arc is given in degrees. The length of the arc is given in a linear unit such as centimeters. We can rewrite this as . EXAMPLE 15 a. Find the diameter of a circle with circumference 24¬ cm. b. Find the circumference of a circle with radius 5 cm. c. Find the circumference of a circle with diameter 9 cm. Solution a. Let the diameter of the circle be d, then the circumference of the circle is C = ¬ · d: 24¬ = ¬ · d d = 24 cm. b. C = 2¬ · 5 = 2¬ · 5 = 10¬ cm c. C = 2¬r = 2r¬ = d · ¬ = 9¬ cm Rule In a circle, the ratio of the length of a given arc AïB to the circumference is equal to the ratio of the measure of the arc to 360°. a A B O r 186 Geometriy 7 a. The length of a semicircle is half of the circumference. arc length of AïB = 2¬r · = 2¬ · 6 · = 2¬ · 6 · = 6¬ cm 1 2 180° 360° 360 o ° b. The length of a 90° arc is a quarter of the circumference. arc length of CïD = 2¬r · = 2¬ · 10 · = 2¬ · 10 · = 5¬ cm 1 4 90° 360° 360 o ° c. arc length of EïF = 2¬r · = 2¬ · 12 · = 2¬ · 12 · = 4¬ cm 1 6 60° 360° 360 o ° d. arc length of GùTH = 2¬r · = 2¬ · 18 · = 2¬ · 18 · = 21¬ cm 21 36 210° 360° 360 o ° Solution Check Yourself 3 1. Find the circumference of the circle with the given radius. a. r = 3 cm b. r = 5 cm c. r = 7 cm d. r = 10 cm 2. Find the radius of the circle with the given circumference. a. 12¬ cm b. 24¬ cm c. 36¬ cm d. ¬ cm 3. Find the length of the minor arc in each figure. Answers 1. a. 6¬ cm b. 10¬ cm c. 14¬ cm d. 20¬ cm 2. a. 6 cm b. 12 cm c. 18 cm d. cm 3. a. cm b. cm c. cm d. 8¬ cm 8 3 ¬ 10 3 ¬ 3 2 ¬ 1 2 O 3 cm A B O 5 cm 120° C D O 8 cm 60° E F O 9 cm 200° G H K L a. b. c. d. EXAMPLE 16 Find the length of each arc. O 6 cm A B O C D 10 cm O E F 60° 12 cm O G H 18 cm 210° T a. b. c. d. 187 Circles 1. Area of a Circle To understand why this property is true, let us divide a circle into 16 equal parts, and rearrange them as follows: O r C 2 = pr r As the number of equal parts increases, the area of the circle gets closer and closer to the area of a parallelogram. The area of a parallelogram is So the area of a circle with radius r is A = ¬r 2 . 2 2 r · · r 2 C 2 r A=r =r = r . A = ¬ · r 2 B. AREA OF A CIRCLE, A SECTOR, AND A SEGMENT Property The area of a circle is ¬ times the square of the radius. O r A b. Let the radius of the circle be r, then A = ¬ · r 2 16¬ = ¬r 2 , r 2 = 16 r = 4 cm. c. The formula for the circumference of a circle is C = 2¬ · r: 10¬ = 2¬ · r r = 5 cm. So the area of the circle is A = ¬ · r 2 = ¬ · 5 2 = 25¬ cm 2 . EXAMPLE 17 a. Find the area of a circle with radius r = 6 cm. b. Find the radius of a circle with area 16¬ cm 2 . c. Find the area of a circle with circumference 10¬ cm. Solution a. Let the area of the circle be A, then A = ¬ · r 2 A = ¬ · 6 2 A = 36¬ cm 2 . 188 Geometriy 7 2. Area of an Annulus O R r = O area of the annulus area of the big circle area of the small circle = r O R A(annulus) = ¬R 2 – ¬r 2 = ¬(R 2 – r 2 ) How can we find the area of an annulus? Look at the diagram. Definition annulus An annulus is a region bounded by two concentric circles. O r R an annulus EXAMPLE 18 Find the area of the annulus bounded by concentric circles with radii 5 cm and 3 cm long. Solution The radius of the big circle is R = 5 cm. The radius of the small circle is r = 3 cm. A = ¬ · R 2 – ¬ · r 2 A = ¬ · (R 2 – r 2 ) A = ¬ · (5 2 – 3 2 ) A = ¬ · (25 – 9) A = 16¬ cm 2 O 5 cm 3 cm 189 Circles EXAMPLE 19 Find the area of each shaded sector. O A B 5 cm 72° O 8 cm 6p P S O A C 15° B 6 cm l r A= 2 or · |AB| r A= 2 ï 2 360 | | = 2 360 | | = 2 360 · · · · · · · · ¬ ¬ ¬ a AB r AB a r A B r a r ï ï ï In the figure, (|AïB| = l ). 3. Area of a Sector For example, in the figure, the smaller region AOB is a sec- tor of the circle. If the degree measure of arc AB is mAïB = a° then the area of sector We can also calculate the area of a sector in a different way: 2 a = . 360 · · AOB p r Definition sector of a circle A sector of a circle is the region bounded by two radii of the circle and their intercepted arc. O r a A B In the above formula the measure of the arc is given in degrees. The length of the arc is given in a linear unit such as centimeters. Rule The area of a sector of a circle is half the product of the length of the arc and the length of its radius. O r a A B l a. b. c. 190 Geometriy 7 4. Area of a Segment O A B a° b r h b h = a° A B O A O B A B a° area of segment = – A = A(sector AOB) – A(AAOB) area of triangle area of sector · · · 2 b h a A= ð r – 360 2 Solution a. r = 5 cm and m/a = 72°. b. r = 8 cm and l = 6¬ cm. c. m/BOC = 2 · m/BAC, so m/BOC = 30° and r = 6 cm. 2 2 2 30 1 (sector )= ð = ð 6 = ð 36=3ð cm . 360 360 12 Z · · · · · · m BOC A BOC r 2 6 8 (sector )= = =24 cm 2 2 · r · r l r A POS 2 2 2 72 1 (sector )= = 5 = 25=5 cm 360 360 5 · r· · r· · r· r a A AOB r Definition segment of a circle A segment of a circle is a region bounded by a chord and its intercepted arc. EXAMPLE 20 Find the area of each shaded segment. O 1 2 c m 120° A B 45° C B A 12 cm O 6 c m A B a. b. c. 191 Circles Check Yourself 4 1. Find the area of a circle with the given radius. a. r = 3 cm b. r = 5 cm c. r = 12 cm d. r = 16 cm 2. Find the area of a circle with the given circumference. a. 4¬ cm b. 12¬ cm c. 20¬ cm d. ¬ cm 3. Find the circumference of a circle with area 36¬ cm 2 . 4. The ratio of the radii of two circles is 5 : 3. What is the ratio of their areas? 5. The area of the shaded region in the figure is 32r cm 2 and R = 9 cm. Find r. O r R Solution a. Since m/AOB = 90°, A(segment) = 9¬ – 18 cm 2 . b. |OH| = 6 cm and |AB| = 12ñ3 cm. A(sector AOB) = = = 48¬ cm 2 . A(AAOB) = A(segment) = 48¬ – 36ñ3 cm 2 . c. A(sector AOC) = A(AAOC) = A(segment) = 9¬ – 18 cm 2 . 2 6 6 36 18 cm 2 2 · = = 2 2 ð ð 36 = =9ð cm 4 4 · · r 2 | | | | 12 3 6 36 3 cm 2 2 · · = = AB OH 1 144 3 ¬· 2 120 ð 12 360 · 2 2 36 ( )= = =18 cm 2 2 A r A AOB 2 2 90 1 (sector )= = 36=9 cm 360 4 r · r · r A AOB r O 12 30° 60° H B A 6 6ñ3 45° C B A 6 6 6 O 192 Geometriy 7 6. Find the area of the shaded region in each circle. r = 5 cm O A B r P Q r O 120° A B r = 8 cm r O 60° A B r = 5 cm r O r = 12 cm A B r O A B r = 5 cm |AB| = 5ñ2 cm r R = 10 cm r = 7 cm O 120° A B r R O A B r r = 8 cm O r R R = 10 cm r = 8 cm Answers 1. a. 9¬ cm 2 b. 25¬ cm 2 c. 144¬ cm 2 d. 256¬ cm 2 2. a. 4¬ cm 2 b. 36¬ cm 2 c. 100¬ cm 2 d. cm 2 3. 12¬ cm 4. 5. 7 cm 6. a. cm 2 b. cm 2 c. 36¬ cm 2 d. cm 2 e. 36¬ cm 2 f. 25¬ cm 2 g. 17¬ cm 2 f. 16 cm 2 25 50 4 ¬- 64 3 ¬ 25 6 ¬ 25 9 4 ¬ 193 Circles 3. In the figure, m/AOB = 120° and r = 6 cm. Find the length of arc AùXB. O 120° 6 cm A B X 4. In the figure, B is the center of a circle and ABCD is a square with |AD| = 5 cm. Find the area of the shaded region. A B C D 6. In the figure, |OB| = 5 cm, m/DOB = 60°, and |BA| = 3 cm. Find the area of the shaded region. B A C D 60° O 7. In the figure, m/BAC = 30° and the radius of the circle is 6 cm. Find the area of the shaded region. 30° A B C 8. In the figure, ABCD is a rectangle and A and B are the centers of two circles. Given |AD| = 6 cm, find the area of the shaded region. A B D C E 5. In the figure, m/AOB = m/COD = m/EOF = 20° and r = 6 cm. Find the sum of the areas of the shaded regions. O 20° 20° 20° F E D C B A EXERCISES 4.3 1. In the figure, m/AOB = 30° and r = 6 cm. Find the area of the shaded region. O 30° A B r 2. In the figure, m/AOB = 45° and r = 10 cm. Find the area of the shaded region. O 45° A B r = 10 r 9. In the figure, A, B, and C are the centers of three congruent tangent circles. If the sum of the circumferences of the circles is 24¬ cm, find the area of the shaded region. A B C 10. B, C, P, and K are the centers of four circles in the figure. Given |AB| = |BC| = |CD| = 4 cm, find the area of shaded region. A P B C K D 11. In the figure, ABCD is a square with perimeter 64 cm. Find the area of the shaded region. A B C D 194 Geometriy 7 13. In the figure, A(AAOB) = 48 cm 2 , |OC| = 8 cm, and [OC] J [AB]. Find r. O A B C 8 cm r 19. In the figure, m/BCD = 130° and m/OAC = 40°. Find m/CBO. O 40° x° C A B 130° D 20. In the figure, m/OAB = 45° and m/OCB = 60°. Find m/AOC. O 45° 60° x° A B C 21. In the figure, m/AOC = 160° and m/ABC = x°. Find m/ABC. O A B C 160° x° 22. In the figure, m/OAD = 40° and m/BOC = 50°. Find m/COD. O A B C D 40° 50° x° 23. In the figure, B is the point of tangency and m/OAB = 30°. Find m/ABC. O x° 30° B C A 14. In the figure, |AB| = |CD| = 8 cm and |OH| = 3 cm. Find the radius of the circle. O 8 cm A B C D H r 15. In the figure, |CE| = 3x – 2, |FB| = x + 4, and |OE| = |OF|. Find x. O C D E A B F 16. In the figure, m/AOB = 60° and |AB| = 5 cm. Find the radius of the circle. O r A B C 17. In the figure, |AB| = 9 cm, |BC| = 8 cm, and |CA| = 5 cm. Find the radius of circle A. C B A 18. In the figure, m/OAB = 50° and m/BCO = 35°. Find m/AOC. O 50° 35° x° A B C 12. In the figure, B and D are the centers of two circles. If ABCD is a square and the shaded area is 16¬ cm 2 , find |DE|. P E F A B C D 195 Circles 25. In the figure, m/BAD = 60° and |AD| = |DC|. Find m/BCD. O 60° x° A B C D 31. In the figure, |OB| = r = 4 cm. Find the area of the shaded region. O A B C 32. In the figure, |AB| = 8 cm and |AC| = 6 cm. Find the area of the shaded region. O 6 c m 8 c m A B C 33. ABCD is a square with sides 10 cm long. Find the area of each shaded region. 26. In the figure, m/AOE = 60° and |OA| = |DC|. Find m/ACE. O x° 60° A B C E D 27. In the figure, m/BAD = 30°. Find m/ACD. O 30° x° A B C D 28. In the figure, m/BAC = 20° and m/DFE = 30°. Find m/COD. O 30° x° 20° A B C D E F 29. In the circle in the figure, |OA| = 6 cm, m/AOB = 50°, m/COD = 30°, and m/EOF = 40°. Find the sum of the areas of the shaded regions. O 40° 30° 50° A B C D E F 30. In the figure, the radius of the circle is 6 cm and the length of arc AùXB is 4r cm. Find the area of the shaded region. O X r = 6 c m A B A B C D A B C D A B C D A B C D A B C D A B C D 24. A and C are points of tan- gency on the circle in the figure. Given m/ABC = 60° and m/BCD = 70°, find m/BAE. x° 70° 60° A E C D B a. b. c. d. e. d. Angles 196 CHAPTER 4 REVIEW TEST 1. In the figure, |OA| = 4 cm and |OC| = 7 cm. What is |BC|? A) 2 cm B) 3 cm C) 4 cm D) 5 cm O l A B C 6. In the figure, line l is tangent to the circle at point C and |OA| = |AB| = 5 cm. Find |BC| = x. A) 4 cm B) 5 cm C) 5ñ3 cm D) 6 cm O 5 5 x l A B C 7. Find m/ABC in the figure. A) 49° B) 50° C) 51° D) 52° O A B C 98° x° 8. In the figure, m/ABD = 60° and m/CED = 80°. Find m/CDE. A) 10° B) 20° C) 25° D) 40° O x° D A B 60° 80° C E 9. In the figure, m/BDC = 70°. Find m/ACB. A) 20° B) 25° C) 30° D) 40° O 70° B A C D 10. In the figure, line l is tangent to the circle at point A and |AB| = |AC|. Find m/CAD. A) 65° B) 55° C) 50° D) 45° O 80° l A D B C 2. Find |AB|in the figure if |CH| = 4 cm. A) 8 cm B) 7 cm C) 6 cm D) 5 cm O A B C D H G 3. In the figure, the radius of the circle is 10 cm and |OH| = 6 cm. Find |AB|. A) 8 cm B) 12 cm C) 16 cm D) 20 cm O A B H 4. In the figure, |OC| = 3ñ2 cm, |AC| = 1 cm, and |BC| = 7 cm. What is the length of the radius? A) 3 cm B) 3ñ3 cm C) 4ñ2 cm D) 5 cm 7 c m O 1 cm 3ñ2 cm C A B 5. In the figure, |OK| = |OH| = 5 cm, |AB| = 2a + 2 cm, and |CD| = a + 13 cm. What is the length of the radius? A) 13 cm B) 12 cm C) 11 cm D) 10 cm O A B C D K H Chapter Review Test 1A 197 11. In the figure, [PE and [PD are tangent to the circle at the points A and B, respectively. Find m/ACB if m/APB = 50°. A) 60° B) 65° C) 70° D) 75° O A B C x° 50° E D P 13. In the figure, m/DCE = 30° and mAïB = 80°. Find the value of x. A) 65 B) 70 C) 75 D) 80 O x° 30° A E B D C 14. In the figure, |AB|= 2 cm and |AC| = 2ñ3 cm. What is the length of the circumference? A) 3¬ cm B) 4¬ cm C) 6¬ cm D) 8¬ cm O 2 cm 2ñ3 cm A B C 15. In the figure, the perimeter of the circle is 10¬ cm and |OH| = 3 cm. Find |AB|. A) 5 cm B) 6 cm C) 7 cm D) 8 cm O 3 cm A B H 12. In the figure, m/APC = 35° and mBïD = 100°. Find m/ADC. A) 15° B) 20° C) 30° D) 40° O 35° x° P A B C D 16. In the figure, m/ABC = 35°, m/ACB = 55°, and |BC| = 4 cm. What is the area of the circle? A) 2¬ cm 2 B) 3¬ cm 2 C) 4¬ cm 2 D) 8¬ cm 2 4 cm 35° 55° A B C 19. In the figure, the circle has radius 6 cm and mZABC = 75°. Find the area of the shaded region. A) 6¬ cm 2 B) 9¬ cm 2 C) 12¬ cm 2 D) 15¬ cm 2 O 6 cm A B C 75° 20. In the figure, ABCD is a square. |BE| = 4 cm, |DF| = 6 cm, and B and D are the centers of two circles. Find the area of the shaded region. A) 40 – 10¬ cm 2 B) 50 – 13¬ cm 2 C) 36 – 12¬ cm 2 D) 64 – 20¬ cm 2 A B C D 6 cm 6 cm 4 cm 4 cm F G E H 17. Find the length of the arc AïB in the figure if the radius is 3 cm and m/ACB = 60°. A) ¬ cm B) cm C) 2¬ cm D) cm 5 2 ¬ 3 2 ¬ O 60° 3 cm A C B 18. In the figure, ABCD is a square with sides 6 cm long. Find the area of the shaded region. A) 9 – 2¬ cm 2 B) 16 – cm 2 C) 36 – 9¬ cm 2 D) 49 – 12¬ cm 2 9 4 ¬ O A B C D H E F G 196 Geometriy 7 1. Because there are no simpler concepts for us to buid on. Therefore, we need to understand these concepts without a precise definition. 3. A ray has closed enpoint but a half line has an open endpoint. 4. 2 5. 3 7. a. size, length, width, thickness b. line c. plane d. skew lines 8. a. true b. true c. true d. false e. true 9. a. 10 b. 21 c. 210 d. 5050 10. lines: HL, HG rays: [LC, [LH, [HL, [HG, [GH half lines: ]LC, ]LH, ]HL, ]HG, ]GH 11. a. line segment CD b. half open line segment PQ c. open line segment AB d. ray KL e. half line MN f. line EF 12. a. l ì (E) = l b. d ì (F) = {C} c. n ì (G) = , 13. ‘M, N and P’, ‘R and S’, and ‘L and K’, are coplanar points. Q P l 14. m d l D E F O 15. (D) ì (E) = m (D) ì (F) = l (E) ì (F) = d m ì d ì l = {O} 16. a. 5 b. (P) ì (Q) = EB, (P) ì (S) = EA, (P) ì (T) = AB, (Q) ì (T) = BC, (Q) ì (R) = EC, (T) ì (R) = DC, (S) ì (R) = ED, (T) ì (S) = AD c. 3 lines pass through point A, B, C, and D, 4 lines pass through point E. EXERCISES 1.1 1. a. , b. , c. {K, O, M} d. , e. {N} f. , g. , h. {P} i. , j. , k. , 2. a. {A} L [CD b. ]AC[ L ]AD[ c. ]CD[ d. ]CE ì ]DF e. ]AB] L [BC[ L ]AH L ]DG 5. a. acute angle b. right angle c. obtuse angle d. straight angle e. complete angle 6. a. 20° b. 12° c. 20° 7. a. 32° b. 20° c. 10° 8. a. 115° b. 65° c. 115° d. 65° e. 115° f. 65° g. 65° 9. 130° 10. 40° 11. 25° 12. 50° 13. 100° 14. 70° 15. x = y + z 16. 160° 17. 35° 18. 140° 19. 80° 20. 90° 21. 35° EXERCISES 2.1 197 Answers to Exercises 1. ADE, DEK, DKF, BDF, CKF, CKE, DEC, ADC, DFC, BDC, CEF, ABC 2. eight triangles: GDT, DTE, ETF, FTG, GDE, GFE, GDF, DEF 3. 51 cm 4. 10 cm 5. 28.2 cm 6. a. B, E, F, C b. F c. segment AC and point E d. segment FC without endpoints 7. 8. 12 9. 7 cm 10. a. b. c. 11. a. Hint: Construct medians for each side. b. Hint: Construct angle bisectors for each angle. c. Hint: Construct altitudes for each vertex. d. Hint: Construct perpendicular bisectors for each side. 14. 15. a. BFC b. CEF, BEF, ABC c. BFC d. ABF e. ABF 17. a. yes b. no c. yes d. yes e. no 18. a. x {4, 9, 14} b. none 19. 20. a. in the interior b. in the interior c. in the interior d. in the interior e. on the triangle f. in the interior g. in the interior h. on the triangle i. in the exterior j. in the interior k. in the interior l. in the exterior 21. a. sometimes b. always c. never d. sometimes e. never f. never g. always h. always 2 121 cm 2 56 168 = ; = 5 13 b c h h A n a r 1 B C r 2 r 2 r r A V a B C A B C h a r 2 r 2 r 1 A B C D A B C E A B C D A B E D F C D E F G T EXERCISES 3.1 22. a. b. c. d. e. f. 23. a. b. isosceles equilateral obtuse triangle isosceles scalene right equilateral 198 Geometriy 7 1. a. 36° b. 114° c. 54° d. 20° e. 100° f. 50° g. 90° h. 64° 2. 40°, 60°, 80° 3. x = 120°, acute angles: 85°, 5° 4. 50° 5. 136° 6. 60° 7. 72° 8. 117° 9. 12° 10. 8° 11. 36° 12. 106° 13. a. b. c. not possible d. e. not possible 14. a. no b. no c. no d. yes 15. 55° 17. a. 2 < a < 14 b. 4 < p < 20 c. 1 < m < 7 18. a. 4 < x < 12 b. 4 < x < 11 c. 3 < x < 10 19. three 20. 9 21. 12 22. 29 23. 36 24. 9 25. five 26. a. yes b. no c. no d. no e. yes f. yes 27. x 28. a. 1 = 2 > 3 b. 1 > 2 > 3 c. 3 > 2 > 1 d. 1 > 3 > 2 29. a. false b. false c. true d. false e. true f. false g. false 30. a. AC b. AC c. DC d. BC 31. a. 5 b. 8 32. 49 33. 25 34. there are no values 35. 11 36. four triangles with side lengths (1, 5, 5), (2, 4, 5), (3, 4, 4), (3, 3, 5) 38. a. A b. D c. D d. D 39. a. A b. B 40. a. A b. B c. A d. A 41. a. A b. B 42. a. D b. A D M F E 80° 40° 20° 80° 40° D M F E 60° 40° 60° 60° 20° D M F E 150° 70° 80° EXERCISES 3.3 2. a. 70° b. 1 c. 60° d. 6 e. 3 3. A K; D L; E N; AD KL; DE LN; AE KN 4. a. 6 b. 20° c. 22° d. 5. a. BC = 3, MN = 8 9. 10 cm 11. 2 12. m(OKM) = 10°, m(OML) = 60°, m(OLK) = 20° 13. 8 14. 15 15. 84° 16. 17. 84° 18. 38° 19. 9.6 cm 20. 20° 21. 22. 3ñ3 23. 8 cm 24. 3ñ3 25. 4 cm 26. 14 27. 15 28. 8ñ3 29. 3 cm 30. 16 31. 2 cm 32. 6 cm 33. 2 cm 34. 2 cm 35. 4ñ3 cm 36. 70° 37. 12 38. 99° 39. 9 cm 40. 3ñ3 – 3 41. 24 cm 42. yes 43. 200 km 44. 25° 45. 7 – ñ5 46. 6 47. 70° 48. 57° 49. 150° 50. 8 cm 51. 16 cm 52. 18° 53. 8 and 12 54. 8 55. 45° 58. 4 cm 59. 10° 60. 3ñ2 60. 72 62. 6 cm 63. 12 cm 64. 12 cm 65. 8 66. 6 cm 67. 6 cm 68. 6 cm 69. 8 cm 70. 9 cm 71. 72. 6 73. 2 74. 18 75. 15 3 2 1 4 5 2 11 8 EXERCISES 3.2 EXERCISES 3.4 1. 2. {21, 31} 3. 4. {–49, 31} 5. 6. 7. a. ò13 b. c. 8. 9. 3x + 4y – 10 = 0 ; 3x + 4y + 30 = 0 10. 53 – 15 12 43 5 2 15 2 4 35 25 {– , } 9 9 16 29 3 – 4 13 10 199 Answers to Exercises EXERCISES 4.1 1. radii: [OF], [OC], [OA], [OB] diameter: [FC] chords: [ED], [FC], [GB] tangent: AH secant: GB center: O 2. Points C, O, and D are in the interior region of the circle. Point E is on the circle. Points A, B, G and F are in the exterior region of the circle 3. a. 9 cm b. 18 cm 4. a. = b. > 5. 24 cm 6. 3 cm 7. cm 8. 4ñ6 cm 9. 6 cm 10. r 1 = 4 cm, r 2 = 7 cm 11. 4 cm 9 2 EXERCISES 4.2 1. 120° 2. 10° 3. 120° 4. 30° 5. 50° 6. 40° 7. 70° 8. 60° 9. 40 10. 60° 11. a = 15° b = 45° 12. 20° EXERCISES 4.3 1. 33¬ cm 2 2. cm 2 3. 4¬ cm 4. cm 2 5. 6¬ cm 2 6. cm 2 7. (6¬ – 9ñ3) cm 2 8. (72 – 18¬) cm 2 9. (16ñ3 – 8) cm 2 10. 12¬ cm 2 11. (256 – 64¬) cm 2 12. (8ñ2 – 8) cm 13. 10 cm 14. 5 cm 15. 3 16. cm 17. 3 cm 18. 170° 19. 10° 20. 150° 21. 100° 21. 30° 22. 60° 23. 50° 24. 120° 25. 20° 26. 120° 27. 80° 28. 12¬ cm 2 29. 12¬ cm 2 30. 4¬ cm 2 31. cm 2 32. a. 50 cm 2 b. (50¬ – 100) cm 2 c. (100 – 25¬) cm 2 d. (50¬ – 100) cm 2 e. cm 2 f. cm 2 ¬ 25 (75 – ) 2 ¬ 25 ( +25) 4 25 - 48 2 ¬ 5 3 3 13 2 ¬ 25 (25 – ) 4 ¬ 25 2 ¬ 200 Geometriy 7 TEST 1 1. D 2. A 3. B 4. C 5. C 6. D 7. C 8. B 9. D 10. D 11. C 12. D TEST 2 1. C 2. A 3. D 4. C 5. B 6. B 7. B 8. B 9. C 10. C 11. D TEST 2A 1. E 2. D 3. A 4. C 5. A 6. E 7. B 8. C 9. C 10. C 11. B 12. C 13. D 14. C 15. E 16. C TEST 2B 1. C 2. A 3. E 4. C 5. A 6. C 7. B 8. A 9. B 10. C 11. C 12. A 13. C 14. C 15. A 16. E TEST 2C 1. E 2. D 3. C 4. B 5. E 6. E 7. D 8. C 9. B 10. A 11. D 12. C 13. D 14. A 15. C 16. C TEST 2D 1. D 2. D 3. A 4. C 5. E 6. B 7. C 8. D 9. A 10. E 11. E 12. C 13. A 14. A 15. C 16. D TEST 2E 1. D 2. E 3. B 4. C 5. C 6. A 7. C 8. C 9. A 10. C 11. D 12. C 13. C 14. C 15. B 16. B Symbol Meaning = is equal to = is not equal to > is greater than ± is greater than or equal to < is less than ± is less than or equal to = is approximately equal to |x| absolute value of x ¬ pi ñ square root /A angle A /A´ exterior angle of A in a triangle m/A measure of angle A in degrees º degrees ´ minutes ´´ seconds right angle m/ABC measure of angle ABC in degrees AïB minor arc with endpoints A and B mAïB measure of minor arc AB in degrees AùCB major arc with endpoints A and B mAùCB measure of major arc ACB in degrees AB line AB, passing through the points A and B [AB] line segment AB or segment AB, with endpoints A and B |AB| length of segment AB [AB ray AB with initial point A, passing through B ]AB half line AB ]AB] half-open line segment AB, excluding point A and including point B ]AB[ open line segment [AB] closed line segment Symbol Meaning = is congruent to =/ is not congruent to is parallel to is not parallel to J is perpendicular to - is similar to AABC triangle with vertices A, B and C h a length of the altitude to side a ÷ is an element of Ç is not an element of L union ì intersection C is contained by A C B A is contained by B A Ç B A is not contained by B A.S.A angle-side-angle S.A.S side-angle-side S.S.S side-side-side A.A angle-angle (E) plane E (int ABC) interior of the triangle ABC (ext ABC) exterior of the triangle ABC A(ABC) area of the triangle ABC P(ABC) perimeter of the triangle ABC ABCD quadrilateral ABCD ABCD paralelogram ABCD O circle with center O C circumference sin o sine o cos o cosine o tan o tangent o cot o cotangent o sec o secant o cosec o cosecant o 202 Geometriy 7 acute angle: An acute angle is an angle with measure greater than 0° and less than 90°. acute triangle: An acute triangle has three acute angles. adjacent angles: Two angles are adjacent if they share a com- mon vertex and side, but have no common interior points. adjacent sides: In a triangle or other polygon, two sides that share a common vertex are adjacent sides. alternate exterior angles: Two angles are alternate exterior angles if they lie outside l and m on opposite sides of t, such as /b and /g. alternate interior angles: Two angles are alternate interior angles if they lie between l and m on opposite sides of t, such as /d and /e. (See figure for alternate exterior angles.) altitude of a triangle: An altitude of a triangle is a segment from a vertex that is perpendicular to the opposite side or to the line containing the opposite side. An altitude may lie inside or outside the triangle. angle: An angle consists of two different rays that have the same initial point. The rays are the sides of the angle and the ini- tial point is the vertex of the angle. angle bisector: An angle bisector is a ray that divides the angle into two con- gruent angles. C O B A [OB is an angle bisector [AN] is an angle bisector A C B vertex: point A sides: [AC and [AB altitude b a c d g e h f l m t a b c Ðb and Ðc are adjacent angles Ða and Ðc are not adjacent angles 50° 60° 70° 20° 85° angle bisector of a triangle: An angle bisector of a triangle is a segment that bisects one of the angles of the triangle. Its endpoints are points on the triangle. angle of depression: The angle formed by the horizontal and the line of sight to an object below the horizontal. angle of elevation: The angle formed by the horizontal and the line of sight to an object above the horizontal. area: The number of square units that cover a given surface. base: The lower face or side of a geometric shape. center of a circle: The center of a circle is the point inside the circle that is equidistant from all the points on the circle. central angle of a circle: A central angle of a circle is an angle whose vertex is the center of the circle. circle: A circle is the set of all points in a plane that are equidistant from a given point, called the center of the circle. circumference of a circle: The circumference of a circle is the distance around the circle. collinear: Points, segments, or rays that are on the same line are collinear. complementary angles: Two angles are complementary if the sum of their measures is 90°. Each angle is a complement of the other. concave polygon: See non-convex polygon. A B C D A, B, C, and D are collinear r O O is the center of the circle horizontal angle of elevation horizon (horizontal) angle of depression A B N C 203 Answers to Exercises concurrent: Two or more lines or segments are concurrent if they intersect at a single point. congruent angles: Two angles are congruent if they have the same measure. congruent arcs: On the same circle or on congruent circles, two arcs are congruent if they have the same measure. congruent polygons: Two polygons are congruent if there is a correspondence between their angles and sides such that corresponding angles are congruent and corresponding sides are congruent. Congruent polygons have the same size and the same shape. congruent segments: Two segments are congruent if they have the same length. consecutive interior angles: Two angles are consecutive interior angles if they lie between l and m on the same side of t, such as /b and /e. convex polygon: A polygon is convex if no line that contains a side of the polygon contains a point in the interior of the polygon. coplanar: Points, lines, segments or rays that lie in the same plane. corresponding angles: Two angles are corresponding angles if they occupy cor- responding positions, such as /a and /e in the figure. concentric circles: Circles that have different radii but share the same center are called concentric circles. cone: A solid figure that has a circular base and a point at the top. cube: A square prism that has six equal square sides. cylinder: A solid with circular ends and straight sides. a d c b g h f e l m t convex polygons a d c b g h f e l m t decagon: A decagon is a polygon that has ten sides. degree: A unit of angle and arc measure. diameter of a circle: A diameter of a circle is a chord that passes through the center. The diameter, d, is twice the radius: d = 2r. diagonal: A line segment joining two non-adjacent vertices of a polygon. equiangular triangle: An equiangular triangle has three congruent angles, each with a measure of 60°. equilateral triangle: An equilateral triangle has three con- gruent sides. exterior angles of a triangle: When the sides of a triangle are extended, the angles that are adjacent to the interior angles of the triangle are the exterior angles. Each vertex has a pair of exterior angles. exterior of an angle: A point D is in the exterior of /A if it is not on the angle or in the interior of the angle. half line: A ray without an endpoint (initial point). half planes: Two halves of a plane that are separated by a line hexagon: A hexagon is a polygon with six sides. hypotenuse: In a right triangle, the side opposite the right angle is the hypotenuse of the triangle. h y p o t e n u s e P P 1 d P = P 1 È P 2 È P P 2 A B ]AB = half line AB D C A B interior angle exterior angle exterior angle d ia g o n a l 204 Geometriy 7 inscribed angle of a circle: An angle is an inscribed angle of a circle if its vertex is on the circle and its sides are chords of the circle. interior of an angle: A point D is in the interior of A if it is between points that lie on each side of the angle. intersecting lines: Coplanar lines which have only one point in common. intersecting planes: Planes which have one common line. isosceles triangle: An isosceles triangle has at least two congruent sides. isosceles trapezoid: A quadrilateral with one pair of parallel sides and at least two sides the same length. kite: A convex quadrilateral with two pairs of equal adjacent sides. legs of a right triangle: Either of the two sides that form a right angle of a right tri- angle. legs of an isosceles triangle: One of the two congruent sides in an isosceles triangle. line: A line is an undefined term in geometry. In Euclidean geometry a line is understood to be straight, to contain an infinite number of points, to extend infinitely in two direc- tions, and to have no thickness. line segment: See segment. leg leg leg leg A B D C ÐAOB is an inscribed angle A O B major arc: On circle P, if m/APB < 180°, then the points A and B together with the points of the circle that lie in the exterior of m/APB form a major arc of the circle. Major arcs are denoted by three letters, as in AùCB. midpoint of a segment: The midpoint of a segment is the point that divides the segment into two congruent segments. minor arc: On circle P, if m/APB < 180°, then the points A and B, together with the points of the circle that lie in the interior of m/APB form a minor arc of the circle. Minor arcs are denoted by two letters, such as AïB. noncollinear: Points, segments, or rays that are not collinear. non-convex polygon: A polygon is non-convex (concave) if at least one line that contains a side of the polygon contains a point in the interior of the polygon. non-coplanar: Not coplanar. oblique lines: Lines are oblique if they intersect and do not form right angles. obtuse angle: An obtuse angle is an angle with measure greater than 90° and less than 180°. obtuse triangle: An obtuse triangle has exactly one obtuse angle. octagon: An octagon is a polygon with eight sides. AïB is a minor arc of the circle P C A B R S T |RS| = |ST| midpoint AùCB is an major arc of the circle P C A B 205 Answers to Exercises parallel lines: Two lines are parallel if they are coplanar and do not intersect. parallel planes: Two planes are parallel if they do not intersect. parallelogram: A quadrilateral with opposite sides parallel, and hence equal in length. pentagon: A pentagon is a polygon with five sides. perimeter of a polygon: The perimeter of a polygon is the sum of the length of its sides. perpendicular lines: Two lines are perpendicular if they intersect to form a right angle. perpendicular line and plane: A line is perpendicular to a plane if it is perpendicular to each line in the plane. plane: A plane is an undefined term in geometry. In Euclidean geometry it can be thought of as a flat surface that extends infinitely in all directions. point: A point is an undefined term in geometry. It can be thought of as a dot that represents a location in a plane or in space. polygon: A polygon is a plane figure formed by three or more segments called sides, such that the following are true: 1. each side intersects exactly two other sides, once at each endpoint, and 2. no two sides with a common endpoint are collinear. l l ^ P P k l k ^ l E A and B are parallel planes postulate: A postulate is a statement that is accepted as true without proof. proof: A proof is an organized series of statements that show that the statement to be proved follows logically from known facts (given statements, postulates, and previously proven theorems). protractor: A device used to determine the measures of angles. pythagorean triple: A set of three positive integers a, b, and c that satisfy the equation a 2 + b 2 = c 2 is a pythagorean triple. prism: A solid figure that has two bases that are parallel, congruent polygons and with all other faces that are parallelograms. pyramid: A solid figure with a polygon base and whose other faces are triangles that share a common vertex. quadrilateral: A polygon with four sides. The sum of the angles is 360°. radius of circle: A radius of a circle is a segment that has the center as one endpoint and a point on the circle as the other endpoint. ray: The ray AB, or [AB, consists of the initial point A and all points on line that lie on the same side of A as B lies. rectangle: A rectangle is a parallelogram that has four right angles. rectangular prism: A solid figure that with two bases that are rectangles and with all other faces that are parallelograms. A B [AB A B [BA 100° 120° 60° 80° 45° 135° 135° 45° c b a a 2 + b 2 = c 2 0° 10° 20° 30° 40° 50° 60° 70° 80° 90° 100° 120° 130° 140° 150° 160° 170° 0° 10° 20° 30° 40° 50° 60° 80° 90° 100° 110° 120° 130° 140° 150° 160° 170° 180° 0 1 2 3 4 5 6 7 8 9 10 180° 110° 70° regular polygon: A polygon whose sides are equal and whose angles are equal. right prism: A prism that has two special characteristics: all lateral edges are perpendicular to the bases and all lateral faces are rectangular. rhombus: A rhombus is a parallelogram that has four congruent sides. right triangle: A triangle with exactly one right angle. scale factor: In two similar polygons or two similar solids, the scale factor is the ratio of corresponding linear measures. scalene triangle: A scalene triangle is a triangle that has no congruent sides. segment: A segment AB, or [AB], consists of the endpoints A and B and all points on the line AB that lie between A and B. similar polygons: Two polygons are similar if their corresponding angles are congruent and the lengths of their corresponding sides are proportional. sine: The ratio of the length of the side opposite an angle to the length of the hypotenuse in a right triangle. surface area: The sum of all the areas of the surfaces of a solid figure. skew lines: Two lines are skew if they do not lie in the same plane. space: The set of all points. sphere: A sphere is the set of all points in space that are a given distance r from a point called the center. The distance r is the radius of the sphere. O d k d and k are skew lines A B C sin a = |AC| |BC| a A B AB or BA A B [AB] or [BA] square: A square is a parallelogram that is both a rhombus and a rectangle; that is, it has four congruent sides and four right angles. straight angle: A straight angle is an angle that measures 180°. supplementary angles: Two angles are supplementary if the sum of their measures is 180°. Each angle is a supplement of the other. tangent: The ratio of the length of the side opposite an angle to the length of the side adjacent to the angle in a right trian- gle. tangent to a circle: A line is tangent to a circle if it intersects the circle at exact- ly one point. theorem: A theorem is a statement that must be proved to be true. transversal: A transversal is a line that intersects two or more coplanar lines at different points. trapezoid: A quadrilateral with exactly one pair of opposite parallel sides. The sum of the angles is 360°. vertex of a polygon: A vertex of a polygon is a common endpoint of two of its sides. vertical angles: Two angles are vertical if their sides form two pairs of opposite rays. volume: The number of cubic units needed to occupy a given space. b c d a Ða and Ðc are vertical angles Ðb and Ðd are vertical angles l k d l is the tranversal B A C A B C tan a = |AC| |AB| a straight angle