Solid mensuration Lecture and exercise

June 21, 2018 | Author: Jake Mauhay | Category: Volume, Sphere, Area, Euclid, Triangle


Comments



Description

MENSURATION9 Mensuration is a branch of mathematics which deals with the measurements of lengths of lines, areas of surfaces and volumes of solids. Mensuration may be divided into two parts. 1. Plane mensuration 2. Solid mensuration Plane mensuration deals with perimeter, length of sides and areas of two dimensional figures and shapes. Solid mensuration deals with areas and volumes of solid objects. After learning this chapter you will be able to * Recognise a cylinder, a cone and a sphere. * Understand the properties of cylinder, cone and sphere. * Distinguish between the structure of cylinder and cone. * Derive the formula to find the surface area and volume of cylinder, cone and sphere. * Solve simple problems pertaining to the surface area and volume of cylinder, cone and sphere. CYLINDER Observe the following figures : 1 Road rollerCircular based storage Tank Cylinder Wheels of a road roller, a circular based storage tank etc will suggest you, the concept of a right circular cylinder. 2 1. The right circular cylinder A right circular cylinder is a solid described by revolution of a rectangle about one of its sides which remains fixed. AP = Radius of the circular plane AB = Axis of the cylinder PQ = Height of the cylinder Features of a right circular cylinder 1) A right circular cylinder has two plane surfaces, circular in shape. 2) The curved surface joining the plane surfaces is the lateral surface of the cylinder. 3) The two circular planes are parallel to each other and also congruent. 4) The line joining the centers of the circular planes is the axis of the cylinder. 5) All the points on the lateral surface of the right circular cylinder are equidistant from the axis. 6) Radius of circular plane is the radius of the cylinder. Two types of cylinders : 1. Hollow cylinder and 2. Solid cylinder A hollow cylinder is formed by the lateral surface only. Example : A pipe You will get a rectangular paper cutting which exactly covers the lateral surface. Wrap the strip around the lateral surface of the cylinder and cut the overlapping strip along the vertical line. Area of the rectangle is equal to the area of the curved surface of the cylinder. l = 2 r Breadth of the rectangle b is the height of the cylinder = h Area of the rectangle A = l x b Lateral surface area of the Cylinder A = 2 r h A = 2 r h sq. Surface area of a right circular cylinder A. units . 2. 4.A solid cylinder is the region bounded by two circular plane surfaces and also the lateral surface. Lateral Surface area : Activity : 1. Take a strip of paper having width equal to the height of the cylinder. Expression for the lateral surface area : (i) (ii) Length of the rectangle is equal to the circumference. (say PQ) 3. Example : A garden roller 2. units. units.Observe : Surface area of a thin hollow cylinder having circumference P and height is ‘h’ = Ph or = 2 r h ( P = 2 r ) B. units 2 [ 2 2 Lateral surface area of a cylinder = 2nrh sq. Given : circumference = 2r = 44 cms and height = h = 10 cms Solution : Lateral surface area of cylinder = 2rh . Total surface area of a cylinder = 2nr (r+h) sq. Remember : Area is always expressed in square units Worked examples : Example 1 : Find the lateral surface area of a cylinder whose circumference is 44 cm and height 10 cm. Total surface area of a cylinder : r2 2rh [ r 2 circular bases of the cylinder = + The total surface area Area of two area of cylinder Lateral surface = r r 2rh = 2r 2rh = 2r (r + h) sq. = 44 x 10 = 440 sq. . cm. Given : Diameter = d = 10 cms d = 5 cms  r = 2 and height = h = 12. What is the height of the tin? Given : The lateral surface area of a cylinder = = 1760 sq. 2rh radius = r = 10 cms Solution : Lateral surface area of a cylinder = 2rh 1760 = 2x 22 7 h = x 10 x h 1760 x 7 2 x 22 x 10 h = 28 cms Exercise : 9. Find the surface area of the pipe. Calculate the curved surface area of the cylinder.5 7 = 550 sq. given that the diameter is 10 cm and height is 12.cm. 2) The circumference of a thin hollow cylindrical pipe is 44 cm and length is 20 mts.Example 2 : Find the total surface area of the cylinder. Solution : Total surface area of a cylinder = 2r (r+h) = 2x = 2x 22 7 22 x5x (5 + 12. cm.5) x5x 17.5 cms. cm and radius is 10 cm.1 1) The radius of the circular base of a cylinder is 14 cm and height is 10 cm. Example 3 : The lateral surface area of a thin circular bottomed tin is 1760 sq. .5 cm. Calculate the total surface area of the cylinder.3) A cylinder has a diameter 20 cm and height 18 cm. . What is the area of the playground? 3. cm and radius is 14 cm.4) Lateral surface area of a cylinder is 1056 sq. Find the cost of painting the lateral surface of the pillars at Rs. 7) A closed cylindrical tank is made up of a sheet metal. 6) The diameter of a thin cylindrical vessel open at one end is 3. . 25 per squaremeter. Volume is expressed in cubic units. How many square meters of sheet metal is required to make? 8) A roller having radius 35 cms and length 1 meter takes 200 complete revolutions to move once over a play ground.5 mts. The height of the tank is 1. 2) Pile up the coins of same size one upon the other such that they form a right circular cylinder of height ‘h’.5 cm and height is 5 cm. 5) A mansion has twelve cylindrical pillars each having the circumference 50 cm and height 3. Volume of a right circular cylinder : Activity : 1) A coin is placed on a horizontal plane. Find the height of the cylinder.3 meters and radius is 70 cm. Volume of a cylinder = Bh The area of the circular base B = r2 [B is the circular base of radius r] Height of the cylinder = h Volume of the cylinder = 1tr2h cubic units Volume of the right circular cylinder of radius ‘r’ and height ‘h’ = 1tr2h cubic units. Calculate the surface area of the vessel. 5)2 h = h = 12 cm  Height of the cylinder h = 12 cms Exercise : 9.Worked examples : Example 1 : Find the volume of the cylinder whose radius is 7 cm and height is 12 cm.  Volume of the cylinder = 1848 cc Example 2 : Volume of the cylinder is 462 cc and its diameter is 7 cm. Given : Volume = V = 462 cc Diameter = d = 7 cm  r = 3.5) x h .5 cm. Find the height of the cylinder.2 2 (3. Solution : Volume of a cylinder V = 462 2 r h = 22 x 7 462 x 7 22 x (3. Given : Radius of the cylinder = r = 7 cm Height of the cylinder = h = 12 cm Solution : Volume of the cylinder V = Bh 2 = r h 22 = 7 x 7 x 7 x 12 49 x 12 22 = 7 x = 1848 cubic cms. cm and height is 10 cm.1) Area of the base of a right circular cylinder is 154 sq. . 2) Find the volume of the cylinder whose radius is 5 cm and height is 28 cm. Calculate the volume of the cylinder. 3) The circumference of the base of a cylinder is 88 cm and its height is 10 cm. Calculate the volume of the cylinder. h O B . 5) A cylindrical vessel of height 35 cm contains 11 litres of juice. Calculate the radius of the cylinder. Calculate the volume of water stored in the well.) 6) Volume of a cylinder is 4400 cc and diameter is 20 cm. 1. What is the height of the tin if the diameter of the tin is 14 cm? (one litre = 1000 cc) THE RIGHT CIRCULAR CONE Observe the following figures : Heap of sandIce cream cone cone A heap of sand. Find the diameter of the vessel (one litre = 1000 cc. Surface area of a right circular cone : A l A right circular cone is a solid generated by the revolution of a right angle triangle about one of the sides containing the right angle. Find the height of the cylinder. 8) A thin cylindrical tin can hold only one litre of paint. 7) The height of water level in a circular well is 7 mts and its diameter is 10 mts. 4) Volume of a cylinder is 3080 cc and height is 20 cm. an Ice cream cone suggests you the concept of a right circular cone . Observe : Radius of the circular section is equal to slant height of the cone = l . 3) The curved surface which connects the vertex and circular edge of the base is the lateral surface of the cone. A. 5) Spread the paper on a plane surface.Properties of a right circular cone : 1) A cone has a circular plane as its base. 3) Cut the paper along the length of slant height say AB 4) Take out the paper which exactly covers the curved surface. 2) The point of intersection of the axis of the cone and slant height is the vertex of the cone (A). 4) The line joining the vertex and the center of the circular base is the height of the cone (AO = h) Remember : The distance between the vertex and any point on the circumference of the base is the slant height. 2) Wrap the curved surface with a piece of paper. The curved surface area : Activity : 1) Take a right circular cone. .....This circular section can be divided into small triangles as shown in the figure say T1. Area of T1 = T2 = T3 = Area of Tn = Area of circular section 2 1 2 1 2 1 2 1 2 = = = = x B1 x l x B2 x l x B3 x l x Bn x l 1 1 B1l + 2 1 2 1 2 1 B2l + ... + Bn) l ( 2r ) 2 1 2 l x 2r [B1 + B2 + .. T2. Tn From the figure lateral surface area of the circular section = sum of the areas of each triangle.....e... + Bn = 2r ] ...... Tn 1 x base x height Area of the Triangle = i..... + 2 Bnl l (B1 + B2 + .. = T1 + T2 + .... T3 .. Area of the curved surface of a cone = n rl sq. units . given radius of the base as 10 cms. Given : Radius of the base Slant height r = 10 cm l = 28 cms Solution : Curved surface area of a cone = nr l x 10 x 28 = 22 7 = 22 x 10 x 4 = 880 sq.5 cm and slant height is 10 cm.5 (3.B. Given : Radius = r = 3.5 + 10) x 3.5 cm Slant height = l = 10 cm Solution : Total surface area of cone = r (r + l) = = 22 7 22 x 3.cm Example 2 : Find the total surface area of the cone whose radius is 3. units Worked Examples : Example 1 : Calculate the curved surface area of cone. and slant height 28 cms. Total Surface area of a cone : nr2 nr l Total surface area of a cone = Area of circular base + Area of the curved surface = nr2 nr l = r (r + l) Total surface area of a cone = r [r + l] sq.5 x 13.5 . 7 = 148. cm .5 sq. 4 cms. Calculate the cost of the canvas used at Rs. Find the total surface area of the cone. 3) The diameter of the cone is 14 cm and slant height is 9 cm.cm.Example 3 : The diameter of a cone is 10 cm and height is 12 cm.6 cms and diameter of the base is 8. Find the radius of the cone. 15 per sq. 7) The height of the cone is 5.3 1) Find the curved surface area of a cone whose circumference of base is 66 cm and slant height is 12 cm. . Given : diameter = d = 10 cm height = h = 12 cm Solution : d = 10 cm  r = 5 cm and l2 = r2 + h2 l  52 122 l  169 l 13 1 = 13 cm Total surface area of cone nr (r + l) = 22 x 5 (5+13) = 7 22 x 5 x 18 = 7 282. meter.cm = Exercise : 9. Find the area of the curved surface. Calculate the total surface area of the cone. 5) The slant height and the diameter of a conical tent are 25 mtrs and 14 mtrs respectively.85 sq. meters and radius is 8 meters. 6) The curved surface area of a conical tomb is 528 sq. 2) The curved surface area of a cone 440 sq. 4) Find the total surface area of conical tomb when the slant height is 8 meters and diameter is 12 meters. and slant height is 10 cm. Find the height of the tomb. .per sq.8) The height of a conical tent is 28 mts and the diameter of the base is 42 meters. 20/. Find the cost of the canvas used at Rs. mts. 1 2 Worked Examples : 1tr h 3 Example 1 : Find the volume of the cone whose base area is 154 sq. Given : Base area = B = 154 sq. 3 1 x Bh [ Volume of a cylinder = Bh] 3 1 = h r 3 [ B = r 2 ] 2 Volume of a cone of radius r and height h = cubic units. 2) Fill the conical cup with water up to its brim. cm. . 3) Pour the water into cylindrical vessel. cm and height is 12 cm. Volume of a cylinder = 3 x volume of a cone having same base and height. Volume of a cone = = 1 of the volume of a cylinder having same base and height. 4) Count how many cups of water is required to fill the cylindrical vessel upto its brim.2. Observe that exactly three cups of water is required to fill the vessel. Volume of a right circular cone Suggested Activity : 1) Take a conical cup and a cylindrical vessel of the same radius and height. and height of the cone = h = 12 cm 1 Solution : Volume of a cone = 3 Bh . Find the radius of the cone.5 cm is melted and recast to form cones of radius 1 cm and height 2.1 = h r 3 1 = Volume of the cone 2 [B = r 2 ] x 154 x 12 3 = 616 cc Example 2 : Find the volume of the cone having radius 7 cm and height 18 cm. Find the volume of the cone. Given : Radius = r = 7 cm Height h = 18 cm 1 Solution : Volume of a cone = 3 Bh 1 2 = h r 3 [ B = r 2 ] 2 = 1 x 22 x 7 3 7 x 18 Volume of the cone = 924 cc Exercise : 9. 22 . 7) A meter long metal rod (cylindrical in shape) of radius 3. Find the number of cones so formed. Find the diameter of the base. 8) A right angled triangle of sides 21 cm 28 cm and 35 cm is revolved on the side 28 cm. 3) The radius of the base is 10 cms and the height is 21 cms. 4) The circumference of the brim of a conical cup is 22 cm and height is 6 cm. Find the height of the cone. Name the solid formed and find the volume.4 1) Area of the base of a cone is 300 sq. 2) Volume of a cone is 550 cc and diameter is 10 cm. 6) The volume and the height of a cone are 2200 cc and 21 cm respectively. How much water does it hold? 5) The volume of a cone is 3080 cc and height is 15 cm.1 cm. Find the volume of the cone. cm and height is 15 cm. THE SPHERE Observe the following figures Shot putBall Sphere A shotput. 1) A sphere has a centre 2) All the points on the surface of the sphere are equidistant from the centre. 1. Remember : A plane through the centre of the sphere divides it into two equal parts each called a hemisphere. will suggest you the concept of a sphere. . 3) The distance between the centre and any point on the surface of the sphere is the radius of the sphere. a ball etc. Surface Area of a sphere : Activity : 1) Consider a sphere of radius r. 2) Cut the solid sphere into two equal halves. Properties of a sphere : A sphere is a solid described by the revolution of a semi circle about a fixed diameter. 4) Starting from the centre point of the curved .3) Fix a pin at the top most point of a hemisphere. Given : Radius of the sphere r = 14 cm. What do you observe from both the activities? It is found that the length of the thread required to cover the curved surface is twice the length required to cover the circular plane surface. wind a uniform thread so as to cover the whole curved surface of the hemisphere. 6) Similarily. . 5) Measure the length of the thread. 7) Starting from the centre. Worked Examples : Example 1 : Calculate the surface area of the sphere whose radius is 14 cm. 9) Compare the lengths.surface of the hemisphere. units. fix a pin at the center of the plane circular surface. 8) Unwind and measure the lengths of the threads. cm. Area of the plane circular surface Curved surface area of a hemisphere Surface area of the whole sphere = r 2 = 2r 2 = = 2 2r 2r 4r 2 2 Surface area of a sphere of radius r = 41tr2 sq. Solution : Surface area of the sphere = 4r 2 = 4x 22 x(14) 2 7 = 4x 22 x196 7 = 2464 sq. wind the thread of same thickness to cover the whole circular surface. Calculate the cost of painting at Rs. 5) The surface area of a sphere is 154 sq. cm. r base = B height = r . 2) The circumference of a globe is 88 cm. Find the diameter of the sphere. 3) Find the total surface area of a hemisphere of radius 14 cm. Given : Surface area = 4r 2 d = ? = 616 sq. Calculate the surface area of the globe.5 1) Find the surface area of a sphere whose radius is 21 cm. cm. Volume of a sphere Observe the following figures. Find the total surface area of both the hemispheres. Solution : Surface area of a sphere 4r 2 22 = 616 = 4 x r2 = r2 r Radius of the sphere = r 616 x 4 = 49 x r2 7 7 22 = 49 = 7 cm d = 2r = 2 x 7 Diameter of the sphere d = 14 cm Exercise : 9. cm. 20 per square meter.Example 2 : The surface area of the sphere is 616 sq.5 cm is cut into 2 halves. 2. Find the diameter of the sphere. 4) The circumference of a hemispherical dome is 44 mts. 6) A solid sphere of radius 10. ....A solid sphere is made up of miniature cones whose height is equal to the radius of the sphere and each having circular base.A.. 2 1 x Bn x r 3 1 = r (B1 + B2 + .S.. + Bn) [ T. of sphere 4 r ] 3 1 = = = 2 r (B) 3 1 [ Surface area of sphere] r x 4 r 2 3 4 3 r 3 4 r 3 Volume of the sphere 3= cubic units Volume of the hemisphere 1 = 2 = 2 3 x 4 3 nr 3 nr 3 .. 1 Volume of each cone = x base x height 3 Volume of cone 1 = 1 x B x r 1 3 1 Volume of cone 2 = Volume of cone n = x B2 x r etc 3 1 x Bn x r 3 Volume of the sphere = Sum of the volumes of all the cones = 1 1 x B x r + 3 3 1 x B x r + .... 2 nr  Volume of a hemisphere 3 =3 cubic untis . Example 2 : A hemispherical bowl has radius 14 cm. Find the diameter of each ball. Given : Volume of 21 balls = 88 cc 88 Volume of each ball = 21 cc Solution : Volume of a Steel ball = .Worked Examples : Example 1 : Find the volume of the sphere whose radius is 21 cm. Given : Radius of the sphere = r = 21 cms Solution : Volume of a sphere = 4 r 3 3 = 4 22 3 x x(21) 3 7 22 x21x21x21 4 = 3x 7 = 4 x 22 x 21 x 21 Volume of the sphere = 38808 cc. r = 14 cms 2 Solution : Volume of a hemisphere = 3 2 = Capacity of the bowl [ nr x 3 3 22 x14 3 7 = 2x 22 = 3 7 x14x14x14 5.75 Litres 1000 cc = 1 litre] Example 3 : Total volume of 21 steel balls in a bearing is 88 cc. How many litres of water does it hold? Given : Radius of the sphere. 3) The depth of a hemispherical water tank is 2. Find the volume of the sphere when the radius of each marble is 2 cm. 2) The diameter of a shot put is 9 cm.6 1) Find the volume of the sphere whose radius is 3 cm.4 r 3 3 88 4 22 3 x 21  3 7 x r 88 3 7 x x 21 4 22 r3 = r3 = 1 r = 3 1 r = 1 cm d = 2r  d = 2 cms Diameter of each ball = 2 cms Exercise : 9. Calculate the volume of the shot-put. 4) Twenty one lead marbles of even size are recast to form a big sphere.1 m at the centre. Find the capacity of the water tank in litres. Remember at a glance : Solid Curved Surface area Total Surface area Volume Cylinder 2rl 2r (r+1) Cone rl r (r+1) Sphere 4r 2 4r 2r 2 3r r h 1 2 r h 3 4 3 r 3 2 3 r 3 Solid hemisphere 2 2 2 . taking the measurements to the scale. You are familiar with the geometrical figures like triangle. trapezium and the formulae for the area of these figures. AB || CD and the distance between AB and CD is 5 cm. 1. rectangle. Represent the irregular shaped field into known geometrical rectilinear figures. .SCALE DRAWING After studying this unit you will be able to * * * * Develop the skill in drawing the polygonal figures to the scale. How to find the area of the land? D 4 R 3 E 5 Q 2 C P 1 B A 1) Irregular shaped field is divided into known geometrical shaped fragments. Area of triangle 1 = 2 1 x base x height = Area of rectangle = length x breadth Area of trapezium = bh = lb 1 2 2 1 x height x (sum of two parallel sides) = 2 h (a+b) Recall : 1) Find the area of triangle when the base is 12 cm and height is 8 cm. 2) Measurements are recorded and a sketch is drawn to the scale. Measurement of the area of a land : Observe the following piece of land. Relate the area of polygonal figure to the area of irregular field. To find the area of triangles and trapeziums. 2) Find the area of trapezium given AB = 8 cm. CD = 6 cm. To. QE and RC 4) Complete the polygon joining the end points. 4) Total area of the land is the sum of the areas of all the right angled triangles and the trapezium. 6) Total area of the land is equal to the sum of the all the triangles and trapezium. 4 Remember : P 30 40 B Area of the land is expressed in hectares. AQE.3) Measurements are recorded in the surveyor’s field book. D 60 50 R E D Q 60 Total Area of the land = Area of  + Area of  + Area of  + 1 100 Area of  + Area of Trapezium. Worked Example : A 2 3 . Surveyor’s mode of recording the measurements of a land is given below. 5) Find out the area of ABP. say 20m = 1 cm 2) Draw the base line AD 3) Draw the perpendiculars to the base line PB. DQE and DRC and also the are of trapezium BPRC.D (in mts) 200 140 To E 60 50 To C 120 40 30 to B From A Activity : To measure the area of the land 1) Take suitable scale. Meter to D 150 70 to C 100 To E 80 80 40 to B 30 From A Solution : Scale 20 mts = 1 cm Observe that.mts .Example 1 : Plan out and find the area of the field from the following notes from the field book. D AM = 30 mts 50 AN = 80 mts 70 P C 20 E N 80 30) = 70 mts ND = (150-80) 50 40 M MP = (100- B 30 = 70 mts PD = (150-100) = 50 mts A 1 1) Area of ABM = = 2 1 2 bh x 30 x 40 = 600 sq. 1 2) Area of Trapezium MDCP = = 2 1 h (a+b) 2 x 70 (70+40) = 3850 sq.22 hectares Observe : 1 hectare =10. Exercise : 9. mts 1 5) Area of NEA = 2 x 80 x 80 = 3200 sq.7 1) Draw a plan and calculate the area of a level ground using the information given below. = 1. Meters to C 220 To D 120 210 120 To E 180 80 From A 200 to B .000 sq.mts 3) Area of DPC = 1 2 x 50 x 70 = 1750 sq. mts Area of the field ABCDE = 600 + 3850 + 1750 + 2800 + 3200 = 12200 sq mts. mts 1 4) Area of DEN = 2 x 80 x 70 = 2800 sq.mts. Meters to D 225 175 To E 90 125 To F 60 100 To G 15 80 20 to C 60 70 to B From A 4) Calculate the area of the field shown in the diagram below : [Measurements are in meters] B C 40 D 35 F 30 E 30 25 G 70 A .2) Plan out and find the area of the field from the data given from the Surveyor’s field book Meters to E 350 To D 100 300 To C 75 250 150 to F 150 To B 50 100 to G 50 From A 3) Sketch a rough plan and calculate the area of the field ABCDEFG from the following data.
Copyright © 2022 DOKUMEN.SITE Inc.