BRILLIANT PUBLIC SCHOOL, SITAMARHI (Affiliated up to +2 level to C.B.S.E., New Delhi) Affiliation No. - 330419 CBSE Board Level IX S.A.- I Maths Chapterwise Printable Worksheets with Solution Session : 2014-15 Office: Rajopatti, Dumra Road, Sitamarhi(Bihar), Pin-843301 Website: www.brilliantpublicschool.com; E-mail:
[email protected] Ph.06226-252314, Mobile: 9431636758, 9931610902 MATHEMATICS (Class-IX) Index: S.A.-I CBSE Chapter-wise Solved Test Papers 1. Number System 001 2. Polynomials 022 3. Introduction to Euclid’s Geometry 040 4. Lines and Angles 045 5. Triangles 079 6. Coordinate Geometry 106 7. Heron’s Formula 119 CBSE TEST PAPER-01 CLASS - IX MATHEMATICS (Number System) 1. Which of the following rational numbers have terminating decimal [1] representation (i ) 3 5 (iii ) 2. 2 13 23 (iv ) 7 (ii ) 40 27 How many rational numbers can be found between two distinct rational [1] numbers? (i)Two 3. 4. (ii) Ten ( )( (iii) Zero (iv) Infinite ) The value of 2 + 3 2 − 3 in [1] (i)1 (ii)-1 (iii) 2 (iv) none of these (27)-2/3 is equal to [1] (i) 9 (ii) 1/9 (iii) 3 (iv) none of these 2×4 3 5. Simplify: 6. Find the two rational numbers between ½ and 1/3. [2] 7. Find two irrational numbers between 2 and 3. [2] 8. Multiply 3 − 5 9. Express 0.8888… in the form p/q. 10. 3 ( ) [2] ( by 6 + 2 ) [2] [3] Simply by rationalizing denominator 7+3 5 7−3 5 [3] 2 11. 1 −1 − 4 Simplify 625 2 12. Visualize 3.76 on the number line using successive magnification. 13. Prove that 1 1+ x b−a +x c −a [3] + 1 1+ x a −b +x c −b + 1 1 1+ x a−c + xb −c =1 [3] CBSE TEST PAPER-01 CLASS - IX MATHEMATICS (Number System) [ANSWERS] Ans 01. (i) Ans 02. (iv) Ans 03. (i) Ans 04. (ii) Ans 05 3 2×4 3 1 1 2 3 ,3 4 The LCM of 3 and 4 is 12 1 4 1 1 ∴ 2 3 = 212 =(24 )12 = 1612 1 3 1 1 3 4 = 312 =(33 )12 = 2712 1 1 1 1 1 2 3 × 3 4 =1612 × 2712 = (16 × 27)12 = (432) Ans 06. 1 12 First rational number 1 1 1 1 3+ 2 5 = + ⇒ ⇒ 2 2 3 2 6 12 1 5 1 Ans , and 2 12 3 Second rational number 1 1 5 1 6 + 5 11 + ⇒ ⇒ 2 2 12 2 12 24 5 11 Ans and 12 24 = Ans 07. Irrational number between 2 and 3 is Irrational number between 2 and 2 6 is 2×3 = 6 8 9x = 8 x= Ans 10. 8 9 7+3 5 7+3 5 × (Rationalizing by denominator) 7−3 5 7+3 5 (7 + 3 5 ) 7 − (3 5 ) 2 2 2 = ( 72 + 3 5 ) 2 + 2× 7 + 3 5 49 − 32 × 3 49 + 9 × 5 + 42 5 49 + 45 + 42 5 = 49 − 27 22 94 + 42 5 94 42 = = + 5 22 22 22 47 21 = + 5 11 11 = Ans11.8888 10 x = 8.8 − − − −(2) 10 x − x = 8. 1 1 −4 − 625 2 2 3 .8888 10 x = 8. (3 − 5 )( 6 + 2 ) = 3(6 − 2 ) − 5 (6 + 2 ) = 18 + 3 2 − 6 5 − 5 × 2 = 18 + 3 2 − 6 5 − 10 Ans 09.1 1 2 × 6 = 22 × 64 = 2 1 1 1 2× 1 4 1 × 64 1 1 = ( 2 2 ) 4 × 6 4 = 4 4 × 6 4 = (24) 4 = 4 24 Ans 08.8888 x=0. Let x = 0.8 − 0.8 − − − (1) 10 x = 10 × 0. x + x b .x + x . x − c 1 1 1 + b −b a −b c −b + c − c a − c b − c −a c −a x .2 1 = 1 2 2 ( 25 ) 1 − ×2 1 4 = 25 1 = 1 625 2 1 = 25 − 1 2 − 1 4 1 = 5 1 2×− 2 = − 1 4 2 1 −m a = am 1 =5 5 −1 Ans12.x + x .x + x .x 1 + x . 1 Ans13.x x .x + x .x + x .x + x . + 1 + 1 1 + x b − a + x c − a 1 + x a −b + x c −b 1 + x a − c + x b − c 1 1 1 + + a −c a −b c −b b −a c −a 1 + x .x −a a b x −a 1 1 1 + −b a + −c a b c b c ( x + x + x ) x ( x + x + x ) x ( x + xb + xc ) a xa xb xc + + ( x a + xb + x c ) ( x a + xb + x c ) ( x a + xb + x c ) = x a + xb + x c =1 x a + xb + x c 4 .x 1 + x .x x .x + x . x + x . Find rationalizing factor of 7.IX MATHEMATICS (Number System) 1. ( ) (3 2 2 −2 3 Express 2. Evaluate ( i ) 3 125 ( ii ) 4 1250 6.236 and ) [3] 2 [3] p q 3 2 [3] 3 = 1. 2.4178 in the from Simplify ( 27 ) It − 2 3 1 − ÷ 9 2 .732 Find the value of 5 2 7 + 5+ 3 5− 3 [5] . [1] [2] [2] 300 Rationalize the denominator 1 5+ 2 and subtract it from 5− 2 [2] 8. Shove that 7 − 3 is irrational [2] 9.CBSE TEST PAPER-02 CLASS .83458456……………is [1] (i) an irrational number (ii) rational number (iii) a natural number (iv) a whole number. 12. Every natural number is (i) not an integer (ii) always a whole number (iii) an irrational number (iv) not a fraction Select the correct statement from the following 7 4 > 9 5 -2 -4 ( iii ) > 3 5 ( ii ) [1] 7. 13. Represent 3 on number line [3] 10.3 5 = 2. Simplify 3 2 + 2 3 11. 68 9 ( ii ) 64 9 ( iii ) 65 9 (iv ) 63 9 0.2 is equal to (i ) 4. 5. [1] 2 3 < 6 9 -5 -3 ( iv ) < 7 4 (i ) 3. (i) 3 125 = ( 5 × 5 × 5 ) 3 = 5 3 (ii) 4 1250 = ( 2 × 5 × 5 × 5 × 5 ) 4 = ( 2 × 54 ) 4 1 ( ) 1 3 =5 1 1 1 1 = 24 × 54 × 4 = 5× 4 2 300 = 2 × 2 × 3 × 5 × 5 6. = 22 × 3 × 52 = 2 × 5 3 = 10 3 Rationalizing factor is 1 5− 2 × 5+ 2 5− 2 7.IX MATHEMATICS (Number System) [ANSWERS] 1. (ii) 2. = 5− 2 ( 5) −( 2) 2 2 = 5− 2 5− 2 = 5−2 3 Difference 5− 2 = 5 − 2 − 3 5 2 + 3 3 5 2 5− − 2 − 3 3 = 5− 2− = = 2 5 2 2 2 − = 3 3 3 ( 6 5− 2 ) 3 . (iii) 3. (i) 5.CBSE TEST PAPER-02 CLASS . (iii) 4. ∴ OB = OA1= 2 = 1. suppose 7 − 3 is rational let 7 − 3 = x (x is a rational number) 7 = x+3 x is a rational no. ( )( ) (3 2 + 2 3 )(3 2 + 2 3 )(3 2 − 2 3 )(3 2 − 2 3 ) = ( 3 2 + 2 3 )( 3 2 − 2 3 )( 3 2 + 2 3 )( 3 2 − 2 3 ) = ( 3 2 ) − ( 2 3 ) ( 3 2 ) − ( 2 3 ) 3 2 +2 3 2 3 2 −2 3 2 2 2 2 2 = [9 × 2 − 4 × 3][9 × 2 − 4 × 3] = [18 − 12][18 − 12] = 6 × 6 = 36 11.178178 q 7 .4178 q p = 2.8.41 Now take A1 B1 = 1unit and ∠A1 =900 In ∆ O B1 A1 [OB12 =OA12 +A1B12 OB12 =( 2) 2 + 12 OB1 = 3 10. 3 is also rational number ∴ x+3 is rational number but 7 is irrational number which is controduction ∴ 9 7 − 3 is irrational number Take OA=AB=1 unit and ∠A = 900 In ∆ OAB=OB2 =12 +12 OB2 =2 OB= 2 . Let p = 2.4178178178 q Multiplying by 10 p 10 = 24. 236 + 5 × 1.472 + 8. ( 27 ) − 2 3 1 − ÷ 9 2.178178 − 24. 000 = 24178.132 = = = 6.566 2 2 = 8 .66 13.178178 q p 10.Multiplying by 1000 p 10. 3 3 2 3 2 ( 3 × 3 × 3) 3 × 3 2 − ( 3 × 3) 2 3 ( ) (3 ) 33 = − 2 3 −2 −m 1 a = a m 1 2 − 3 × 32 1 2 1 32 33 1 = = = 1 1+ 3 3 3 3 1 1 = 4 =3 81 33 2 7 + 5+ 3 5− 3 13.178178 q q p 9999 = 24154 q p 24154 = q 9999 12. 000 = 1000 × 24. 2 ( ( ) ( 5 + 3) = 2 3 )( 5 − 3 ) 5− 3 +7 5+ 5 −2 3 +7 5 +7 3 5−3 9 5 + 5 3 2 × 2.178178 q p p 10000 − = 24178.732 = 2 2 4. 8 is q 8 8 1 (i ) (ii) (iii) 10 100 8 3 The value of 1000 is The (i) 1 4. [2] 5 Evaluate 112 3 9.414 9 [3] [3] [5] .CBSE TEST PAPER-03 CLASS .236 and 2 = 1. find the value of 5 2+ 3 Visualize 2. Simplify ( 5 + 2) 2 [2] 8.IX MATHEMATICS (Number System) 1. Find two rational numbers between 7 and 5.414 and 3 = 1. 13. Find the value of 3 7 + . 112 Find three rational numbers between 2. [2] 7. If 5+ 2 5− 2 5 =2. 12. Give an example of two irrational numbers whose [3] (i) sum is a rational number (ii) product is a rational number (iii) Quotient is a rational number. 11.3 [3] 10. [1] p form of the number 0. If 2 = 1.4646 on the number line using successive magnification . [1] (ii) 10 (iv) 1 [1] (iii) 3 (iv) 0 The sum of rational and an irrational number [1] (i) may be natural (ii) may be irrational (iii) is always irrational (iv) is always rational 5.2 and 2.732. 3. A terminating decimal is (i) a natural number (ii) a rational number (iii) a whole number (iv) an integer. Show that 5 + 2 is not a rational number. [2] 6. 2. =7.. 3.. 112 3 am m ∵ a n = a − n 5 3 − 2 = 112 112 5− 3 2 = 11 3 = 11 2 = 11 9. 4. ( 5+ 2 ) ( ) ( ) 2 = 2 5 + 2 2 +2 5× 2 = 5 + 2 + 2 10 = 7 + 2 2 5 8.CBSE TEST PAPER-03 CLASS . 10.5 1 1 13 [ 7 + 6] = ×13 = 2 2 2 = 6 and 13 2 let 5 + 2 is rational number.212341365. and 2. (i) 2 + 2. (ii) 2.2321453269.. (i) 5. First rational number = (ii) 1 12 [ 7 + 5] = = 6 2 2 Second rational number = 6. Three rational numbers between 2.. and 2 − 2 Sum 2 2 + 2 − 2 = 4 rational number (ii) 3 2 and 6 2 Product 3 2 × 6 2 = 18 × 2 = 36 rational (iii) 2 125 and 3 5 Quotient 2 125 2 12525 2 2 10 = = 25 = × 5 = 3 5 3 3 3 3 5 10 (iii) .IX MATHEMATICS (Number System) [ANSWERS] 1.2 and 2. Say 5 + 2 = x 2 = x−5 x is a rational number 5 is rational number ∴ x − 5 is also rational number But 2 is irrational number Which is a contradiction ∴ 5 + 2 Is irrational number 7.. 6 ..3 are 2. 656 = 28. ) ( 2) − ( 3) 5 = ( 2− 3 2 2 = (Rationalising by denominator) 5 2− 3 2−3 = −5 [1.016 = 11 .414 − 1.414 = 3 = 22. 3 = = ( ( ) ( 5 + 2) 2 )( 5 − 2 ) 5− 2 +7 5+ 3 5 −3 2 + 7 5 +7 2 5−2 10 5 + 4 2 3 10 × 2.236 + 4 × 1.318 = 1.732] = −5 × −0.36 + 5.5 2− 3 × 2+ 3 2− 3 11. 3 7 + 5+ 2 5− 2 13.59 12. 7.4178 in the form Multiply 3 by 3 3 4 and 7 7 36 (i ) 100 [2] ( ii ) 2 11 [2] a b [2] 5 Rationalize the denominator of [2] 1 4+2 3 [3] 10. 6.37 on the no. [1] 3 2 and is 5 3 (B) irrational number (D) integer [1] 600 is ( B )100 6 ( D )10 6 60 3 Find four rational numbers between Writ the following in decimal form Express 2.CBSE TEST PAPER-04 CLASS . Visualize the representation of 5. line 3 decimal places [3] 11. 9. Simplify 2+ 5 2− 5 + 2− 5 2+ 5 [5] 12 . Shone that 5 2 is not rational number [3] 12.123 is equal to 122 990 122 (C) 99 (A) 3.IX MATHEMATICS (Number System) 1. The rational number not lying between 49 75 47 (C ) 75 50 75 46 (D) 75 ( A) 2. 8. ( B) [1] 0. Simplify 3 3 250 + 7 3 16 − 4 3 54 [3] 13. (B) ( The number 1 + 3 ) [1] 2 The simplest form of ( A )10 ( C ) 20 5. 122 100 (D) None of these is (A) natural number (C) rational number 4. 178178178. .178178178. 10. (i) 7. (D) 5.178 ______(1) 10 x = 24.e.178178. .4178 = 24154 12077 = 9990 4995 13 .4178 (ii) 2 = 0. (B) 2. (A) 3. 000 x − x = 24178. 1000 ×10 x = 1000 × 24. . 3 31 32 33 34 35 .36 100 6. (B) 4.. 7 7 70 70 70 70 70 36 = 0. ...178 _______(2) Eq (2) – eq(1) 10. 7 70 70 70 70 70 4 7 3 10 30 4 10 40 × = and × = 7 10 70 7 10 70 Take any four rational numbers between 30 40 and i.... .IX MATHEMATICS (Number System) [ANSWERS] 1. rational numbers between 70 70 3 4 31 32 33 34 35 and are . 000 x = 24178.18 11 10 x = 24. x = 2.178 − 24.178 9990 x = 24154 x= 24154 9990 2.. .CBSE TEST PAPER-04 CLASS . 10000 x = 24178.. 14 . 1 4−2 3 × 4+2 3 4−2 3 = 4−2 3 ( 16 − 2 3 ) 2 = ( 4−2 3 4−2 3 = 16 − 4 × 3 16 − 12 = 4−2 3 2 2− 3 = 4 42 = 2− 3 2 ) 10.8. 3 and 3 5 1 2 Or 3 and 5 1 3 LCM of 2 and 3 id 6 1 1 3 × 3 = ( 33 ) 6 = ( 27 ) 6 1 1 2 × 2 = ( 52 ) 6 = ( 25 ) 6 32 = 32 53 = 53 1 1 1 1 1 1 1 3 × 3 5 = ( 27 ) 6 × ( 25 ) 6 = ( 27 × 25 ) 6 1 = 675 6 = 6 675 9. 5 is rational no. let 5 2 is rational no x=5 2 ( x is rational) x = 2 5 x is rational no. 3 3 250 + 7 3 16 − 4 3 54 = 33 5 × 5 × 5 × 2 + 7 3 2 × 2 × 2 × 2 − 4 3 3× 3× 3× 2 = 3× 5 3 2 + 7 × 2 3 2 − 4 × 33 2 = 15 3 2 + 14 3 2 − 12 3 2 = [15 + 14 − 12 ] 3 2 = 17 3 2 15 . ∴ x is rational no. 5 But x = 2 and 5 2 is irrational no.11. Which is a contradiction ∴ 5 2 is irrational number 12. 7 7 7 7 6 7 Find 6 rational no.CBSE TEST PAPER-05 CLASS . 12.142857. Simplify 3 3 + 2 2 2 3 + 3 2 8. find the value of . . 10. [1] is (B) [1] 9 8 ( D ) None of these [1] 8 is an (A) natural number (C) integer Find the value of (B) rational number (D) irrational number 2+ 5 it 5 5 = 2. 13.23 + 0. 5. 9.236 and 10 = 3. [1] The value of 0.45 ( C ) 0.22 is ( A ) 0.IX MATHEMATICS (Number System) 1.25 into rational 7. Convert 0.44 ( D ) 0.45 2. between and 5 5 [3] Show how [3] If 4 can be represented on the number line Find a and b if 3− 6 = a 6 −b 3+ 2 6 [3] [5] 16 .9 2 Simplify 1 9 2 Simplify 3 48 − [3] 5 1 +4 3 2 3 1 2 3 4 = 0. ( B ) 0.44 1 3 The value of 2 × 2 − 4 3 1 2 ( D ) None of these ( A) 2 ( B) (C ) 3 3. ( )( 3 − [2] [2] ) [2] [2] 4 9 2 . 16 13 ÷ 9 52 is equal to 3 9 8 (C ) 9 ( A) 4.162 6. 11. 25. 0.. = 6.22 = 0. −4 1 4 − 3 23 × 2 3 = 23 1− 4 3 =2 −3 =23 1 2 16 13 ÷ 9 52 16 13 16 13 16 8 1 = × = × 9 524 9 2 9 52 = 4.. 0... ( i ) x = 0..162 + 5 8..CBSE TEST PAPER-05 CLASS .232323..22 = 0.162 = = 1.23 + 0.454545. 2 −1 = 3.6324 5 5 let x = 0.. 0. = 0. 8 9 8 is an irrational number ∵ 4× 2 = 2 2 2+ 5 5 10 + 5 × = 5 5 5 5... 3...222222.45 1 2... 17 ..23 = 0.IX MATHEMATICS (Number System) [ANSWERS] 1.252525. . 3 + 6.25 − 0.. 9 2 .9 2 9 1 2 3 4 − 2 92 = 9 a m ..25 99 x = 25 7....a n = a m + n 1 2 −1 = 9 1 = 92 = 1 9 9..Multiplying both sides by 100 100 x =25.25. ( ii ) Eq(ii) . x= 25 99 (3 3 +2 2 2 3 +3 2 )( ( ) ) ( 3 3 2 3 +3 2 +2 2 2 3 +3 2 ) = 6 × 3 + 9 3..eq(i) 100 x − x = 25.. 2 + 4 2. 2 2 92 1 = 92 = 3 48 − 1 1 a −m = a m 1 1 + 2 92 1 9 5 1 +4 3 2 3 = 3 16 × 3 − 5 3 +4 3 2 3× 3 5 1 = 3× 4 3 − .. 3 + 4 3 2 3 = 12 3 − 5 3+4 3 6 18 .252525… 100 x = 25.2 = 18 + 9 6 + 4 6 + 12 = 30 + ( 9 + 4 ) 6 = 30 + 3 6 −4 3 8. 285714 3 = 3 × 0.3 6. .428571 4 = 4 × 0. . . . .142857 7 = 0.571428 11.5 6. . . . . . Take OA = AB = 1 unit B And ∠A = 90 In ∆OAB 2 3 B1 B2 4 OB 2 = 12 + 12 = OB 2 = 2 OB = 2 OB = OA1 = 2 o 19 A A1 A2 A3 . .7 6.8 6. 50 50 50 50 50 50 12.6 6. .5 5 = 12 − + 4 3 = 16 − 3 6 6 = 10. between 6 7 and 5 5 61 62 63 64 65 66 .4 6.2 6.142857 7 = 0.142857 7 = 0. 91 3 6 1 = 0.142857 7 2 = 2 × 0.9 7 .1 6. is divided into 5 5 10 equal parts than the value of each part 6. . 6 7 . 5 5 5 5 5 5 5 5 5 5 ∴ 6 rational no. on denominator 3− 6 3− 2 6 × = a 6 −b 3+ 2 6 3− 2 6 9 − 6 6 − 3 6 + 2×6 ( 3 − 2 6 2 ) 2 = a 6 −b 9 − 9 6 + 12 = a 6 −b 9 − 24 −9 6 + 21 = a 6 −b −15 −/93 6 21 7 + = a 6 −b −/ 155 −155 3 7 6 − = a 6 −b 5 5 Comparing coefficient on both sides 3 7 a = . +b = + 5 5 20 .H. 3− 6 = a 6 −b 3+ 2 6 Rationalizing L.S.Take A1 B1 = 1 And ∠A1 = 90 In ∆OB1 A1 = OB12 = ( 2) 2 + 12 OB1 = 3 OB1 = OA2 = 3 Take a2b2 = 1 unit ∠A2 = 90 OB2 2 = ( 3) 2 + 12 = OB2 2 = 3 + 1 OB2 = 4 = 2 13. Without actual division. prove that the polynomial 2x3 + 4x2 + x – 34 is exactly [3] 3 [3] divisible by (x – 2) 13.CBSE TEST PAPER-01 CLASS . If x2 – bx + c = (x + p) (x – q) then factorise x2 – bxy + cy2 21 [5] . Which of the following expression is a polynomials (a) x3 − 1 (b) x +2 1 x2 (d) t + 5t − 1 (c) x 2 − 2. Without actually Calculating the cubes. 1 If x – 3 and x − are both factors of px2 + 5x + r . Find the value of K it x – 2 is factor of 4x3 + 3x2 – 4x + K [2] 7.IX Mathematics (Polynomials) 1. 3 5 Using suitable identify expand x + 4 4 11. Factorise the polynomial x3 + 8y3 + 64 z3 – 24xyz [2] 8. g(x) = x + 4 [1] 6. find the value of (-12)3 + (7)3 + (5)3 [2] 9. 4. A polynomial of degree 3 in x has at most (a) 5 terms (b) 3 terms (c) 4 terms (d) 1 term The coefficient of x2 in the polynomial 2x3 + 4x2 + 3x + 1 is (a) 2 (b) 3 (c) 1 (d) 4 The monomial of degree 50 is (a) x50 + 1 (c) x+50 [1] [1] [1] [1] (b) 2x50 (d) 50 5. 3. then show that P = r 3 [2] 10. Divide f(x) by g(x) & verify that the remainder f(x) = x3 + 4x2 – 3x – 10. Using factor theorem factorise f(x) = x2 – 5x + 6 [3] 12. IX Mathematics (Polynomials) 1. (b) 3. (d) 4. (a) 2.CBSE TEST PAPER-01 CLASS . x3 + 8 y 3 + 64 z 3 − 24 xyz x+4 x3 + (2 y )3 + (4 z )3 − 3.3x + 4x2 – 12 + 2 = x3 + 4x2 – 3x – 10 Verified 6. (b) 5.3) + 2 = x3 .(2 y )(4 z ) = ( x + 2 y + 4 z )[ x 2 + (2 y + 2 + 4 z ) 2 − x × 2 y − 2 y × 4 z − x × 4 z ] = ( x + 2 y + 4 z )( x 2 + 4 y 2 + 16 z 2 − 2 xy − 8 yz − 4 xz ) 8. Dividend = x3 + 4x2 – 3x – 10. a 3 + b3 + c3 = 3abc a+b+c = 0 (−12)3 + (7)3 + (5)3 = 3 × −12 × 7 × 5 = −1260 ∵ −12 + 7 + 5 = −12 + 12 = 0 22 x2 − 3 x3 + 4 x 2 − 3 x − 10 x3 + 4 x 2 − − − 3x − 10 − 3x − 12 + + 2 . divisor = x + 4 Quotient = x3 – 3 Remainder = 2 Dividend = Divisor × quotient + Remained = (x + 4) (x2 .x. x – 2 is factor of 4x3 + 3x2 – 4x + k x–2=0 ⇒ x=2 ∴ 4(2)3 + 3(2)2 − 4 × 2 + k = 0 32 + 12 − 8 + k = 0 44 − 8 + k = 4 36 _ k = 0 k = −36 7. 3 5 x+ 4 2 3 (a + b)3 = a 3 + b3 + 3ab(a + b) 3 3 3 3 5 3 5 35 3 5 x + = x + 3× x × x + 4 2 4 2 42 4 2 = 125 x3 27 45 5 3 + + x x + 8 64 8 2 4 125 x3 27 225 2 135 + + x + x 8 64 16 4 125 x3 225 2 135 27 = + x + x+ 8 16 4 64 = 11. f(x) = x2 – 5x + 6 Put x = 1 f (1) = 12 − 5 × 1 + 6 = 2 ≠ 0 Put x = 2 f (2) = 22 − 5 × 2 + 6 = 4 − 10 + 6 = 0 ∴ x − 2 is factor of f(x) 23 . x = 3 2 zero of Px + 5 x + r ∴ P (3)2 + 5 × 3 + r = 0 9 P + 15 + 4 = 0 9 P + r = −14 − − − − − − − − − −(1) 2 1 1 P + 5× + r = 0 3 3 P 5 + +r =0 9 3 P + 15 + 9r =0 9 P + 9r = −15 − − − − − − − − − (2) 9 P + r = P + 9r By (1) & (2) 9 P = 9r p=r Hence prove 10.9. 1 ∵ x − 3 and x − are factors of 3 1 Px 2 + 5 x + r ∴ x = 3. We get x 2 + ( p − q ) xy − pqy 2 x 2 + pxy − qxy − pqy 2 x ( x + py ) − py ( x + py ) ( x + py )( x − qy ) 24 . We have x 2 − bx + c = ( x + p )( x − q ) x 2 − bx + c = x 2 + ( p − q ) x − pq Equating coefficient of x and constant −b = p − q and c = − pq Substituting there values of b and c in x 2 − bxy + cy 2 . Let f ( x ) = 2 x 3 + 4 x 2 + x − 34 x − 2 Is factor of f ( x ) x = 2 Zero of f ( x ) f ( z ) = 2 × 23 + 4 × 2 2 + 2 − 34 = 16 + 16 + 2 − 34 = 34 − 34 = 0 13.x −3 x − 2 x2 − 5x + 6 x2 − 2 x − + −3 x + 6 −3 x + 6 + − 0 ∴ x 2 − 5 x + 6 = ( x − 2)( x − 3) 12. Find the values of m and n if the polynomial 2x3 +mx2 + nx – 14 has x – 1 and x + 2 [2] as its factors. 4.2 when polynomial x3 + 3x2 + 3x + 1 is divided by x + 1. Using remainder theorem find the remainder when f(x) is divided by g(x) [2] f(x) = x24 – x19 – 2 g(x) = x + 1 8. then the value of P(1) + P(-1) is (a) 0 (b) 1 (c) 2 (d) . Factorise 1 – a2 – b2 – 2ab [3] 11. Show that 5 is a zero of polynomial 2x3 – 7x2 – 16x + 5 [2] 7. Which of the following expression is a monomial (a) 3 + x (b) 4x3 (c) x6 + 2x2 + 2 (d) None of these A linear polynomial [1] [1] (a) May have one zero (b) has one and only one zero (c) May have two zero (d) May have more than one zero If P(x) = x3 – 1. Factorise (2x – 3y)3 + (3y – 4z)3 + (4z – 2x)3 25 [5] . [3] 1 1 Expand a − b + 1 3 2 2 12. 3. the remainder is (a) 1 (b) 0 (c) 8 (d) – 6 [1] [1] 5. 10. Verify each of the following identifies (i) x3 + y3 = (x + y) (x2 – xy + y2) (ii) x3 – y3 = (x – y) (x2 + xy + y2) [3] 13. Find K if x + 1 is a factor of P(x) = Kx2 – 2 x + 2 [2] 9. 2.CBSE TEST PAPER-02 CLASS . Factorise x2 + y – xy – x [1] 6.IX Mathematics (Polynomials) 1. x2 + y – xy – x x2 – x + y – xy = x2 – x – xy + y = x (x – 1) – y (x – 1) = (x – 1) (x – y) 6. (d) 5.(2) − m = 13 ⇒ m = −13 26 4. (b) .CBSE TEST PAPER-02 CLASS . x – 1 and x + 2 is factor of 2x3 + mx2 + nx – 14 x = 1.IX Mathematics (Polynomials) 1. x = -2 ∴ 2(1)3 + m(1) 2 + n(1) − 14 = 0 2 + m + n − 14 = 0 m + n − 12 = 0 m + n = 12 − − − − − (1) 2(2)3 + m(2)2 + n(2) − 14 = 0 16 + 4m + 2n − 14 = 0 4m + 2n + 2 = 0 4 m + 2 n = −2 2m + n = −1 − − − − − −(2) soving eq. When f(x) is divided by g(x) Then remainder f(-1) F(-1) = (-1)24 – (-1)19 – 2 = 1 – (-1) – 2 = 1+ 1 – 2 = 0 8. Here P ( x) Kx 2 − 2 x + 2 ∵ x + 1 is factor of P(x) ∴ P(−1) = 0 K (−1) 2 − 2(−1) + 2 = 0 K + 2 +2=0 K = −(2 + 2) 9.(1) and eq. (b) 3. (a) 2. Put x = 5 in 2x3 – 7x2 – 16 x + 15 2 × 53 − 7 × 52 − 16 × 5 + 5 = 250 − 175 − 80 + 5 = 255 − 255 = 0 ∴ x = 5 is zero of polynomial 2x3 – 7x2 – 16x + 5 7. H.H .S = x( x 2 + xy + y 2 ) − y ( x 2 + xy + y 2 ) = x3 + x 2 y + xy 2 − yx 2 − xy 2 − y 3 = x3 − y 3 = L.S . c = 4 z − 2 x then a + b + c = 2 x − 3y + 3y − 4z + 4z − 2x = 0 ∴ a3 + b3 + c3 = 3abc 3 3 3 ( 2 x − 3 y ) + ( 3 y − 4 z ) + ( 4 z − 2 x ) = 3 ( 2 x − 3 y )( 3 y − 4 x ) 2 ( 2 z − 2 x ) = 3 ( 2 x − 3 y )( 3 y − 4 x ) 2 ( 2 z − 2 x ) = 6 ( 2 x − 3 y )( 3 y − 4 x )( 2 z − x ) 27 .S .S . L.S .H .S . b = 3 y − 4 z . = R. Verified 13. Let a = 2 x − 3 y.H .H . Verified x3 − y 3 = ( x − y )( x 2 + xy + y 2 ) R.Put m = −13 in (1) − 13 + n = 12 n = 12 + 13 = 25 10.H .H .S ( x + y )( x 2 − xy + y 2 ) = x( x 2 − xy + y 2 ) + y ( x 2 − xy + y 2 ) = x3 − x 2 y + xy 2 + yx 2 − xy 2 + y 3 = x3 + y 3 = L.H . 1 1 a − b + 1 3 2 2 −1 −1 1 1 1 1 = a + − b + 12 + 2 × a × b + 2 × b ×1 + 2 × a × 1 2 3 3 2 2 3 2 2 a b ab 2b = + +1− − + a 4 9 3 3 2 12.S = R. (i) 2 x3 + y 3 = ( x + y )( x 2 − xy + y 2 ) Taking R. (ii) L. 1 − a 2 − b 2 − 2ab 1 − (a 2 + b 2 + 2ab) = 12 − (a + b) 2 = (1 + a + b)(1 − a − b) 11. Find the integral zeroes of the polynomial x3 + 3x2 – x – 3 [1] 6. Check whether 7+ 3x is a factor of 3x2 + 7x [2] 7.CBSE TEST PAPER-03 CLASS . Factorise: 12(y2 + 7y)2 – 8 (y2 + 7y) (2y – 1) – 15(2y – 1)2 [5] 28 . Factorise [2] 2 2 4 x −x− x 3 8. Without actually calculating the cubes. Find m and n it x – 1 and x – 2 exactly divide the polynomial x3 + mx2 – nx + 10 [2] 10. Find the value of (26)3 + (-15)3 + (11)3 [3] 13. Factorise (i) 64y3 + 125z3 (ii) 27m3 – 343n3 [3] 12.IX Mathematics (Polynomials) 1. Evaluate (101)2 by using suitable identify [2] 9. Using identify (a + b)3 = a3 + b3 + 3ab (a + b) drive the formula a3 + b3 = (a + b) [3] (a2 – ab + b2) 11. The value of K for which x – 1 is a factor of the polynomial 4x3 + 3x2 – 4x + K is (a) 0 (b) 3 (c) – 3 (d) 1 The factors 12x2 – x – 6 [1] [1] (a) (3x – 2) (4x + 3) (b) (12x + 1) (x – 6) (c) (12x – 1) (x + 6) (d) (3x + 2) (4x – 3) x3 + y3 + z3 – 3xyz is [1] (a) (x + y – z)3 (b) (x – y + z)3 (c) (x + y + z)3 – 3xyz (d) (x + y + z) (x2 + y2 + z2 – xy – yz – zx) The expended form of (x + y – z)2 is [1] (a) x2 + y2 + z2 + 2xy + 2yz + 2zx (b) x2 + y2 – z2 + 2xy – 2yz – 2xz (c) x2 + y2 + z2 + 2xy – 2yz – 2zx (d) x2 + y2 + z2 + 2xy + 2yx + 2xz 5. 3. 2. 4. x = 1 Zeros of polynomial -1. x = −1.IX Mathematics (Polynomials) 1. (d) 5. 1. (c) . -3. 3 2 4 x −x− 2 3 3 −4 × = −2 2 3 We factorise by splitting middle term −2 + 1 = −1 3 2 4 x − 2x + x − 2 3 3 4 4 = x x − + 1 x − 2 3 3 4 3 = x + 1 x − 3 2 29 4.CBSE TEST PAPER-03 CLASS . Given polynomial P( x) = x3 + 3 x 2 − x − 3 p( x) = x 2 ( x + 3) − 1( x + 3) = ( x + 3)( x 2 − 1) For zeros p ( x) = 0 ( x + 3)( x 2 − 1) = 0 ( x + 3( x + 1)( x − 1) = 0 x = −3. (c) 2. Let p ( x) = 3 x 2 + 7 x 7 + 3x is factor of p(x) Remainder = 0 7 Remainder = P − 3 2 7 7 = 3 − + 7 − 3 3 49 49 = 3× − 9 9 =0 Hence 7 + 3x is factor of p(x) 7. (d) 3. 6. (101)2 = (100 + 1) 2 (a + b) 2 = a 2 + 2ab + b 2 here a = 100. we get − m = −2 m=2 put m = 2 in eq. b = 1 (101)2 = (100 + 1) 2 = 100 2 + 2 ×100 ×1 + 12 = 10000 + 200 + 1 = 10201 9. (2) form (1). Solution given (a + b)3 = a 3 + b3 − 3ab(a + b) ∴ a 3 + b3 = (a + b)3 − 3ab(a + b) = (a + b)[(a + b) 2 − 3ab] = (a + b)[a 2 + b 2 + 2ab − 3ab] = (a + b)(a 2 + b2 − ab) = (a + b)(a 2 − ab + b2 ) 30 .8. (1). we get 2 − n = −11 − n = −11 − 2 + n = +13 n = 13 m=2 10. Let p ( x) = x3 + mx 2 − nx + 1 x – 1 and x – 2 exactly divide p(x) ∴ p (1) = 0 and p(2) = 0 p(1) = 13 + m ×12 − n ×1 + 10 = 0 1 + m − n + 10 = 0 m − n + 11 = 0 m − n = −1 − − − − − − − − − (1) p (2) = 23 + m × 2 2 − n × 2 + 10 = 0 8 + 4m − 2n + 10 = 0 4m − 2n = −18 2m − n = −9 − − − − {dividing by 2} − + + − − − − − − − − (2) subtracting eq. Solution (i) 64y3 + 125z3 (4 y )3 + (5 z )3 (4 y + 5 z )[(4 y ) 2 − 4 y × 5 z + (5 z ) 2 ] ∴ a 3 + b3 = (a + b)(a 2 − ab + b 2 ) = (4 y + 5 z )(16 y 2 − 20 yz + 5 z 2 ) (ii) 27m3 – 343n3 = (3m)3 − (7 n)3 = (3m − 7 n)[(3m) 2 − 3m × 7 n + (7 n)2 ] [∵ a 3 − b3 = (a − b)(a 2 + ab + b 2 )] (3m − 7 n)(9m 2 − 21mn + 49n 2 ) 12. c = -11 a + b + c = 26 – 15 – 11 = 0 Then a3 + b3 + c3 = 3abc (26)3 + (−15)3 + (−11)3 = 3 × 26 × −15 × −11 = 12870 13. Solution Let a = 26. b = -15. Let a = y 2 + 7 y. b = 2 y − 1 Then 12( y 2 + 7 y )2 − 8( y 2 + 7 y )(2 y − 1) − 15(2 y − 1)2 = 12a 2 − 8ab − 15b 2 = 12a 2 − 18ab + 10ab − 15b 2 = 6a (2a − 3b) + 5b(2a − 3b) = (2a − b)(6a + 5b) put a = y 2 + 7 y and b = 2 y − 1 = [2( y 2 + 7 y ) − 3(2 y − 1)][6( y 2 + 7 y ) + 5(2 y − 1) = [2 y 2 + 14 y − 6 y + 3][6 y 2 + 42 y + 10 y − 5] = (2 y 2 + 8 y + 3)(6 y 2 + 52 y − 5) 31 .11. [3] 11.CBSE TEST PAPER-04 CLASS . 2. 2. The value of 1023 is (a) 1061208 (c) 1820058 [1] (b) 1001208 (d) none of these (a-b)3 + (b-c)3 + (c-a)3 is equal to [1] (a) 3abc (b) 3(a-b) (b-c) (c-a) (c) 3a3b3bc3 (d) [a-(b+c)]3 The zeroes of the polynomial P(x) = x (x-2) (x+3) are (a) 0 (b) 0. Evaluate 993 [2] 7. if x-1 is factor of P(x) and P(x) = 3x2+kx+ 2 [2] 8. q = -1 (c) p = 2. 13. Prove that x2+6x+15 has no zero. Find the value of k. 2. 2 Expand x + 1 3 9. -3 (d) none of these If (x+1) and (x-1) are factors of Px3+x2-2x+9 then value of p and q are (a) p = -1. 3 (c) 0.5(x+y) + 2 [3] 12. 3. [3] 10. Find the values of m and n so that the polynomial x3-mx2-13x+n has x-1 and x+3 as factors. q = 1 (d) p = -2.IX Mathematics (Polynomials) 1. The volume of a cuboid is given by the expression 3x3-12x. q = -2 [1] [1] 5. Factorise 3 (x+y)2 . Factorise 8a3-b3-12a2b+6ab2 [2] 6. 4.z6+6x2y2z2 [5] 32 .find the possible [3] [2] 3 expressions for its dimensions. Factorise x6 + 3y6 . q = 2 (b) p = 2. Let polynomial be P( x) = x3 − mx 2 − 12 x + n x − 1 is factor of P(x) ∴ P(1) = 0 33 . (c) Ans04. Ans05.CBSE TEST PAPER-04 CLASS . x-1 is factor of P(x) P(1) = 0 ∴ 3 × 1+k × 1+ 2 = 0 3+k+ 2 = 0 k = -(3+ 2) Ans08. (a) Ans03. 993 = (100 − 1)3 We know that (a − b)3 = a 3 − 3a 2b + 3ab 2 − b3 Take a = 100. b = 1 (100-1)3 = 1003 − 3 × 1002 × 1 + 3 × 100 × 12 − 13 = 1000000-30000+300-1 = 1000300-30001 = 970299 Ans07. b =1 3 3 3 2 8 3 4 2x 2 2 2 3 x + 1 + x2 + x + 1 = x + 1 + 3 × x × 1 x + 1 = 3 3 3 3 3 3 27 Ans09. (b) 8a 3 − b3 − 12a 2b + 6ab 2 = (2a)3 − b3 − 6ab(2a − b) = (2a)3 − b3 − 3(2a ) b (2a − b) = (2a-b)3 = (2a-b) (2a-b) (2b-b) Ans06. (a + 3)3 = a 3 + b3 + 3ab(a + b) 2 a = x. (a) Ans02.IX Mathematics (Polynomials) [ANSWERS] Ans01. m = 3 Put m = 3 in eq (1) m = 3 and n = 15 Ans13. 3( x + y ) 2 − 5( x + y ) + 2 Let x + y = z 3z 2 − 5 z + 2 = 3 z 2 − 3 z − 2 z + 2 = 3z (z-1) -2 (z-1) = (3z-2) (z-1) Put z= x+y ∴ 3(x+y) 2 − 5( x + y ) + 2 = [3( x + y ) − 2] [x+y-1] = [3x+3y-2] [x+y-1] Ans12. zero. x6 + 8 y 6 − z 6 + 6 x2 y 2 z 2 = (x 2 )3 + (2 y 2 )3 + (− z 2 )3 − 3( x 2 ) (2y 2 ) (-z 2 ) = [x 2 + y 2 − z 2 ] [(x 2 )3 + (2 y 2 ) 2 + (− z 2 ) 2 − x 2 × 2 y 2 − 2 y 2 (− z 2 ) − x 2 × (− z )2 = (x 2 + 2 y 2 − z 2 ) (x 4 + 4 y 4 + z 4 − 2 x 2 y 2 + 2 y 2 z 2 + x 2 z 2 ) 34 . The volume of cuboid is given by 3x 3 − 12 x = 3 x( x 2 − 4) = 3x (x+2) (x-2) Dimensions of the cuboid are given by 3x. zero.x 2 6 x + 15 Ans10. Ans11. (2) from (1) 8m = 24. = x 2 + 2 × 3 x + 32 + 6 = (x+3) 2 + 6 ( x + 3) 2 is positive & 6 is positive ∴ (x+3) 2 + 6 has no. (x+2) and (x-2) P(1) = 13 − m × 12 − 13 × 1 + n = 0 = 1-m-13+n = 0 = -m+n =12 (1) x+3 is factor of P(x) ∴ P(-3) = 0 P(-3) = (3)3 − m(−3) 2 − 13 × (−3) + n = 0 = -27-9m+39+n = 0 = -9m+n 12 =0 (2) = -9m+n = -12 Subtracting eq. x 2 + 6 x + 15 has no. 11.2) (b) ( x + 2 2) (x+ 2) (c) ( x − 2 2) (x+ 2) (d) ( x − 2 2) (x. Find [3] the possible expressions for the dimensions of the cuboid. If x+y+x = 0. is (a) 3 3. If y 3 + ay 2 + by + 6 is divisible by y – 2 and leaves remainder 3 when divided by [3] y – 3. [1] (b) 2 (c) 1 (d) 0 Factors of x2 + 3 2 x + 4 are [1] (a) ( x + 2 2) (x. Using factor theorem check whether g(x) is factor of p(x) [2] P(x) = x3 − 4 x 2 + x + 6. Factorise 27 x3 + y 3 + z 3 − 9 xyz [2] 6. (c) 3xyz (d) 0 The value of (x-a)3 + (x-b)3 + (x-c)3 – 3 (x-a) (x-b) (x-c) when a + b + c= 3x. 13. find the values of a and b. Factorise x6 – 64 [3] 12.2) The degree of constant function is (a) 1 [1] (b) 2 [1] (c) 3 (d) 0 5. g(x) = x-3 8. The volume of a cuboid is given by the algebraic expression ky2-6ky+8k. 2 Expand x − y 3 3 [2] 9.CBSE TEST PAPER-05 CLASS .IX Mathematics (Polynomials) 1. 1 −5 3 3 2 3 Factorise: ( 2 x + 5 y ) + y + z − z + x 27 4 4 3 3 3 35 3 [5] . 2. Evaluate 105 × 95 [2] 7. Using remainder theorem factorise x3 − 3x 2 − x + 3 [3] 10. then x3 + y 3 + z 3 is (a) xyz (b) 2xyz 4. (b) Ans05. X-3=0 Put x=3 in p (x) P (3) =33 -4 × 32 +3+6 = 27+9 -4 × 9 =36-36 =0 Remainder =0 ∴ By factor theorem g(x) is factor of P (X) Ans08.IX Mathematics (Polynomials) [ANSWERS] Ans01. Given g(x) =X-3. (b) Ans02. X3 -3X2 – X+3 Coefficient of X3 is 1 Constant =3 3×1 = 3 ∴ We can Put X= ± 3 and ( X ) and check Put= X=1 36 Ans04.( y )3 − 3 X × Y ( X − Y ) 3 3 3 3 8 3 2 =X3 − y − 2 XY ( x − Y ) 27 3 8 3 4 =X3 y − 2x 2Y + Xy 2 27 3 Ans09. (c) Ans03.CBSE TEST PAPER-05 CLASS . x−( 2 3 y) 3 ∴ (a − b)3 = a 3 -b3 -3ab(a-b) Hence a=X . 27X3 +Y3 +Z3-9XYZ = (3X) 3 +Y3 +Z3 -3(3X) (Y) (Z) = [3X+Y+Z] [(3X) 2 +Y2 +Z2 -3XY –YZ-3X2] = (3X+Y+Z) (9X2+Y2+Z2 – 3XY –YZ -3X2) Ans06. (d) . b= 2 y 3 2 2 2 2 ∴ (x.y )3 = x 3 . 105 × 95 = (100+5) (100-5) =1002-52 [ ∴ (a+b) (a-b) =a2-b2] =10000 -25 = 9975 Ans07. 1+3 1–3–1+3=0 Remainder =0 ∵ X − 1 is factor of x3 − 3 x 2 − x + 3 x 2 -2x-3 x − 1 X 3 − 3x 2 x3 − x2 -2x 2 .13 – 3 × 12 . Let P(y) = y3 + ay2 + by + 6 P(y) is divisible by y – 2 Then P (2) = 0 23 + a × 2 2 + b × 2 + 6 = 0 8 + 4a + 2b + 6 = 0 4a+2b = -14 2a+b = 7 If P (y) is divided by y-3 remainder is 3 ∴ P (3) 33 + a × 32 + b × 3 + 6 = 3 9a + 3b = -30 3a + b = -10 (ii) eq (i) .eq (ii) -a = 3 and Put a = -3 in eq (i) (i) a = -3 2 × -3 + b = -7 -6 + b = -7 b = -7 + 6 b = -1 37 .x -2x 2 +2x -3x + 3 -3x + 3 0 ∴ x 3 − 3 x 2 x + 3 = ( x − 1) (x 2 − 2 x − 3) = (X-1) (x 2 − 3 x + x − 3) = (X-1) [X(X-3) +1 (X-3)] = (X-1) (X-3) (X+1) Ans10. Given volume of cuboid ky 2 − 6ky + 8k = k [y 2 − 6 y + 8] k [ y 2 − 4 y − 2 y + 8] = k [y (y-4) -2 (y-4)] = k (y-2) (y-4) Thus dimension of cuboid k. (y-2) and (y-4) Ans13. 2 1 3 3 −5 (2 x + 5 y )3 + y + z − z + x 3 27 4 4 3 3 5 −3 2 1 −5 y + z z − x = 3 (2 x + 5 y ) 4 4 3 3 3 5 3 2 −5 = − (2 x + 5 y ) y + z z + x 4 4 3 3 −20 y + 15 z 9 z + 8 x = − (2 x + 5 y ) 12 12 1 (2 x + 5 y ) (20 y − 9 z ) (9 z + 8 x) = 144 38 3 . Given expression can be written as 3 3 3 3 2 1 5 3 (2 x + 5 y ) + − 3 y + 4 z + − 4 z − 3 x 1 5 −5 Let (2 x + 5 y ) = a.Ans11. y+ z =b 3 3 4 −3 2 and z− x=C 4 3 2 5 5 5 3 2 a+b+c = x+ y− y+ z − z − x = 0 3 3 3 4 4 3 3 3 3 ∴ a + b + c = 3abc 3 Thus. x 6 − 64 = (x 2 )3 .(2 2 )3 = (x 2 -22 ) [x 4 +4x 2 +16] = (x+2) (x-2) (x 4 +4x 2 +16) Ans12. If a point C lies between two points A and B such that AC=BC. 39 [3] . – Distinct points. If A.CBSE TEST PAPER-01 CLASS . The edges of a plane surface are. In fig AB =CD prove that AC=BD [2] 8. 2 Prove that An equilateral triangle can be constructed on any given line segment. 1 AB Explain by drawing the figure. [1] [1] [1] (b) two (d) four. Given four points such that No three of them are collinear. then prove that AC [3] = 10. How would you write Euclid’s fifth postulate [2] 9. then prove that. then prove that AB=CD [2] 7. The wholes are – (a) equal (b) not equal (c) Doubled (d) none of these. (a) Line (b) points. 3. [1] [2] AC. 5. If equals are added to equals. B and C are three points on a line and B is between A and C. (c) Angles (d) planes. 2. (a) 2 lines (b) 4 lines (c) 6 lines (d) 5 lines One and only one line passes through. (a) one (c) Three 4. then there exists.BC = AB 6. In fig AC = BD.IX Mathematics (Introduction to Euclid’s geometry) 1. [3] 12. (iii) A terminated line can be produced indefinitely on both sides. then their radii are equal. If AB= PQ and PQ =XY then prove that AB = XY.11. Which of the following statements are true and which are false Explain. then AB=XY 40 [5] . Explain by drawing the fig. (I) Only one line pass. Through a single point (ii) There are an infinite number of line which passes through two distinct points. Give a definition for each of the following are there other terms which need to [3] be defined first? What are they? 13. (iv) If two circles are equal. (v) In fig AB=PQ and PQ=XY. Solution: AC=BD [given] Also BC=BC [Things which coincide with one another are equal to one another] ∴ AC-BC = BD-BC AB=CD [If equals are subtracted from equal then remaining are equal] Ans07. (a) Ans05. (b) Ans04. there exists a unique line m passing through p and parallel to L. 41 .IX Mathematics (Introduction to Euclid’s geometry) [ANSWERS] Ans01.CBSE TEST PAPER-01 CLASS . Solution. (A) Ans02. For every line L and for every point p not lying on L. Solution: AB=CD [given] Also BC=BC [things which coincide with one another are equal to one another] ∴ AB+BC=CD+BC [it equals are added from equal then remaining are equal] Ans08. in fig AB coin cider with AC –BC So AC-BC=AB [things which coincide with one another are equal to one another] Ans06. (C) Ans03. 42 . (ii) Two lines are perpendicular to each other if they make a right angle. (i) AB=PQ and PQ =XY AB =XY [things which are equal to the same things are equal to one another] Two lines in a plane are parallel if they are equidistant. Angle and right angle are two terms to be defined before the above definition. Solution: using Euclid’s third postulate. distance should be defined before the above definition.Ans09. Solution: In fig point C is between A and B AC+CB = AB [things which coincide with one another are equal to one another] But AC=BC ∴ AC+AC=AB 2AC=AB 1 AC = AB 2 Ans10. Given Ans12. Solution. Now. Draw circles with centre d at the points A and B and radius equal to the length of the segment AB cutting each other at C. AB=AC [radii of the same circle] BC=AB [radii of the same circle] ∴ AC=BC Ans11. Ans13. (i) False: An infinite number of lines can pass through a single point. (ii) False: There is a unique line joining two distinct points. (iii) True: There days a terminated line is called a line segment but Euclid called it a terminated line. So a line segment can be produced indefinitely on both sides and line segment AB when produced on both sides becomes line AB. (iv) True: In equal circles, when the region bounded by one circle in superimposed on the other then the circles coincide i.e. their centre and boundaries coincide. As such their radii are equal. (v) True: by Euclid’s axiom, things which are equal to the same things are equal to one another. 43 CBSE TEST PAPER-01 CLASS - X Mathematics (Lines and angles) 1. 2. (ii) 90 (ii) 90 (iii) 5. [1] (iii) 180 (iv) none of these [1] In fig if x= 30 then y= (i) 4. (ii) between 0 and 90 (iv) between 180 and 360 The sum of angle of a triangle is (i) 0 3. [1] Measurement of reflex angle is (i) 90 (iii) between 90 and 180 150 (iv) 180 210 It two lines intersect each other then [1] (i) vertically opposite are equal (ii) corresponding angle are equal (iii) alternate interior angle are equal (iv) none of these In fig lines x y and m n intersect at 0 If [2] ∠ poy =90 and a:b =2:3 find c 6. In fig find the volume of x and 7. What value of x would make AOB a line if ∠ AOC=4x and ∠ BOC=6x+30° 44 y then Show that ABIICD [2] [2] 8. In fig POQ is a line. Ray OR is perpendicular [3] to line PQ. OS is another ray lying between rays OP and OR. Prove that 1 ∠ ROS= (∠ QOS- ∠POS) 2 9. In fig lines P and R intersected at 0, if x = 45 find x, y and u 10. Prove that sum of three angles of a triangle is 180 11. It is given that ∠ XYZ =64 and X Y is produced to point P, draw a fig from the given information. If ray Y Q bisects ∠ Z Y P, find ∠ XYQ and reflex ∠ QYP. 12. In fig if PQIIST, ∠ PQR = 110 and ∠ RST= 130 find ∠ QRS. 13. In fig the side AB and AC of △ A B C Are produced to point E And D respectively. If bisector BO And CO of ∠ CBE And ∠ BCD respectively meet at point O, then prove 1 that ∠ BOC = 90 − ∠ BAC 2 45 [2] [3] CBSE TEST PAPER-01 CLASS - X Mathematics (Lines and angles) [ANSWERS] Ans01 (iv) Ans02. (iii) Ans03. (iii) Ans04. (i) Ans05. Given in fig. ∠ POY= 90 a: b: 2: 3 Let a=2x and b =3x a + b + ∠ POY= 180 (∵ xoy is a line ) 2x+3x+ 90 = 180 5x= 180 -90 5x= 90 90 = 18 5 ∴ a= 36 , b=54 x= MoN is a line. b+C= 180 54 +C = 180 C = 180 − 54 = 126 Ans C = 126 Ans06. 50 + x = 180 ( by liner pair ) x = 180 − 50 x = 130 y = 130 (∵ vertically opposite angles are equal ) 46 R. X = 45 ∴ Z=45 X+y=180 ∵ vertically opposite angles are equal ( By linear pair ) 45 + y =180 y=180 − 45 y=135 y=u ( vertically opposite angles ) u=135 Ans10.S Hence proved = Ans09. given ∠AOC=4x And ∠ BOC =6x+30 ∠ AOC+∠ BOC =180 4 x + 6 x + 30 = 180 ( By linear pair ) 10 x = 180 − 30 10 x = 150 = x = 15 Ans08.Ans07. given . △ABC To prove : ∠A + ∠B + ∠C = 180 construction : through A draw XYIIBC Proof : ∵ XYIIBC 47 .S = 1 ( ∠QOS − ∠POS ) 2 1 ( ∠ ROS+∠ QOR -∠ POS) 2 1 → (1) = ( ∠ ROS +90 − ∠ POS) 2 ∵ ∠POS+ ∠ROS=90 ∴ By 1 1 = ( ROS + ∠POS + ∠ROS − ∠POS ) 2 1 = × 2 ∠ROS=∠ROS 2 =∠ L.H.H . ∠4. Through point R Draw line KLIIST ∵ PQIIST STIIKL ∴ PQIIKL ∵ PQIIKL ∴ ∠PQR+∠1=180 ( Sum of int erior angle on the same side of 110 + ∠1 = 180 Similarly ∠1=70 ∠2+∠RST=180 ∠2+130 = 180 ∠2=50 48 transverral is 180 ) .∴ ∠2=∠4 → (1) ∵ Alternate interior angle And ∠3=∠5 → ( 2 ) Angles are equal Adding eq (1) And eq (2) ∠2+∠3=∠4+∠5 Adding both sides ∠1 ∠1+∠2+∠3=∠1+∠4+∠5 ∠1+∠2+∠3=180 (∵ ∠1. and ∠5 forms line ) ∠A+∠B+∠C=180 ( linear pair ) Ans11. ∠XYZ + ∠PYZ = 180 ⇒ 64 + ∠PYZ = 180 ( given∠XYZ = 64 ) ∠PYZ=180 − 64 ∠pyz =116 1 116 ∠ZYQ= ∠ZYP = = 58 2 2 ∠XYQ= ∠XYZ+∠ZYQ =64 + 58 = 122 Also reflex ∠QYP= ∠XYQ + struight ∠XYP =122 + 180 =302 Ans12. Solution : Ray BO bisects ∠CBE 1 ∠CBE 2 1 = (1 8 0 − y ) (∵ ∠ C B E + y = 1 8 0 ) 2 y =90 − → (1 ) 2 S im ila r ly r a y C o b i s e c ts ∠ B C D ∴ ∠CBO = 1 ∠BCD 2 1 = (1 8 0 − Z 2 Z =90 − 2 ∠BCO= In △ BOC ∠BOC+∠BCO+∠CBO=180 Z Y + 90 − = 180 2 2 1 ∠BOC= (Y + Z ) 2 X+Y+Z=180 ∠BOC+90 − But Y+Z=180 − X 1 X 180 − X ) = 90 − ( 2 2 1 ∠BOC= 90 − ∠BAC 2 ∠BOC= 49 ) .∠1+∠2+∠3=180 70 + 50 + ∠3 = 180 ∠3=180 − 120 ∠3=60 ∠QRS=60 Ans13. Find the angles. 7. sides QP and RQ of ∆PQR are produced to points S and T respectively. Find the other two angles of the triangle. If two angles of a triangle is 30o and 45o what is measure of third angle (1) 3. The measure of Complementry angle of 63o is (a) 30o (b) 36o o (d) none of there (c) 27 (1) 2. one is twice the smallest and another is three (2) times the smallest. (a) 95o (b) 90o (c) 60o (d) 105o The measurement of Complete angle is (a) 0o (b) 90o (c) 180o (d) 360o The measurement of sum of linear pair is (a) 180o (b) 90o (c) 270o (d) 360o The exterior angle of a triangle is 110o and one of the interior opposite angle is (1) (1) (1) 35o. 4. If ∠SPR = 135o and ∠PQT=110o . 5. 6. In fig. Of the three angles of a triangle. 50 (2) . the (2) triangle is right angled. 8.X Mathematics (Lines and angles) 1. find ∠PRQ . Prove that if one angle of a triangle is equal to the sum of other two angles.CBSE TEST PAPER-02 CLASS . 9. (5) 13. AB // CD. Determine ∠a . then prove that ∠MPN = 1 ( ∠Q − ∠R ) 2 51 . CE is drawn parallel to AB show (3) that ∠ACD = ∠A + ∠B. Also prove that ∠A + ∠B + ∠C = 180o . In fig the bisector of ∠ABC and ∠BCA intersect each other at point O prove 1 that ∠BCO = 90o + ∠A 2 (2) 10. In fig. In given fig. The side BC of ∆ABC is produced from ray BD. ∠Q > ∠R and M is a point on QR such that PM is the bisector of ∠QPR if (5) the perpendicular from P on QR meets QR at N. 12. then each pair of (3) alternate interior angles is equal. 11. Prove that if a transversal intersect two parallel lines. Proof: ∠A + ∠B + ∠C = 180o ∠A + ∠C = ∠B From (1) and (2) ∠B + ∠B = 180o (1) [Sum of three angle of a ∆ ABC is 180o] (2) 2∠B=180o ∠B=90o 52 . ∴ ∠ACD=∠A+∠B 110=∠A+35o ∠A=110o − 35o ∠A=75o ∠C=180-(∠A+∠B) ∠C=180-(75o + 35o ) ∠C=70o Ans06. Let the smallest angle be xo Then other two angles are 2xo and 3xo xo+2xo+3xo=180o [sum of three angle of a triangle is 180o] 6xo +=180o 180 x= 6 =30o angles are 30o. The exterior angle of a triangle is equal to the sum of interior opposite angles. Given in ∆ABC ∠B=∠A+∠C To prove: ∆ABC is right angled. (c) Ans02. (d) Ans03. (a) Ans05. (d) Ans04.CBSE TEST PAPER-02 CLASS .X Mathematics (Lines and angles) [ANSWERS] Ans01. 60o and 90o Ans07. A ∆ABC such that the bisectors of ∠ABC and ∠BCA meet at a point O 1 To Prove ∠BOC = 90o + ∠A 2 Given Proof: In ∆BOC ∠1 + ∠2 + ∠BOC = 180o In ∆ABC ∠A + ∠B + ∠C = 180 (1) ∠A + 2∠1 + 2∠2 = 180o ⇒ ∠A + ∠1 + ∠2 =90o 2 ∠1+∠2=90o − Ans10. ∠PQT + ∠PQR = 180o 110o + ∠PQR = 180o ∠PQR = 180o − 110o ∠PQR=70o ∠SPR=∠PQR+∠PRQ Also 135o = 70o + ∠PRQ ∠PRQ=135o − 70o ∠PRQ=65o Ans09. we get ∠A + ∠B + ∠C = ∠C + ∠ACD ∠A + ∠B + ∠C = 180o 53 .Ans08. [BO and CO bisects ∠B and ∠C ] [Divide by 2] ∠A 2 AB // CE and Ac intersect them (1) [Alternate interior angles] ∠1 = ∠4 Also AB//CE and BD intersect them (2) [Corresponding angles] ∠2 = ∠5 Adding eq (1) and eq (2) ∵ ∠1 + ∠2 = ∠4 + ∠5 ∠A + ∠B = ∠ACD Adding ∠C on both sides. Given: line AB//CD intersected by transversal PQ To Prove: (i) ∠2 = ∠5 (ii) ∠3 = ∠4 Proof: (i) [Vertically Opposite angle] ∠1 = ∠2 (ii) [Corresponding angles] ∠1 = ∠5 By (i) and (ii) ∠2 = ∠5 Similarly ∠3 = ∠4 Hence Proved Ans12.Ans11. ∵ PM bi sec ts ∠QPR ∵ ∠QPM = MPR In ∆PQN ∠PQN + ∠QPN + ∠PNQ = 180o ∠PQN+∠QPN+90 = 180 o ⇒ [By angle sum property of ∆] o ∠PQN = 90o − ∠QPN (i) In ∆ PNR ∠PRN + ∠NPR + ∠PNR = 180o ∠PRN+∠NPR+90 = 180 o [By angle sum property] o ∠PRN=90o − ∠NPR Subtracting eq (2) from (1) (2) ∠PQN -∠PRN=(90o − ∠QPN ) − ( 90° − ∠NPQ ) 54 . Through O draw a line l parallel to both AB and CD Clearly ∠a = ∠1 + ∠2 ∠1 = 38o ∠2 = 55o [Alternate interior angles] ∠a = 55 + 38 o o ∠a = 93o Ans13. = ∠NPR -∠QPN =(∠NPM+∠MPR)-(∠QPM-∠NPM) =2∠NPM+(∠MPR-∠QPM) =2∠NPM+(∠MPR-∠MPR) [ ∵ PM bisects ∠QPM ∴∠QPM = ∠MPR] Hence ∠PQN − ∠PRN = 2∠NPM ∠Q − ∠R = 2∠NPM 1 or ∠NPM= [∠Q − ∠R] 2 1 or ∠MPN= [∠Q − ∠R] 2 Hence Proved 55 . 350 (b) 700 . 56 [1] [2] . The complement of ( 90 − a ) is (b) ( 90 + 2a ) (a) .1100 Given two distinct point P and Q in the interior of ∠ABC . 2. [1] 0 0 0 (d) a 0 The number of angles formed by a transversal with a pair of lines is (a) 6 (b) 3 (c) 8 (d) 4 In the given figure ∠POR and ∠QOR from a linear pair if a – b = 800 . Find the value of ‘a’ and ‘b’.a 0 (c) ( 90 − a ) 4. 5. then AB will be [1] [1] (a) in the interior of ∠ABC (b) in the interior of ∠ABC (c) on the ∠ABC (d) on the both sides of BA 3. 650 (d) 700 . The difference of two complementary angles is 400 .IX Mathematics (Lines and angle) 1.CBSE TEST PAPER-03 CLASS . 300 (c) 250 . The angles are (a) 650 . ∠APQ = 500 and ∠PRD = 127 0 find x and y . In ∆ ABC ∠B = 450 . If ray OC stands on a line AB such that ∠AOC = ∠BOC . find [3] 57 . In the given figure show that AB//EF [2] 8. then stands that [2] ∠AOC = 900 7. In the given figure ∆ ABC is right angled at A. [2] 9. AD is drawn perpendicular to BC. ∠C = 550 and bi sec tor ∠A meets BC at a point D.6. In figure if AB//CD. Prove that ∠BAD = ∠ACB [3] 10. (a) ∠BOD (b) ∠AOD (c) ∠AOC (d) ∠BOC The side BC of a ∆ ABC is produced to D. prove that ∠ ABC+ ∠ ACD=2 ∠ALC In fig M and N are two plane mirrors perpendicular to each other. > > 13. Find the value of x hence find 12. 58 [3] .∠ADB and ∠ADC 11. It ∠BOD = x 0 and ∠AOD = (45 − x)0 . prove that the incident ray CA is parallel to reflected ray BD. In figure two straight lines AB and CD intersect at a point O. the bisector of ∠ A meets B in L as shown if fig. q (1) And (2)] 0 Put a=1300 in eq (1) 1300 + b = 1800 b=1800 − 1300 = 500 Ans06. (D) Ans04.IX Mathematics (Lines and angle) [ANSWERS] Ans01. a + b = 1800 → (1) [By lineas pair] a-b=1800 2a= 260 a=130 0 → (2) [ Adding e. (C) Ans03.CBSE TEST PAPER-03 CLASS . ∠ BCD=∠ BCE+∠ ECD = 360 + 300 = 660 = ∠ ABC ∴ AB ∏ CD [A lte rn a te in te rio r a n g le s a re e q u a l 59 . (C) Ans05. (A) Ans02. ∠AOC = ∠BOC [Given] ∠AOC+∠BOC=1800 [BY lines pair] ∠AOC+∠AOC=1800 2∠AOC=1800 ∠AOC=900 Ans07. AB ∏ CD and PQ is a transversal ∠ APQ =∠PQD [ Pair of alternate angles] 500 = X Also AB ∏ CD and PR is a transversal ∠APR=∠PRD 500 + Y = 127 0 Y= 127 0 − 500 = 77 0 Ans09.Again ∠ECD=300 and ∠FEC=1500 ∴ ∠ECD+∠FEC=300 + 1500 = 1800 Hence EF ∏ CD [Sum of consecutive interior Angle is 1800 ] AB ∏ CDand CD ∏ EF then AB ∏ EF Ans08. ∵ AD ⊥ BC ∴∠ADB = ∠ADC = 900 from ∆ABD ∠ABD+∠BAD+∠ADB=1800 ∠ABD+∠BAD+900 = 1800 ∠ABD+∠BAD=900 ∠BAD=900 − ∠ABD → (1) But ∠A+∠B+∠C=1800 in ∆ABC ∠B+∠C=900 ∴ ∠A=900 ∠C=900 − ∠B → (2) from (1) and (2) ∠BAD=∠C ∠ BAD=∠ACB Hence proved Ans10. In ∆ABC ∠A+∠B+∠C=1800 [Sum of three angle of a ∆is 1800 ] 60 . ∠ADB = ∠AOD + ∠DOB BYlinear pair 1800 = 4 x − 5 + x 1800 + 5 = 5 x 5x=185 185 x= = 37 0 5 ∴ ∠AOD=4x-5=4 × 37-5=148-5 =1430 ∠BOC=1430 ∠BOD=x=37 ∴ ∠AOD and ∠BOC 0 verti cally copposite Angles ∠BOD = ∠AOC = 37° 61 . We have ∠1+∠B+∠ADB=1800 ⇒ 400 + 450 + ∠ADB = 1800 ⇒ ∠ADB=1800 − 850 = 950 Also ∠ADB+∠ADC=1800 950 + ∠ADC = 1800 ∠ADC=1800 − 950 = 850 Ans ∠ADB=950 and ∠ADC=850 Ans11.⇒ ∠A+450 + 55 = 1800 0 ∠A=1800 − 1000 = 800 ∴ AD bisects ∠A 1 1 ∠1=∠2= ∠A = × 800 = 400 2 2 Now in ∆ ADB. > 62 .Ans12. In ∆ABC we have ∠ACD=∠B+∠A → (1) [ Exterior angle property] ⇒ ∠ACD=∠B+2L1 [ ∴ A∠ is the bisector of ∠A =2L1] In ∆AB∠ ∠ALC=∠B+∠BA∠ [Exterior angle property] ∠A∠C=∠B+∠1 ⇒ 2∠A LC = 2∠B+2L1 → (2) Subtracting (1) from (2) 2∠ALC-∠ACD=∠B 2∠ALC=∠B+∠ACD ∠ACD+∠ABC = 2∠ALC Draw AP ⊥ M and BQ ⊥ N ∴ BQ ⊥ N and AP ⊥ M and M ⊥ N ∴ ∠BOA = 900 ⇒ BQ ⊥ AP In ∆ BOA ∠2+∠3+∠BOA=1800 [ By angle Sum property ] ⇒ ∠2+∠3+900 = 1800 ∴ ∠2+∠3=900 Also ∠1=∠2 and ∠4=∠3 ⇒ ∠1+∠4=∠2+∠3=900 ∴ (∠1+∠4)+ (∠2+∠3) =900 + 900 = 1800 ⇒ (∠1+∠2)+(∠3+∠4)=1800 or ∠CAB+ ∠DBA=1800 ∴ CA ∏ BD [ By sum of interior angle of same side of transversal] > > Ans13. CBSE TEST PAPER-04 CLASS - IX Mathematics (Lines and angle) 1. In fig L1 ∏ L2 And ∠1 = 520 the measure of ∠ 2 is. [1] (A) 380 (B) 1280 (C) 520 (D) 480 2. In fig x= 300 the value of Y is [1] (A) 100 (B) 400 (C) 360 (D) 450 3. Which of the following pairs of angles are complementary angle? [1] (A) 250 , 650 (B) 700 ,1100 (C) 300 , 700 (D) 32.10 , 47.90 4. In fig the measure of ∠ 1 is. [1] (A) 1580 (B) 1380 (C) 420 (D) 480 5. Prove that if two lines intersect each other then vertically opposite angler are [2] equal. 6. The measure of an angle is twice the measure of supplementary angle. Find its measure. 63 [2] 7. In fig ∠ PQR = ∠ PRQ. Then prove that ∠ PQS= ∠ PRT. [2] 8. In the given fig ∠ AOC = ∠ ACO and ∠ BOD = ∠ BDO prove that AC DB [2] 9. In fig lines XY and MN intersect at O If ∠ POY= 90 and a:b=2:3 find ∠ C [3] 10. In fig PT is the bisector of ∠ QPR in ∆ PQR and PS ⊥ QR, find the value of x [3] 64 11. The sides BA and DC of a quadrilateral ABCD are produced as shown in fig show [3] that ∠ X+ ∠ Y = ∠ a+ ∠ b 12. In the BO and CO are Bisectors of ∠ B and ∠ C of ∆ ABC, show that ∠ BOC= 90 + 13. [3] 1 ∠ A. 2 In fig two straight lines PQ and RS intersect each other at o, if ∠ POT= 750 Find the values of a,b and c 65 [3] CBSE TEST PAPER-04 CLASS - IX Mathematics (Lines and angle) [ANSWERS] Ans01. (b) Ans02. (b) Ans03. ( a ) Ans04. (c) Ans05. Given: AB and CD are two lines intersect each other at O. To prove: (i) ∠1 = ∠2 and (ii) ∠3 = ∠4 Proof: ∠1 + ∠4 = 1800 Ans06. → (i ) [ Bylinearpair ] ∠4 + ∠2 = 1800 → ∠1 + ∠4 = ∠4 + ∠2 ∠1 = ∠2 Similarly, ∠3 = ∠4 (ii ) [ Bylinearpair ] [ By eq (i ) and (ii )] Let the measure be x 0 Then its supplement is 1800 − x 0 According to question x 0 = 2 (1800 − x 0 ) x 0 = 3600 − 2 x 0 3 x = 3600 x = 1200 The measure of the angle is 1200 . 66 ∠AOC = ∠BOD [ given] [vertically opposite angles ] ∠AOC = ∠BOD and ∠BOD = ∠BDO ∠ACO = ∠BDO ⇒ ∴ AC BD Ans09. ∴∠C = ∠XON = ∠MOY [vertically opposite angle] = ∠a + ∠POY But . ∠PQR = ∠PRQ ∠PQS = ∠PRT ∴ Ans08. ∠POY = 900 ∴ ∠C = ∠a + 900 → (1) Also. [Given] ∠AOC = ∠ACO and ∠BOD = ∠BDO But . [ By alternate int erior angle property ] Lines XY and MN intersect at O.Ans07. a:b = 2:3 2 a = × 900 5 = 360 [Given] → From (1) and (2) we get ∠C = 360 + 900 = 1260 67 (2) . ∠PQS + ∠PQR = ∠PRQ + ∠PRT [ Bylinearpair ] But . ∠POX = 1800 − ∠POY = 1800 − 900 = 900 ∴ a + b = 900 But . ∠ABC + ACB + ∠A = 1800 ∴ ∠ABC + ACB = 1800 − ∠A 1 1 [ ∠ABC + ACB ] = 900 − ∠A 2 2 From (1) and (2) we get 1 ∠1 + ∠2 = 900 − ∠A 2 68 → (2) → (3) . and ∴ 1 ∠1 = ∠ABC 2 1 ∠2 = ∠ACB 2 1 ∠1 + ∠2 = ( ∠ABC + ∠ACB ) 2 → (1) But .Ans10. = 100 Join BD In∆ABD ∠b = ∠ABD + ∠BDA [ Exterior angle theorem] In∆CBD ∠a = ∠CBD + ∠BDC ∠a + ∠b = ∠CBD + ∠BDC + ∠ABD + ∠BDA = ( ∠CBD + ∠ABD ) + ( ∠BDC + ∠BDA ) = ∠x + ∠y ∠a + ∠b = ∠x + ∠y Ans12. ∠QPR + ∠Q + ∠R = 1800 [ Angle sum property of ∆] ∠QPR = 1800 − 500 − 300 = 1000 1 ∠QPT = ∠QPR 2 1 = × 1000 = 500 2 ∠Q + ∠QPS = ∠PST = 90 [ Exterior angle theorem] 0 ∠QPS = 900 − ∠Q = 900 − 500 = 400 x = ∠QPT − ∠QPS = 500 − 400 Ans11. a = 84 . ∠BOC + ∠1 + ∠2 = 1800 [ Angle of a ∆] ∠BOC = 1800 − ( ∠1 + ∠2 ) 1 = 1800 − 900 − ∠A 2 1 = 900 + ∠A 2 Ans13. PQ intersect RS at O ∴ ∠QOS = ∠POR a = 4b Also. ∠QOR and ∠QOS form a linear pair ∴ a + 2c = 1800 u sin g (2) 840 + 2c = 1800 2c = 1800 − 840 2c = 960 960 c= = 480 2 0 Hence.But . [vertically opposite angles ] → (1) a + b + 750 = 1800 ∴ a + b = 180 − 75 0 [∵ POQ is a straight lines ] 0 = 1050 using (1) 4b + b = 1050 5b = 1050 or 105 = 210 5 a = 4b a = 4 × 21 a = 84 b= ∴ Again. b = 210 and c = 480 69 . In figure if lines PQ and RS intersect at ∠PRT = 400 .IX Mathematics (Lines and angle) 1. 4. ∠RPT = 950 and ∠TSQ = 750 . 2. One angle is five times its supplement. The measure of ∠8 is [1] (a) 1200 (b) 600 (c) 300 (d) 450 5. 750 (b) 300 . Such that [2] . (b) The ray AB is the same as the ray BA.1500 (c) 360 . 70 point T. (d) Two lines are coincident if they have only one point in common.1440 (d) 1600 . In figure the measure of ∠a is (a) 300 (c) 150 [1] (b) 150 (d) 500 0 The correct statement is- [1] (a) A line segment has one end point only.CBSE TEST PAPER-05 CLASS . find ∠SQT . 3. The angles are- (a) 150 . 400 [1] In figure if m n and ∠1: ∠2 = 1: 2. (c) Three points are collinear if all of them lie on a line. find ∠ROT . In figure ray OS stands on a line POQ. ∠TQR = 400 and ∠SPR = 500 find x and y. if QT ⊥ PR. In figure lines PQ and RS intersect each other at point O. ray OR and ray OT are angle bisector of ∠POS and ∠SOQ respectively. [3] 10. 8. If ∠POS = x. [2] Find all the angles. In figure sides QP and RQ of ∆PQR are produced to points S and T respectively [2] if ∠SPR = 1350 and ∠PQT = 1100 . If a transversal intersects two lines such that the bisectors of a pair of [3] 71 . If ∠POR : ∠ROQ = 5 : 7 . find ∠PRQ. In figure. 9.6. [2] 7. corresponding angles are parallel. If the bisectors of ∠PQR and ∠PRS meet at point T. An incident ray [3] AB striker the mirror PQ at B. 11. 72 [5] . then prove that the two lines are parallel. Then prove that ∠QRT = 12. [3] 1 ∠QPR 2 In figure PQ and RS are two mirror placed parallel to each other. In figure the sides QR of ∆PQR is produced to a point S. If ray YQ bisects ∠ZYP. Find ∠XYQ and reflex ∠QYP . Prove that AB CD. the reflected ray moves along the path BC and strikes the mirror RS at C and again reflects back along CD. Draw a figure from the given information. 13. It is given that ∠XYZ = 640 and XY is produced to point P. (C) Ans03. (B) Solution. (A) Ans05. Ans02. In ∆TQR 90 + 400 + x = 1800 [ Angle sum property of ∆] ∴ x =500 Now. Y=∠SPR+X ∴ Y= 300 + 500 = 800 Ans07. In ∆PRT ∠P+∠R+∠1=1800 [ By angle sum property ] 950 + 400 + ∠1 = 1800 ∠1=1800 − 1350 ∠1=450 ∠1=∠2 [vertically opposite angle] ∠2=∠450 In ∆TQS ∠2+∠Q+∠S=1800 450 + ∠Q + 750 = 1800 ∠Q+1200 = 1800 ∠Q=1800 − 1200 ∠Q=600 ∠SQT=600 Ans06. Solution. Solution.CBSE TEST PAPER-05 CLASS . (B) Ans04.IX Mathematics (Lines and angle) [ANSWERS] Ans01. 1100 + ∠2 = 1800 [by linear pair] ∠2= 180 − 1100 0 ∠2 = 700 ∠1 + 1350 = 1800 ∠1=1800 − 1350 ∠1=45 73 . ∠1+∠2+∠R=1800 [By angle sum property] 450 + 700 + ∠R = 1800 ∠R=1800 − 1150 ∠R=650 ∠PRQ=650 Ans08. Solution. ∠SOQ =∠POR= 750 [vertically app angle] Solution. Given AD is transversal intersect two lines PQ and RS To prove PQ RS BE bisects ∠ ABQ 1 ∠ = ∠ABQ → (1) 2 Similarity C G bisects ∠ BCS Proof : 74 . RayOS s tan ds on the line POQ ∴ ∠POS+∠SOQ=1800 But ∠POS=X ∴ x+ ∠SOQ=1800 ∠SOQ=1800 − X Now ray OR bisects ∠POS. 1 therefore ∠ROS= × ∠POS 2 1 Similarly ∠ SOT = × ∠SOQ 2 ∠ROT = ∠ROS + ∠SOT Ans10. 1 x ×x = 2 2 1 x ⇒ = × (1800 − X ) = 90 − 2 2 x x ⇒ = + 90 − = 900 2 2 ⇒ = Solution. ∠POR + ∠ROQ = 1800 but [ linear pair of angle] ∠POR: ∠ROQ=5:7 [Given] 5 ×1800 =750 12 7 Similarly ∠ROQ= × 1800 =1050 12 Now ∠POS=∠ROQ=1050 [vertically opposite angle] ∴ ∠ POR= And Ans09. ∠1=∠2 → (1) [Angle of incident ] And ∠3=∠4 → (2) [is equal to angle of reflection] ∵ MB∠Q=∠NCS=900 ∴ MB NC [ By corresponding angle property] ∴ ∠2=∠3 → (3) [ alternate interior angle] BY e. Draw MB ⊥ PQand MC ⊥ RS .q.q. Solution. Solution. (2) And (3) ∠1=∠4 75 . In ∆PQR ∠PRS=∠Q+∠P [By Exterior angle thearem] ∠4+∠3=∠2+∠1+∠P → (1) 2∠3=2∠1+∠P ∴ QT and Rt are bisectors of ∠Q and ∠PRS In ∆ QTR ∠3= ∠1+∠T → (2) [By exterior angle thearem] BY e.1 ∠BCS → (2) 2 But BE CG and AD is the transversal ∴ ∠1=∠2 1 1 ∴ ∠ABQ = ∠BCS [By (1) and (2)] 2 2 ⇒ ∠ABQ=∠BCS ∴ PQ RS [ ∵ corner ponding angle are equqal] ∴∠2 = Ans11. (1). (1) and (2) we get 2[∠1+∠T ] =2∠1+∠P 2∠1+2∠T = 2∠1+∠P 1 ∠T= ∠P 2 1 ∠QTR= ∠QPR Hence proveel 2 Ans12. ∴ YQ bi sec cts ∠ ZYP ∴ ∠1=∠ 2 ∠1+∠ 2+∠ 64 0 = 180 0 [YX is a line] ∠1+∠1+64 0 = 180 0 2∠1=180 0 − 64 0 2∠1=116 0 ∠1 = 580 ∴ ∠ XYQ=64 0 + 580 = 122 0 ∠ 2+∠ XYQ =180 0 ∠1 = ∠ 2 = ∠ QYP = 580 ∠ 2+122 0 = 180 0 ∠ 2=150 0 − 122 0 ∠ QYP = ∠ 2 = 580 Re flex ∠ Q YP=360 0 − ∠ QYP =360 0 − 580 = 302 0 76 . Solution.∠1+∠2=∠4+∠3 ⇒ ∠ABC=∠BCD > ∴ AB CD [By alternate interior angles] > Ans13. two parallel lines. In fig 1.1. IN fig 1. If DA and CB are equal perpendiculars to a line segment AB. then the true statement is (A) AB ≠ AC (B) AB=BC (C) AB=AD (D) AB=AC [1] [1] 5. In quadrilateral ACBD. the bisector AD of ∆ ABC is ⊥ to the opposite side BC at D.2. then ∠ ACB is equal to [1] (A) ∠ ABD (B) ∠ BAD (C) ∠ BAC (D) ∠ BDA 2. if ABCD is a quadrilateral in which AD= CB. 8. AC=AD and bisects ∠ A. show that ∆ ABC ≅ ∆ CDA. show that ∆ ABC is isosceles? 77 [2] . 7. In fig 1. L and M. 4. then [1] ∠ CAB is equal to (A) ∠ ACD (B) ∠ CAD (C) ∠ ACD (D) ∠ BAD 3.3. If o is the mid – point of AB and ∠ BQO = ∠ APO. Show that CD [2] bisects AB. show ∆ ABC ≅ ∆ ABD? [2] 6. then ∠ OAP is equal to (A) ∠ QPA (B) ∠ OQB (C) ∠ QBO (D) ∠ BOQ IF AB ⊥ BC and CB = ∠ c. are intersected by Another pair of parallel lines P [2] and C. and ∠ D= ∠ B.CBSE TEST PAPER-01 CLASS .IX Mathematics (Congruent Triangles) 1.AB=CD. if AD =BC and ∠ BAD = ∠ ABC. [3] 13.9. [3] the triangle is right angled: 11. 2 Prove that is one angle of a triangle is equal to the sum of the other two angles.5. IF fig 1. [3] 1 ∠A. PQ ∏ SR. AD= AE and D and E are point on BC such that BD=EC prove AB=AC.4. 12. the bisector of ∠ ABC and ∠ BCA intersect each other at the point o prove that ∠ BOC = 90 + 10. If ∆ ABC. Prove that sum of the quadrilateral is 3600? [5] 78 . if PQ ⊥ PS. If in fig 1. ∠ SQR = 28 and ∠ QRT = 65 . then find the values [3] of X and Y. ... (ASA rule) ∴ AB=AC ..... (Given) ∴ ∠ CAD= ∠ACD. (AB bisects ∠ A) And AB= AB …………....... (SAS axiom) Ans06........ AD=BC ……….. (Each=90 ) and ∠ AOD ≅ ∠BOC. In ∆ AOD and ∆ BOC... (AAS rule) ∴ OA = OB .. (vert opp. (D) Ans05. LS) AC=CA ....(common) ∆ABD ≅ ∆ACD . (ASA rule) Ans08... LS) ∴ ∆AOD= ∆BOC..LS) P ∏ Q and AC cuts them . (D) Ans02. IN ∆ ACB and ∆ ADB.... (ALT.... (AD is the bisector of ∠A) And ∠ADB= ∠ADC=90.. (C.... (C.... (Given) ∠A=∠ B..... (Common) ∴∆ ABC ≅ ∆ ABD ………......T) Hence CD bisects AB..CBSE TEST PAPER-01 CLASS .............. (ALT... L ∏ M and AC cuts them ... (C) Ans03.(Given) ∴ ∠ACB= ∠CAD...IX Mathematics (Statistics) [ANSWERS] Ans01....T) Hence ∆ ABC is isosceles... 79 . Ans07. AC=AD……… (Given) ∠ BAC= ∠ BAD…….( AD ⊥ BC ) ∴ AD = AD.C...P. (common) ∴ ∆ ABC ≅ ∆CDA ... In ∆ ABD and ∆ ACD ∠1 =∠2.........C... (C) Ans04.P... Ans09. we have ∠1+∠2+∠BOC= 180 → (1) In ∆ABC . PQ ∏ SR and QR is the transversal. ∴ ∠ PQR=∠QRT [pair of alternate angles] or ∠ PQS+∠SQR =∠QRT or x+28 =65 ∴ x = 65 − 28 =37 Also in ∆ PQS. ∠A + ∠B + ∠C =180 [ sum of three angles of o ∆is 180 ] → (1) Given that : ∠A+∠C=∠B → (2) from (1) and (2) ∠B+∠B=180 180 = 90 2 Hence ∆ ABC is right angled. ∠SPQ+∠PSQ+∠PQS=180 ⇒ 90 + Y + X = 180 or 90 + Y + 37 =180 Y=53 80 . we have ∠A+∠B+∠C=180 ∠A + 2(∠1) + 2(∠2) = 180 ∠A ⇒ + ∠1 + ∠2 = 90 2 90 − ∠A ⇒ ∠1+∠2= 2 substituting this value of∠1+∠2 in (1) ∠A 90 − +∠BOC =180 2 ∠A ∠ BOC =90 + 2 So ∠A ∠ BOC= 90 + 2 ⇒ Ans10. ⇒ ∠B = Ans11. In ∆ BOC. 81 .Ans12. BD= CE AD=AE ∠ADB=∠AEC ∴ ∆ABC ≅ ∆ACE [BY SAS] ⇒ AB=AC [CPCT] Ans13.e. ∠ADE+∠ADB=180 [liner pair] Also. In ∆ ADE . ∠ BAD+∠ABD+∠BDA+∠CBD +∠BGCD+∠BCD+∠CDB=360 or ∠BAD + (∠ABD+∠CBD)+∠BCD+(∠CDB+∠BDA)=360 or ∠BAD+∠ABC+∠BCD+∠CDA=360 i. Join B and D to obtain two triangles ABD ∆ BCD. AD=AE [Given] ∴ ∠ADE =∠AED [ angles oppasrte to equal side are equal] Now. ∠A+∠B+∠C+∠D=360 So. ∠AED+∠AEC=180 [" " ] ⇒ ∠ADE+∠ADB=∠AED+∠AEC But ∠ADE= ∠AED Now in ∆ ABD and ∆ ACE. Sum of quadrilateral is 360 Hence proved. ∠BAD+∠ABD+∠BDA=180 [ sum of three angles of ∆ is 180 ] → (1) And ∠CBD+∠BCD+∠CDB=180 [ sum of three angles of ∆ is 180 ] → (2) Adding (1) and (2). [3] 82 . In ∆ ABC is an isosceles triangle and ∠ B = 650 . Prove that ∆ AEB ≅ ∆ ADC [2] 6. Prove that ∆ ABC is isosceles if altitude AD bisects ∠ BAC. AC=BC. [2] 9. In the given figure. Prove that ∆ DBC and ∆ EAC are congruent and hence DC=EC. Prove that ∆ ABC is an isosceles triangle. If AE=AD and BD=CE. [2] 8. In ∆ ABC. (a) 500 (b) 600 (c) 700 (d) none of these What is the sum of the quadrilateral:- [1] (a) 2600 (b) 3600 (c) 1800 (d) 900 The sum of the triangle will be:- [1] (a) 3600 (b) 2700 (c) 1800 (d) 900 5. 4. show that ∆ ABC ≅ ∆ ABD. find ∠ A [1] 3. In quadrilateral ABCD.CBSE TEST PAPER-02 CLASS . [2] What can you say about BC and BD? 7.IX Mathematics (Congruent triangle) 1. AC=AD and AB bisects ∠ A. If AB=AC and ∠ ACD= 1200 . find x. the median AD is ⊥ to BC. ∠ DCA= ∠ ECB and ∠ DBC= ∠ EAC. (b) 700 (a) 600 (c) 500 (d) none of these [1] 2. prove that ∠ BAD=3 ∠ ADB. From the following figure 2. ABCD is a quadrilateral in which AD=BC and ∠ DAB= ∠ CBA. Prove that. prove that AD//BC. Prove that AC=BD and AC//BD. [3] 12. 83 [5] . [3] (a) ∆ ABD ≅ ∆ BAC (b) BA=AC (c) ∠ ABD= ∠ BAC 13.1. O is the mid-point of AB and CD.10. ∆ ABC is an isosceles triangle with AB=AC. [3] 11. AD bisects the exterior ∠ A. IX Mathematics (Congruent triangle) [ANSWERS] Ans01. (b) Ans03. AE=AD ∠ EAB= ∠ DAC AB=AC ∆ AEB ≅ ∆ ADC Ans06. (c) Ans02.∵ AD ⊥ BC ] [ By SAS ] [CPCT ] 84 . We have. In ∆ ABC and ABD.CBSE TEST PAPER-02 CLASS . AE=AD and CE=BD ⇒ AE+CE=AD+BD (i) ⇒ AC=AB Now. (b) Ans04. AC=AD ∠ CAB= ∠ DAB AB=AB ∆ ABC ≅ ∆ ABD ∴ BC=BD [given] [common] [from (i)] [by SAS] [given] [AB bisects ∠ A] [common] [SAS criterion] [CPCT] In ∆s ABD and ACD. (c) Ans05. BD = CD [ D is mid − po int of BC ] AD = AD [common] ∠ADB = ∠ADC ∆ABD ≅ ∆ACD ∴ AB = ACE [ Each 900 . Ans07. in ∆ AEB and ∆ ADC. In ∆s AOC and BOD AO = OB [O is the mid − po int of AB] ∠AOC = ∠BOD [vertically opposite angles ] CO = OD [O is the mid − po int of CD] ∆AOC ≅ ∆BOD [ BySAS ] AC = BD [CPCT ] ⇒ ∠CAO = ∠DBO [CPCT ] Now. In ∆s ABD and ACD.e. AD ⊥ BC ] ∠BAD = ∠CAD [ AD bi sec ts ∠BAC ] AD = AD [common] ∆ABD ≅ ∆ACB [ ByAAS ] ⇒ AB = AC [CPCT ] Thus. ∠DCA = ∠ECB [ given] ⇒ ∠DCA + ∠ECD = ∠ECB + ∠ECDE ⇒ ∠ECA = ∠DCA [ adding ∠ECD on both sides ] → (i ) Now. in ∆s DBC and EAC ∠DCB = ∠ECA BC = AC [ from(i )] [ given] ∠DBC = ∠EAC [ given] ∆DBC ≅ ∆EAC [ ByASA] DC = EC [CPCT] ⇒ Ans10. alternate angle are equal. 85 . Let ∠ADC = Q ⇒ ∠CAD = Q Exterior [∵. CA = CD] ∠ACB = ∠CAD = Q + Q = 2Q ∠BAC = 2Q ⇒ [∵ BA = BC ] Hence∠BAD = ∠BAC + ∠CAD = 2Q + Q = 3Q = 3∠ADC = 3∠ADB Ans11. ∠ADB = ∠ADC [ Each 900 . ∆ABC is an isosceles triangle. We have. Ans09. AC and BD are two line s in ter sec ted by a transversal AB such that ∠CAO = ∠DBOi.Ans08. AD = BC [ given] ∠DAB = ∠CBA [ given] AB = AB ∆ABD ≅ ∆BAC (i ) ∵ Ans13. In ∆s ABD and BAC . 1 EAD= ∠EAC 2 1 1 = 1800 − ∠1 = 900 − ∠1 2 2 → (i ) ∴∠1 + ∠EAC = 1800 ( Linear pair ) ∠1 + ∠2 + ∠3 = 1800 ⇒ ∠1 + ∠2 + ∠2 = 1800 [∴ AB = AC ] 2∠2 = 1800 − ∠1 ⇒ 1 0 ∠ = − ∠1 2 90 But 2 Hence from (i ) and (ii ) ∠EAD = ∠2 = ∠ABC But these are corresponding angles ∴ AD / / BC 86 (i) . [ SAScriterion] BD = AC (ii ) ⇒∴ (iii ) ⇒ [common] [CPCT ] Also∠ABD = ∠BAC [CPCT ] Since AD bisects the exterior A.Ans12. Show that these altitudes are equals. 5. [1] 3.CBSE TEST PAPER-03 CLASS . AB=AD and ∠BAD = ∠EAC. [2] 7.IX Mathematics (Congruent triangle) 1. find x. (A) 120 (B) 60 (C) 45 (D) 90 Which of the following pairs of angle are supplementary? (A) 30 . ABC is An isosceles triangle in which altitudes BE and CF are drawn to side AC [1] [1] [2] and AB respectively. An angle is 4 time its complement. show that BC =DE.30 (D) None of these. Find measure. Find its measure. (A) 42 (B) 32 (C) 52 (D) 62 [1] 2. 6. (A) 62 (B) 72 (C) 52 (D) 42 Find the measure of angles which is equal to its supplementary. An angle is 14 more than its complement. 4. It AC= AE.120 (B) 45 . In the given figure. Line ∠ is the bisector of an angle ∠ A and B is any point on ∠ BP and BQ are [2] ⊥ from B to the arms of ∠ A show that : (i) ∆ APB ≅ ∆ AQB (ii) BP = BQ or B is A equidistant from the arms of ∠A 8. 87 [2] . ∆ ABC is an isosceles triangle and ∠ B = 65 .135 (C) 120 . It AB and XY intersect each other at P. Two side AB and BC and median AM of one triangle ABC are respectively equal [3] to side PQ and QR and median PN of ∆ PQR. 13. (ii) ∆ ABC ≅ ∆PQR In the given figure. show(I) ∆ ABM ≅ ∆ PQN 12. prove that (i) ∆ APX ≅ ∆ BPY. the bisector of ∠ B and ∠ C intersect each [3] other at o. show that: (i) OB=OC (ii) AO bisects ∠ A. 10. with AB =AC. join A to o.9. 11. 88 [5] . In an isosceles ABC.AX and By are equal two equal line – segments drawn on [3] opposite side of line AB such that AX ∏ BY. Show that ∠ BCD is a right angle. Prove that ∠ ABD = ∠ ACD. ABC and DBC are two triangle on the same base BC such that [3] AB=AC and DB=DC. (ii) AB and XY bisect each other at P. ∆ ABC is an isosceles triangle in which AB=AC side BA is produced to d such that AD=AB. AB is a line – segment. IX Mathematics (Congruent triangle) [ANSWERS] Ans01. ∠BAD = ∠EAC [ given] [ given] ∴ ∠BAC + ∠DAC = ∠EAC + ∠CAD ⇒ ∠BAC = ∠EAD ∴ ∆ BAC ≅ ∆ DAE [ SAS criterior ] ⇒ BC = DE [CPCT ] Ans07. In ∆s BAC and DAE . ∠BAP = ∠BAQ ∠APB = ∠AQB = 90 AB = AB (i ) ∴ ∆ABP ≅ ∆ABQ (ii ) BP = BQ 0 [ given] [common] [common] [ AAS rule] [CPCT ] 89 . ∠A = ∠A [common] ∠AEB = ∠AFC = 900 AB = AC [ given] ∴ ∆ABE ≅ ∆ACF [ AAS rule] ⇒ BE = CF [CPCT ] Ans06. (C) Ans02. (B) Ans03. (D) Ans04. In ∆s ABE and ACF . (B) Ans05.CBSE TEST PAPER-03 CLASS . AB = AD [ given] AC = AE Also. In ∆s ABP and ABQ. Ans10. ∠A + ∠B + ∠C = 1800 x + 130 = 1800 x = 500 Ans09.e. AB = AC ⇒ ∠B = ∠C [ Anglesoppositetoequalsidesareequal ] But ∠B = 650 ∴ ∠B = ∠C = 650 So. (i ) In ∆ ABC.Ans08. AO bisecls ∠A 90 . In ∆s APX and BPY . [ Alternate angle] ∠1 = ∠2 ∠3 = ∠4 AX = BY [Vertically opposite angle] [ given] ∴∆APX ≅ ∆BPY [ ByAAS ] ⇒ AP = BP and PX = PY [CPCT ] ⇒ AB and XY bi sec ts each other at P. AB=AC [given] ∠ ACB=∠ABC [angles oppaside to equal side] 1 1 ∠ACB= ∠ABC 2 2 or ∠OCB =∠OBC ∴ ⇒ OB=OC (ii ) [ side oppasite to equal angle] In ∆AOB and ∆ AOC AB = AC ∠ ABO=∠ACO OB=OC ∴ ∆ AOB ≅ ∆ AOC ⇒ ∠BAO=∠CAO [ given] [ Halves of equals] [ proved] [SAS rule] [CPCT] i. In ∆s ABC . (i ) In ∆s ABMand PQN. ∠ABC = ∠ACB Also.Ans11. ∠CAD= ∠ABC+∠ACB [ exterior angles of ∆ ABC] = 2∠ACB And ∠ BAC= ∠ACD+∠ADE [Exterior amgles of ∆ ADC] = 2 ∠ACD ∴ ∠BAC+ ∠CAD=2 (∠ACD+∠ACB) =2∠BCD i.e. ∠ACD=∠ADC Now. in ∆s ABC and PQR. AB=AC [ Given] ∴ ∠ ABC= ∠ACB [ angles oppaosite to equal side are equals] Similarly in ∆DBC. ∠BAC+∠CAD=180 [ Angles oposite to equal side] ["" " " " " " "] [ Linear Pair] Also. AB=PQ [ Given] BM=QN AP=PN [ Halves of equal] [ Given] ∴ ∆ ABM ⇒ ∆ PQN [SSS rules] (ii ) ⇒ ∠B=∠Q Now. 2∠BCD=180 or ∠ BCD =90 91 . In ∆ABC . DB=DC [Given] → (1) ∴ ∠DBC= ∠DCB → (2) Adding (1) and (2) : ∠ ABC+∠DBC= ∠ACB+∠DCB or ∠ ABD= ∠ACD Ans13. AB=PQ BC=QR [ Given] [Given] ∠B=∠Q [Proved] ∴ ∆ ABC ≅ ∆PQR [SAS ruls] Ans12. CBSE TEST PAPER-04 CLASS . Prove that the Angle opposite of the greatest side of a triangle is greater that [3] 92 . is AB=AC and ∠A = 90 . Prove that RS<QS. [2] 6. If PQ= PR and S is any point on side PR. [2] 8. Find the measure of each exterior angle of an equilateral triangle. [2] 9. prove that AD>EC. Which is the longest side. is AB=AC and ∠ = 70 . (B) 100 (A) 110 (C) 120 (D) 150 [1] 2. [1] 3. [2] 7. 4. (A) 70 (B) 80 (C) 95 (D) 60 In an ∆ ABC. (A) BC (B) AC (C) CA (D) None of these. is ∠ B= ∠ C= 45 . Prove that MN+NO +OP+>2MO. In an ∆ ABC. Prove that MN+NO+OP>PM. Find ∠ B. Find ∠ A. In an isosceles ∆ ABC. (A) 40 (B) 50 (C) 45 (D) 60 [1] [1] 5. If ∠ E> ∠ A and ∠ C> ∠ D.IX Mathematics (Congruent triangle) 1. 93 [5] . Prove that ∠ A> ∠ C and ∠ B> ∠ D. ∠ A= ∠ C and AB =BC. Prove that ∆ ABD ≅ ∆ CBE.third of a right angle i.e. It the bisector of a vertical angle of a triangle also bisects the opposite side. 11. [3] prove that the triangle is an isosceles triangle. great her than 60 10. 13. In the given figure. AD is the bisector of ∠ A of ∆ ABC. Prove that AB>BD and [3] AC>CD. AB and CD are respectively the smallest and the largest side [3] of a quadrilateral ABCD. In the given figure. where D lies on BC.two. 12. (a) [Gven] [ Side opposite to greater angle is l arg er ] → (i ) Similarly in ∆ BCD. In ∆ MON . MN + NO + OP + MO > MO + PM In ∆ ABC. MN + NO > MO [ Sum of any two side of ∆ is greater than third sides ] → (i) Similarly in ∆ MOQ. Ans05. ∠C > ∠D [Given] ⇒ ∠BD > ∠BC → (ii ) Adding (i ) and (ii ) AB + BD > EB + BC Or AD > EC Ans06. In ∆ PQR.IX Mathematics (Congruent triangle) [ANSWERS] Ans01. PQ=PR [Given] ⇒ ∠PRQ = ∠PQR [ Angle opposite to equal side are equal ] Now. MN + NO > MO [ Sum of any two side of ∆ is greater than third sides ] → (i) Similarly in ∆ MPQ. (a) Ans04. ∠E > ∠A ⇒ AB > EB (a) Ans03. ∠SQR<∠PQR [ ∠SQR is a part of ∠PQR ] ∴∠SQR < ∠PRQ OR ∠SRQ ⇒ RS <QS [Side opposite to smaller angle in∆SRQ] Ans07. MO + OP > PM → (ii) Adding (i) and (ii) Or Ans09. In ∆ ABE . (c) Ans02. AB > BC [Given] 94 . OP + PM > MO → (ii) Hence from (i) and (ii) MN + NO + OP + PM > 2MO Ans08. In ∆ MON .CBSE TEST PAPER-04 CLASS . ∠C > ∠A Similarly. we can show that ∠B > ∠D 95 . Ans11. In ∆ ABC. CD > AD [CP is the largest side of quadrilateral ABCD] ∠2 > ∠4 [Angle opposite to larger side is greater] → (ii) Adding (i) and (ii) ∠1 + ∠2 > ∠3 + ∠4 Or ∠A > ∠C Similarly. by joining BD. Join AC. ∠C > 600 Ans10. ∠1 = ∠2 ∴ ∠ 4 > ∠2 ⇒ AC > CD [Side opposite to greater angle is larger]. BC > AB [AB is the smallest sides of quadrilateral ABCD] ⇒ ∠1 > ∠3 [ Angle opposite to larger side is greater] → (i) In ∆ ADC. 3∠C > (∠A + ∠B + ∠C) 3∠C > 1800 [Sum of three angles of ∆ is 1800 ] Or . [Angle opposite to large side is greater] → (i) AB > AC ∴ ∠C > ∠B → (ii) Adding (i) and (ii) 2∠C > (∠A + ∠B) Adding ∠C to both sides. In ∆ ADC. ∠4 > ∠1 [Exterior opposite of ∆ is greater than each of interior opposite angle] But. ∠3 > ∠2 [Exterior angles of ∆ is greater than each of the interior opposite angles] But ∠2 = ∠1 [AD bisects ∠A] ∴ ∠3 = ∠1 ⇒ AB > BD [ Side opposite to greater angle is larger] In ∆ ABD. In ∆s AOE and COD. ∠AEO + ∠OEB = 1800 And ∠CDO + ∠ODB = 180 [Linear pair] 0 [Linear pair] ⇒ ∠AEO + ∠OEB = ∠CDO + ∠ODB ⇒ ∠OEB = ∠ODB ⇒ ∠CEB = ∠ADB Now. DC = DB [Given] AD = ED [By construction] ∠ADC = ∠EDB [Vertically opposite angle] ∴∆ADC ≅ ∠EDB [By SAS] ⇒ AC = EB and ∠DAC = ∠DEB [CPCT] But ∠DAC = ∠BAD [ ∵ AD bisects ∠A] ∴ ∠BAD = ∠DEB ⇒ AB = BE But BE = AC [Proved above] ∴ AB = AC Ans13. in ∆s ABD and CBE. ∠A = ∠C ∠ADB = ∠CEB AB = CB ∆ABD ≅ ∆CBE 96 [Using (i)] → [Given] [From (ii)] [Given] [By AAS] (ii) . In ∆s ABC and EDB.Ans12. ∠A = ∠C ∠AOE = ∠COD ∴ [Given] [Vertically opposite angle] ∠A + ∠AOE = ∠C + ∠COD ∵ ∠A +∠AOE + ∠AEO = 1800 and 0 ∠C +∠COD + ∠CDO = 180 → (i) ⇒ 1800 − ∠AEO = 1800 − ∠CDO ⇒ ∠AEO = ∠CDO Now. CBSE TEST PAPER-05 CLASS . ∆ ABC is an equilateral triangle and ∠B = 600 . ∆ ABC is an isosceles triangle and ∠B = 450 . if ∠A = 450 and ∠B = 700 . Determine the shorteat sides of the triangles. Find third angles. 3. (a) AC (b) CA (c) BC (d) none of these The sum of two angles of a triangle is equal to its third angle. find ∠A . [2] 97 . (a) AC (b) BC (c) CA (d) none of these In an ∆ ABC. find ∠B . find ∠C . AB = AC and ∠ACD = 1200 . 4. In the figure 1. (a) 900 (b) 450 (c) 600 (d) 700 [1] 5. (a) 900 (b) 450 (c) 600 (d) 700 Two angles of triangles are 650 and 450 respectively. In a ∆ ABC. [1] 6. Find the third [1] angles. determine the longest sides of the [1] [1] triangle. 2. If ∠A = 450 and ∠B = 700 .1.IX Mathematics (Congruent triangle) 1. [2] 7. find ∠A [2] 9. Prove that in a right triangle.8. [3] 13. In the given figure. PQ>PR. (ii) AD bisects ∠A In the given figure. PR>PQ and PS is the bisector of ∠QPR. [3] 10. In the given figure 1. Prove that SQ>SR.1. QS and RS are the bisectors of the ∠s Q and R [3] respectively. 12. hypotenuse is the longest (or largest) side. AD is an altitude of an isosceles triangle ABC in which AB = AC that: [3] (i) AD bisects BC 11. Prove [5] that ∠PSR > ∠PSQ . Draw AP ⊥ BC to show that ∠B = ∠C. 98 . ABC is an isosceles triangle with AB = AC. ∠C = ∠B = 600 Ans08. ∠A + ∠B + ∠C = 1800 [Sum of three angles of a ∆ is 1800 ] ∠A + 600 + 600 = 1800 ⇒ ∠A = 1800 − 1200 ∠A = 600 99 . Ans07. (a) Ans04. ∠C = 600 [Angles opposite to equal sides are equal] 0 Since in ∆ ABC. (c) Ans03. In ∆ ABC. ⇒ AB = AC ∠B = ∠C But ∠B = 450 = ∠C [Angle opposite to equal sides ae equal] and ∠A + ∠B + ∠C = 1800 ∠A + 900 = 1800 ∠A = 900 Ans06. (b) Ans02. AB = AC ⇒ ∠B = ∠C [Angles opposite to equal are equal] Also. ∠ACB + ∠ACD = 180 ⇒ 0 [Linear pair] ∠ACB = 1800 − 1200 and . AB = AC ⇒ ∠B = ∠C But ∠B = 60 So. In ∆ ABC.IX Mathematics (Congruent triangle) [ANSWERS] Ans01. In ∆ ABC.CBSE TEST PAPER-05 CLASS . (d) Ans05. the side opposite to ∠B is the hypotenuse is the longest side. In right ∆s APB and APC. [CPCT] AD bisects ∠A. since ∠B = 900 ∴ ∠A + ∠B + ∠C = 1800 ∠A + ∠C = 1800 − 900 = 900 i.Ans09.e. Ans11. AD bisects BC (ii) Also. Side AD = Side AD [common] Hypotenuse AB = Hypotenuse AC [Given] ∴ ∆ ABD ≅ ∆ ACD [By RHS] ⇒ BD = CD [CPCT] Also. 100 . ∠BAD = ∠CAD i. [Common] Hypotenuse AB = Hypotenuse AC [Given] ∴ ∆ APB ≅ ∆ APC [RHS rule] ⇒ ∠B = ∠C [CPCT] (i) In right triangle ABD and ACD.e. Since PQ>PR ∠R > ∠Q 1 1 ⇒ ∠R > ∠Q ∴ 2 2 ⇒ ∠SRQ > ∠SQR ⇒ SQ > SR [Angle opposite to larger side is garden] [side opposite to greater angle is larger] Given a right angled triangle ABC in which ∠B = 900 ∴ AC is its hypotenuse. ∠B = ∠A + ∠C ⇒ ∠B > ∠A and ∠B > ∠C Hence. Ans12. AP = AP Ans10. Now. In ∆ PQR.(∠1 + ∠4) or.Ans13. ∠6 .∠5 > 0 ∠6 > ∠5 or ∠PSR > ∠PSQ 101 . ∠5 = ∠2 + ∠4 But ∠2 = ∠1 ∴ ∠5 = ∠1 + ∠4 [PS bisects ∠QPR] → (iii) Subtracting (iii) from (ii) ∠6 . ∠6 > ∠1 + ∠3 [Exterior angle theorem] → (ii) Similarly.∠4 = 0 From (iv) and (v) ∠6 .∠5 = ∠3 . → (iv) ∠3 > ∠4 ⇒ → (v) ∠3 . PR > PQ [Given] ⇒ ∠3 > ∠4 [Angle opposite to larger side] → (i) Also.∠4 Now.∠5 = (∠1 + ∠3) . B (-2. 3) and (-4. 3. (2) 8. The point of intersection of X and Y axes is called (a) zero point (b) origin (c) null point (d) none of these The distance of the point (-3. 5. 102 (3) . 0). (5. Take a triangle ABC with A (3. (0.X Mathematics (Co-ordinate geometry) 1. -5) and (6.CBSE TEST PAPER-01 CLASS . Write the Co-ordinates of a point which lies on y-axis and is at a distance of (2) 3units above x-axis. 10. 5). 0). Find its mirror image. 5). -2) from x-axis is (a) 2units (b) 3units (c) 5units (d) 13 units The distance of the point (-6. 7. 4. (-3. (-3. 1). 2). Write the mirror image of the point (2. 12) is (A) abscissa 12 and ordinate (B) abscissa 8 and ordinate 12 (C) abscissa 0 and ordinate 20 (D) none of these Write the name of each part of the plane formed by Vertical and horizontal (1) (2) lines. -2) from y-axis is (a) 6units (b) (c) 2units (d) 8units (1) (1) (1) 38 units The abscissa and ordinate of the point with Co-ordinates (8. Draw its graph. 1). 5). C (2. Write the Co-ordinates of a point which lies on the x-axis and is at a distance of (2) 4units to the right of origin. 9. (2. 2. Represent on the graph. -6) with respect to x-axis. 6. 1) in the (3) Cartesian plane. Locate the points (5. 2) and (-3. 4). Also write Co-ordinate of point of AC and BD. (3) (1. formed. . 0). (3) In fig. In which quadrant or on which axis do each of the points (-2. See fig. write the Co-ordinates of the points and if . . and write the following (i) (5) The Co-ordinates of B (ii) The Co-ordinates of C (iii) On which axes point L lies. (vii) The ordinate of the point H (Viii) The Co-ordinates of the point M (ix) The point identified by the Co-ordinate (2. we join the points write the name of fig. -5) lie? Verify your answer by locating them on the Cartesian plane. -4) (x) The point identify by the Co-ordinates (-3.11. 13. 12. . (2. (iv) The abscissa of the point D (v) The Co-ordinates of point L (vi) In which axes point M lies. (-1. -1). -5) 103 . (a) Ans04.X Mathematics (Co-ordinate geometry) [ANSWERS] Ans01. 3). Ans08.-6) with respect to x-axis. 0) . -3) with respect to x-axis. Ans06. -6) is (-4. (a) Ans03. (b) Ans02. (b) Ans05. (ii) The mirror image of (-4. (i) The mirror image of point (2. Ans07.CBSE TEST PAPER-01 CLASS . (4. 104 . 3) is (2. Horizental line is called X-axes. The Co-ordinates of the point which lies on y-axis and at a distance of 3units above x-axis is (0. (i) (ii) Vertical line is called Y-axes. -1). -1) respectively. . Mirror image of A (3. . . 0). 0) and D is (-2. 0). B is (2. B (-2.Ans09. Ans11. . B’ (-2. 0). C is (-2. 1) and (2. . . C’ (2. . 1) are A’ (-3. 105 . . (ii) It we joined them we get square. . Ans10. 0). (i) The Co-ordinate of point A is (0. 2). (iii) Co-ordinate of point of AC and BD is (0. 0). . 4) lies in II quadrant. -5) lies in III quadrant. . -1) lies in IV quadrant. (-1. 0) (ix) G (x) E 106 . 0) lies on –ve x-axis. (vii) The ordinate of point H is -3 (viii) (-3. . (-2. 2) (ii) (5. (-3. Ans13. (v) (0. -5) (iii) Y-axis. (1. (iv) The abscissa of point D is 6. 5) (vi) Point M lies on X-axis. . (i) (-5.Ans12. (2. . 2) lies in I quadrant. 0) (iv) (-4. y) (C) (o.1) (ii) (-7.-1) lien in [1] (A) 1st quadrant (B) 2nd quadrant (C) 3rd quadrant (D) 4th quadrant The point (3. o) (B) (o. State the quadrant in which each of the following points lie: [2] 7. o) (D) none of thence.3) from y axis’s (A) 2 units (B) 3 units (C) 5 units (D) 13 units The point (-2. [1] (A) (X. 3.-4) [2] 6.-5) Which of the following points belongs to 2nd quadrant (i) 8.11) (iii) (-6.0) lies in (A) +ve x axis (C) + ve y axis [1] [1] (B) – ve x axis (D) –ve y axis 5. (i) (2.3) (ii) (-3.X Mathematics (Co-ordinate Geometry) The co-ordinate of origin in 2. Write abscissa and ordinate of point (-3.2) What is the name of horizontal and vertical lines drawn to determine the [2] [2] position of any point in the Cartesian plane? 9. 4. (2. In fig of vertices find co-ordinates of △ ABC 107 [3] .-4) (iv) (-5. The distance of the point (2.CBSE TEST PAPER-02 CLASS .2) (iii) (2. Locate the points [3] (A) (-3. 13. Plot the following ordered pairs of number (x. 11.3) and (D) (-3.-4) (B) (-5.2) [3] (C) (-3. y) as points in the Cartesian [3] plane. 12.4) find its mirror image with respect to y.10. Find some ordered pairs of the linear equation 2x+y=4 and plot them ‘how many such ordered pairs can be found and plotted? 108 [5] .axis. 4) and (C) (0. 4) (B) (3. 0) in a Cartesian plane write the name of figure which is fumed by joining them. Take a quadrilateral ABCD (A) (-5. 109 .-5) (I) Quadrant (ii) Quadrant (iii) Quadrant (iii) Quadrant Ans07. 0) Ans10. 3) C’ (3. (A) 2 units Ans03. 4) (B) (-5. The mirror image of point.axis Ans05.X Mathematics (Co-ordinate Geometry) [ANSWERS] Ans01. (c) (-2. 1) (-7. (C) (0. (A) +ve x. Ans08. (C) 3rd quadrant Ans04. 0) Ans02. (A) (0. 3) and (D) (-3. (-4. Abscissa -3 ordinate -4 Ans06. (2. 4). (B) (1. The name of horizontal line is x –axis The name of vertical line is y – axis Ans09. 2) belongs to (ii) quadrant.CBSE TEST PAPER-02 CLASS . 3) and D’ (3. 2). 11) (-6. 3) B’ (5.-4) (-5. A’ (+5. (A) (-5. 4) with respect to y-axis are. 2). 2) (C) (-3. The point (-3. 4) Ans11. -1) ‘(-1. Solution the given equation is 2x+y=4 When x=0 y=4 i. 2) When x=2 y=0 i.-2) Some more ordered pairs satisfying the equation 2x+y=4 are given by (4. (2.e.e.e.Ans12. 110 . 4) When x=1 y=2 i. Ans13. 6) (-3. (3. Thus an infinite number of ordered pairs can be found and plotted. 10) -------By plotting the points in the Cartesian plane. (0.e. (1. 0) When x=3 y = -2 i. 2. 0) from x. 4) (d) (3. -3) (c) (0. Name the points of the plane which do not belong to any of the quadrants. 8. 3) (c) (0. 4) (d) (-4. -2) Find Co-ordinates of vertices of rectangle ABCD 111 (2) (2) (3) . 9. 0) Which of the following points belongs to 1st quadrant (a) (3. (a) (2.axis (c) –ve x.axis (b) +ve y.CBSE TEST PAPER-03 CLASS . 5) lies in (1) (a) 1st quadrant (b) 2nd quadrant (c) 3rd quadrant (d) 4th quadrant The distance of the point (3. Which of the following points belong to the x. 0) (b) (3. The distance of the point (3.axis? (2) 7. 3) (b) (-1.axis is (a) 3 units (b) 4 units (c) 5 units (d) 6 units The point (0. 5) from x.axis is (a) 3 units (b) 0 units (c) 9 units (d) none of these (1) 5. 1) (d) (-2.X Mathematics (Co-ordinate Geometry) 1. 4. 3. (2) 6.axis (d) –ve y-axis The point (-2. 2) (c) (-3. 4) Which of the following points belongs to 3rd quadrant (a) (1. -5) lies on (1) (1) (a) +ve x. 0) (b) (1. Plot each of the following points in the Cartesian Plane (a) (3. Do you get a straight line by joining these points? 112 (5) . Find its (3) mirror image with respect to x. (b) (-3. 12. 2) and D (-2. 4) 13. 2). -5) (e) (2. Take a rectangle ABCD with A (-6. -4) (c) (0.axis.10. C (-2. 11. The following table gives measures (in degrees) of two acute angles of a right (3) triangle X 10 20 30 40 50 60 70 80 Y 80 70 60 50 40 30 20 10 Plot the point and join them. B (-6. 4). -5) (3) (d) (2. 0) The following table given the relation between natural numbers and odd natural numbers X 1 2 3 4 5 6 7 Y 3 5 7 9 11 13 15 Plot the points and join them. 4). -4) and B (-6. -2) and D (2. 4) belongs to 1st quadrant. 113 . -2) belongs to 3rd quadrant. B (-2. Ans04. 4) is D’ (-2. 4) is A’ (-6. 2) is C’ (-2. Ans05. . C (-2. The Co. Ans08. -2). Ans02. . The mirror image of A (-6. . 0) and (-2. Ans09. . 0) belongs to x.X Mathematics (Co-ordinate Geometry) [ANSWERS] Ans01. (b) 2nd quadrant. C (-2. Ans03. . (2. . Ans06. -3) and (-4. 2). 2).ordinate of vertices of rectangle A (2.CBSE TEST PAPER-03 CLASS . 0). (c) 5 units. -2) and D (-2. .axis. -4) Ans11. Ans10. (d) –ve y-axis. Ans07. (-1. 2) is B’ (-6. -2). . 2) and (3. (a) 3 units. The points in a plane which do not belong to any one of the plane are (0. (1. . . . . 114 . . .Ans12. we get a straight line by joining them. . . . . . Ans13. . Yes. each piece measuring 20cm 50cm and 50cm. BC=4cm. Find the area of the field. Measure of one side of an equilateral triangle is 12cm. has [3] one diagonals 38cm. find the area of rhombus of Its [2] diagonals BD measures 12cm 7. Find the area of the triangle to the nearest cm. the measures of whose adjacent sides are 28cm and 42cm. Find Its altitude on the side 42cm. A field in the shape of a trapezium whose parallel side are 25m and 10m. the measure of one of Its equals side being [2] ‘b’ and the third side ‘a’. the length of whose side is 60m and 25m has one diagonal 65m [3] long. 11. (B) 18 3 sq cm (C) 27 3 sq cm 3 cm2 is [1] [1] (D) 36 3 sq cm Two adjacent side of a parallelogram are 74cm and 40cm one of Its diagonals is [1] 102cm. [5] 115 . A parallelogram. (B) 1224 sq m (A) 10cm (B) 8cm (C) 6cm [1] (D) 5cm An umbrella is made by stitching 10 triangles pieces of cloth of two different [2] colour. CD= 4cm. From a point in the interior of an equilateral triangle perpendiculars drawn to the three sides are 8cm. DA=5cm and AC= 5cm.CBSE TEST PAPER-01 CLASS . [3] 13. 10cm and 11cm respectively. The measure of each side of an equilateral triangle whose area is (A) 8cm (B) 2cm (C) 4cm (D) 16cm 2. Find the area of a quadrilateral ABCD in which AB=3cm. area of the ||gram is (A) 612 sq m 4. How much cloth of each colour is required for the umbrella? 6.X Mathematics (Heron’s formula) 1. [2] 8. If one of its diagonals is 16cm. (C) 2448 sq m (D) 4896 sq m The area of rhombus is 96 sq m. Find the area of the parallelogram. The non. 12. Find area of triangle with two sides as 18cm & 10cm and the perimeter is 42cm.73 ) [3] 10. 9. then length of Its side is 5.parallel side are 14m and 13m. Its area is given by (A) 9 3 sq cm 3. (use 3 =1. A parallelogram. The perimeter of a rhombus ABCD is 40cm. Find the area of in isosceles triangle. (d) Ans03. a=20cm.X Mathematics (Heron’s formula) [ANSWERS] Ans01 (b) Ans02. (c) Ans04. BD=12cm and DA=10cm 10 + 12 + 10 cm =16cm 2 ∴ by herons pormula ∴ S= area of △ABD = 16(16 − 10) (16-12) (16-10) = 16 × 6 × 4 × 6 =48sq cm ∴ areaofr hom bus ABCD=2 × area of △ABD = 2 × 48sq cm =96sq cm 116 . ∴ AB = BC = CD = DA = 40 cm 10cm 4 now in △ABD.CBSE TEST PAPER-01 CLASS . AB=10cm. (a) Ans05. b=50cm ∴ cloth required for each colour = 5 × Area of one triangle piece a 4b 2 − a 2 4 20 = 5× 4(50) 2 − (20) 2 sq cm 4 =25 × 40 6 sq cm =5 × =1000 × 6 sq cm thus. (1000 6) sq cm cloth of each colour is required Ans06. the measure of whose each side of x 3 2 x sq cm → (ii) 4 fro m (i) an d (ii) = 3 2 29 x = x 4 2 ∴x = 4 × 29 58 = 3×2 3 29 58 841 3 × = sq cm 2 3 3 841× 1.73 = = 485 sq cm. C=14cm a + b + c 18 + 10 + 140 ∴ S= = = 21cm 2 2 new area of triangles = 21(21 − 18) (21-10) (21-14) = 21× 3 × 11× 7 =21 11 sq cm Here S = Ans08. b=10cm Perimeter =42cm ∴ a + b + c = 42cm so.Ans07. Let a=18cm. a+b+c a + 2b units = units 2 2 ∴ area of △ = ( = ( = a 4 a + 2b a + 2b a + 2b a + 2b )( − a)( − b)( − c) 2 2 2 2 a + 2b 2b − a a a )( ) × sq units 2 2 2 2 4b 2 − a 2 sq units Let the each side of the equilateral ∆ ABC measure be x cm. Let OD= 11cm. OB and OC Now ar (△ ABC ) = ar (△OBC ) + ar (△OCA) + ar (△OAB) Ans09. 3 ∴ area of ∆ABC = 117 . OE =8cm and OF=10cm Join OA. 1 1 1 = x × 11 + x × 8 + x × 10 sq cm 2 2 2 29 = x sq cm → (1) 2 But area of equilateral ∆ . S= = 5 4 × 1 6 × 2 6 × 1 2 sq cm .w e g e t 21h=144 3 144 3 48 3 = cm 21 7 48 3 th u s . Ans11. BC=AD= 25m and AC=65m Area of 11gm ABCD= Area of ∆ ABC + area of ∆ ACD = 2 Area of ∆ ABC [∴ ar ∆ ABC =ar ∆ ABD] now 60 + 65 + 25 S= m = 75m 2 ∴ area of ∆ABC = S ( s − a)( s − b)( s − c) = 75(75 − 60)(75 − 65)(75 − 25) sq m =(5 × 3 × 5 × 2 × 5) sq m =750 sqm → (II) from (i) and (ii). w h a re h c m is a ltitu d e 2 = 21h sq cm ∴ fro m (i) a n d (ii). AB=DC=60cm. = 1 4 4 3 sq cm ∴ a re a o f ∆ A B D = 1 4 4 3 s q c m 1 a g a in a re a o f ∆ A B D = b a s e × a ltitu d e 2 1 = × 4 2 × h s q c m . re q u ire d a ltitu d e = cm 7 h= 118 .Ans10. we get area of 11gm ABCD =2 × 750=1500sq m. AB=DC=42cm=C BC=AD= 28cm =b And BD=38cm=a Let A be the area of ∆ ABD 38 + 28 + 42 = 54cm 2 A = 5 4 (5 4 − 3 8 )(5 8 − 2 8 )(5 4 − 4 2 ) now. AB=25m.Ans12. a=15m. area of II gm AECD=Base × height 56 =10 × m=112 sq m → (ii) 5 thus area of trap ABCD=area of II gm + area of∆ BCE =112 sq m+84 sq m [using(i) and (ii)] =196 sq m 119 . S= 5+4+5 =7cm 2 ∴ area of ∆ACD= 7(7 − 5)(7 − 4)(7 − 5) = 7 × 2 × 3 × 2 = 9. ∴ Area of quad ABC=area of ∆ ABC + area of ∆ ACD (i) 3+ 4+ 5 For ∆ ABC.2 sq cm Ans13. (ii) and (iii).2 sq cm → (iii) ∴ from (i). b=14m and C=13 m a+b+c 15 + 14 + 13 = 2 2 =21m ∴ area of ∆BCE= ∴S = 21(21 − 15)(21 − 14)(21 − 13) = 21× 6 × 7 × 8 sq m =84 sq m → (i) 1 Also area of∆BCE = × BE × CP 2 56 2area of ∆BCE 2 × 84 = m= m [using(i)] CP= BE 15 5 Now. CD =10m AD=13m and BC=14 Draw ECIIAD and CP ⊥ EB Now CE =AD=13m and EB=AB-AE= (25-10) m= 15m In ∆ BCE.2) =15. we get area of quad ABCD =(6+9. S = = 6cm 2 ∴ Area of ∆ABC= 6(6 − 3)(6 − 4)(6 − 5) = 6 × 3 × 2 × 1 sq cm =6sq cm → (ii ) For ∆ ACD. If its sides are in the ratio 1:3:2. 5cm (2) and 5cm.24] The triangular side’s walls of a flyover have been used for advertisements. If its sides are in the ratio 1:3:2. (b) 48 sq cm (c) 50 sq cm (d) 30 sq cm One of the diagonals of a rhombus is 12cm sides area is 54 sq cm. (b) 12 (c) 18 (d) none of these If the perimeter of a rhombus is 20cm and one of the diagonals is 8cm. The perimeter of a triangle is 60cm. then its smallest side is (a) 15 (b) 5 (c) 10 (d) none of these.X Mathematics (Heron’s formula) 1. (1) 2. (3) 10. The advertisement yield on earning of Rs 5000 per m2 per year. If the sides of the wall area 15m. One of its sides wall has been painted in some colour (2) with a message “KEEF THE CITY GREEN AND CLEAN”. Find the area painted in colour. 70m and 90m at Rs 8 per square meter. 8. The (2) sides of the walls are 122m. the perimeter (1) of the rhombus is (a) 72 cm 5. (d) (2) 5 = 2. 9. 22m and 120m. then its (1) largest side is (a) 6 3. The perimeter of a triangle is 36cm. 11m and 6m. Find the perimeter and area of a triangle whose sides are of length 2cm. [use 6. How much rent did it pay? 7. The area (1) of the rhombus is (a) 24 sq cm 4. Find (3) 120 . (b) 3 10 cm (c) 6 10 cm 12 10cm Find the cost of leveling the ground in the form of a triangle having its sides are 40m.CBSE TEST PAPER-02 CLASS . A company hired one of its walls for 4 months. The perimeter of a triangle is 450m and its sides are in the ratio of 13:12:5. Find the base of an isosceles triangles whose area is 12cm and the length of one of the equal side is 5cm. There is a slide in a park. 11. 12. Then they cleaned the area enclosed within their lanes. Verify your result by using Heron’s formula. If AB=9m. BC=40m. If its hypotenuse is 10 cm. The sides of a triangle are 39cm. One group walked through the lanes AB. 13. 121 (5) .the area of the triangle. find the other two sides. A parallelogram stands on the (3) greatest side of the triangle and has the same area as that of the triangle. They walked through the lanes in two groups. CD and DA [fig1. while group cleaned more area and by how much? Find also the total area cleaned by the students. The perimeter of a right triangle is 24 cm.1]. DA=28m and ∠B =90o. The students of a school staged a rally for (3) cleanliness campaign. 42cm and 45cm. Find the height of the parallelogram. CD=15m. BC and CA whiles the other group through the lanes AC. Find its area by using the formula area of a right triangle. (c) Ans03. Rent = Rs 5000/sq m per year 5000 × 1320 × 4 ∴ Rent for 4 month = Rs 12 = Rs 22. As.X Mathematics (Heron’s formula) [ANSWERS] Ans01.000 122 .24 ) sq m = 1344 sq m ∴ Cost of leveling the ground = Rs ( 8 × 1344 ) =Rs 10752 Ans06.00. Here 40 + 70 + 90 m 2 = 100 m S= ∴ Area of a triangular ground = = 100 (100 − 40) (100-70) (100-90) sq m 100 × 60 × 30 ×10 sq m = (10 × 10 × 6 5) sq m = ( 600 × 2. (d) Ans05. 1202 + 222 = 14400 +484 = 14884 = (122)2 ∴ Walls are in the form of right triangles 1 Area of one wall = × Base × height 2 1 = × 120 × 22 sq m 2 = 1320 sq m. 22m and 120m. (a) Ans04. (c) Ans02. The lengths of the sides of the walls are 122m.CBSE TEST PAPER-02 CLASS . b = 5cm and c = 5cm ∴ Perimeter = a+ b+ c = (2 + 5 + 5) = 12 cm S = semi perimeter 12 = = 6 cm 2 using Heron’s formula. we have a 2 (100 − a 2 ) = 144 × 16 or or = 2304 a − 100a + 2300 = 0 ( a 2 − 64 ) ( a 2 − 36 ) = 0 4 2 ∴ Either a2 = 64 i. ∴ The sides of the wall is in the triangular from with sides. Here.9 sq cm Ans08.e. b = 6 m and c = 11 m 15 + 6 + 11 m ∴S= 2 = 16 m ∴ Area to be painted in colour = Area of the side wall = s ( s − a) (s-b) (s-c) sq cm = 16 (16 − 5) (16-6) (16-11) sq m Ans09. = 4 50 sq m = 20 2 sq m ∴ Area of an isosceles triangle = a 4b 2 − a 2 4 a or 100 − a 2 4 ∴ a 4b 2 − a 2 4 = 12 = 12 Squaring both sides. ∴ Area of triangle = = s ( s − a) (s-b) (s-c) sq cm 6 (6-2) (6-5) (6-5) sq cm = 24 sq cm = 4. A = 15 m. a = ± 8 or a2 = 36 i.e.Ans07. a = 2cm. a = ± 6 ∴ Required base = 8 cm or 6 cm 123 . and 5 × 15 I. Let the sides of the triangle be 13x. Area of //gm BCDE = Area of ∆ ABC h × BC = 756 ∴ or 45h = 756 756 h = 45 h = 16. 180 m and 75 m a+b+c 450 S= = = 225 m ∴ 2 2 ∴ Area of the triangle = Ans11. 195 m. To find the area of ∆ ABC 45 + 42 + 39 cm S= 2 = 63 cm ∴ Area of ∆ ABC = 63 (63-45) (63-42) (63-39) sq cm = 63 ×18 × 21× 24 sq cm = 9 × 7 × 2 × 3 ×2 sq cm = 756 sq cm Let h be the height of the parallogram Now. right angle ∆ ABC.Ans10. 12x and 5x Perimeter of a triangle = 450 m ∴ 13x + 12x + 5x = 450 m or 30x = 450 x = 15 ∴ ∴ The sides are 13 × 15. AC2 =AB2 + BC2 AC2 = 92 + 402 AC2= 1680 AC = 41 124 .8 cm Ans12. height of the //gm = 16. ∴ We have. s ( s − a) (s-b) (s-c) sq m = 225 (225-195) (225-180) (225-75) sq m = 225 × 30 × 45 × 150 sq m = (15 × 15 × 2 × 3 × 5) sq m = 6750 sq m.8 cm Hence. 12 × 15.e. BC = y cm and AC = 10 cm ∴ By the given condition. 28 + 15 + 41 S= 2 = 42 m ∴ Area of ∆ ACD = 42 (42-28) (42-15) (42-41) sq m = 42 × 14 × 27 × 1 sq m = 7 × 3 × 2 × 7 × 2 × 9 × 3 sq m = 126 sq m ∴ First group cleaned more = (180 – 126) sq m = 54 sq m ∴ Total area cleaned by students = (180 + 126) sq m = 306 sq m Ans13. Let x and y be the two lines of the right ∠ . DC=15m and AC=41 Hence. 1 Area of ∆ ABC = × 40 m × 9 m 2 = 180 sq m The second group has to clean the area of ∆ ACD which has AD=28 m. ∴ 100 + 2xy =196 96 2 = 48 sq cm from (2) xy = (3) 125 . Perimeter = 24 cm Or x + y = 14 (1) By Pythagoras theorem. ∴ AB = x cm. x 2 + y 2 = (10)2 = 100 (2) x 2 + y 2 = (14)2 Or x 2 + y 2 + 2xy = 196 From (1).The first group has to clean the area of ∆ ABC which is right triangled Now. (x-y) 2 = (x+y) 2 – 4xy = (14)2 . x-y = 2 and x+y = 14. X – y = -2 and x + y = 14. 8 cm and 10 cm 24 S= = 12 cm 2 Area of ∆ ABC = 12 (12-6) (12-8) (12-10) sq cm = 12 × 6 × 4 × 2 sq cm = 24 sq cm Which is same as found in (4) Thus. then 2x = 16 or x = 8. y = 6 (ii) When. 126 . y = 8 Verification by using Heron’s formula:Sides are 6 cm.1 xy sq cm 2 1 = × 48 sq cm 2 =24 sq cm Area of ∆ ABC = (4) again. then 2x = 12 or x = 6. the result is verified.4 × 48 or x-y = ± 2 (i) When. the [1] area of the rhombus is 4. (A) 40sq cm (B) 24sq cm (C) 20sq cm (D) 13sq cm. Its area is. One of the diagonals of a rhombus is 12 cm and Its area is 54sq cm. 14m? [2] 6. The side of an equilateral triangle is 4 3cm . and the [2] perimeter is 40 cm. A traffic signal board indicating ‘school ahead’ is an equilateral triangle with side ‘a’ find the area of the signal board using herons. 5. (A) 10 cm (B) 8 cm (C) 6 cm (D) 12 10 cm.CBSE TEST PAPER-03 CLASS . Find the area of isosceles triangle whose side is 14 m. 9. at Rs 7 per square meter. what will be Its area? 127 [3] . (A) 12 3 sq cm 12 [1] 12 (B) 10 sq cm (D) 6 [1] 6 sq cm 10 sq cm. 50 cm. 16 cm. find the area of the rhombus of Its [2] diagonal BD measures 16 cm? 7. The perimeter of a rhombus ABCD is 60 cm. 12 m. It the perimeter of a rhombus is 20sq cm and one of the diagonals is 8 cm. 8.X Mathematics (Heron’s formula) 1. and 20 cm. Its area is (A) 100sq cm (C) 2. Find the area of a triangle two side of which are 18 cm. (C) 3. (B) 90sq cm 96sq cm (D) 120sq cm. Find the cost of leveling the ground in the from of a triangle having Its side as 70 [2] cm. Its perimeter is 180 cm. the [1] perimeter of the rhombus is. and 12 cm. The side of a triangle is 12 cm. and 60 cm. and 60m find the cost of leveling the [3] field Rs 18 per meter. 13. and 40m has one diagonal [3] 75m long. A floral design of a floor is made up of 16 tiles which are triangular. 12. 56m. find the cost of polishing the tiles. 128 [5] . is a space of 4cm is to be left for entry gate. Find the area of the parallelogram? 11. at RS 50 paisa/sq cm. The side of a triangular find is 52m.1 find the total area of the paper wed. Radha made a picture of an aero plane with colored paper as shown in fig 1. and 35 cm.10. 28 cm. The side of [3] the triangle being 9 cm. An parallelogram the length of whose sides are 80m. (B) Ans04.6 = 2 0 3 .6 sq m = 75.X Mathematics (Heron’s formula) [ANSWERS] Ans01 (C) Ans02.2 sq cm 129 .6 sq cm area o f rh o m b u s = 2 × 1 0 1 .6 sq m Ans06.CBSE TEST PAPER-03 CLASS . ∴ AB=BC=CD=D A= 60 = 1 5 cm 4 in ∆ A B D . (A) Ans03.7 = 1 0 1 . 14 + 12 + 14 = 20cm 2 Area of isosclese triangle = s ( s − a)( s − b)( s − c) = 20(20 − 14)(20 − 12)(20 − 14) sq m = 20 × 6 × 8 × 6 sq m = 6 160 =6 × 12. (D) S= Ans05. As side of rhombus are equal. S = 15 + 1 5 1 6 = 23 cm 2 so A rea o f ∆ A B C = 2 3( 23 − 1 5)( 2 3 − 1 5)( 2 3 − 1 6 ) = 23 × 8 × 8 × 7 = 8 23 × 7 = 8 × 1 2 . 56 sq cm Ans09. p erim eter = 1 8 0 c m 180 ∴ e a c h s id e = = 60cm 3 = 3 (6 0 ) 2 sq c m 4 = 900 3 sq cm ∴ A re a o f s ig n a l b o a rd = Ans10. S= 70 + 50 + 60 180 = = 90m 2 2 ∴ area of triangle = 90(90 − 70)(90 − 50)(90 − 60) = 90 × 20 × 40 × 30 = 1469. S= 80 + 40 + 75 = 97.7) = 10287. b=12 cm and C=? So a+b+c=40 cm 18+12+C=40 C=(40-30) cm =10 cm 18 + 12 + 10 =20 cm 2 ∴ area of triangle = 20(20 − 18)(20 − 12)(20 − 10) ∴ S= = 20 × 2 × 8 × 10 sq cm = 56. S= a+a+a 3a units units 2 2 ∴ A r e a o f t r i a n g le = = 3a 3a 3a 3a ×( − a )( − a )( − a) 2 2 2 2 3a a a a × × × 2 2 2 2 a2 3 s q u n its 4 N o w .7 sq m ∴ cost of levelling the ground=RS (7 ×1469. AB = DC = 80cm BC = AD = 40cm and AC=75 cm In ∆ABC.9 Ans08 Let a=18 cm.5 2 130 .Ans07. 5(97.5 × 57.5 − 40)(97.Area of triangle = s( s − a)( s − b)( s − c) = 97. we have 35 + 28 + 9 cm=36 cm 2 ∴ Area of Each tile = s ( s − a )( s − b)( s − c) S= = 36(36 − 35)(36 − 28)(36 − 9) sq cm = 36 6 sq cm ∴ Area of 16 tile= 16 × 36 3 sq cm 1 ∴ cost of polishing = Rs [ × 16 × 36 6] =Rs=288 6 2 = Rs(288 × 2.75 sq m Area of 11gm ABCD=2 × Area of ∆ACD = 2 × 1485.94 =1485.7 = 2971.60 131 .5 × 22.5 − 75) sq m = 97.5 − 80)(97. 56m and 60m.4 sq m Ans11.5 × 17.5 sq m = 2207460. side of the field are 52m. 52 + 56 + 60 =84m 2 ∴ Area of field = 84(84 − 52)(84 − 56)(84 − 60) sq m ∴ S= = (7 × 12)(2 × 16(4 × 7)(12 × 2) sq m = 7 × 7 ×12 × 12 × 2 × 2 × 4 × 16 sq m = 1344 sqm ∴ total cost of levelling the field = Rs5 ×1344 = Rs 6720 space to be left for entry gate =4 cm ∴ space to be fenced =168-4=164m cost of fencing 1 m =Rs 18 ∴ total cost of fencing the field = Rs 18 × 164 = Rs 2952 Ans12.45) =Rs 705.75 sq m = 1485. for Each triangalar tile. 196 × (1) 2 sq cm = or sq cm 4 4 4 = 1.5+1.3+9) sq cm = 19.5 sq cm Area (III) =Area of trapezium = 3 × Area of equilational ∆ with side =1 cm 3 3 ×1.5cm and b=1 cm = 6.Ans13.3 sq cm (approx) =3 × 1 Area of (IV+V) =2 × × 6 ×1.5+6. Area (1) = area of isosceles triangle with a=1 cm and b= 5 cm a 4b 2 − a 2 4 1 99 = 100 − 1 = sq cm (approx) 4 4 Area (II) = area of rectangle with = L= 6.5 ×1 sq cm =6.3 sq cm 132 .5 sq cm = 9sq cm 2 ∴ total area of the paper used =Area (I+II+III+IV) = (2.732 5. An equilateral triangle has a side 50cm long. 56m and 70m? [2] 6. After growing the vegetables in it. The lengths of the side of a triangular park are 90m. And AC=7 cm. The perimeter of a rhombus ABCD is 80 cm.X Mathematics (Heron’s formula) 1. The [3] side of the walls are 122 m. DA [3] =7cm. The measure of one side of a right triangle is 42m. 11. (A) 1340 sq m (B) 1344 sq m (C) 1440 sq m (D) 1444 sq m [1] 2. find the measure of two unknown side? [3] 10. Find the area of isosceles triangle whose equal side is 6 cm. Find the area of the garden which he has to look after daily. The triangular side walls of a flyover have been used for advertisements. Find the area of triangle whose side is 42m. He wants to divide it in seven equal parts And look after each part once a week. (A) 404. find the area of rhombus if Its [3] diagonal BD measures 12 cm. 22m And 120m. Find the area of equilateral triangle whose side is 12 cm. 133 [3] . Find the area of an isosceles triangle . [1] (A) 625 3 sq cm (B) 625 6 sq m (C) 256 6 sq m (D) 625 10 sq m 3.9 sq cm (B) 405. The area of an isosceles triangle is 12 sq cm.CBSE TEST PAPER-04 CLASS . Shashi Kant has a vegetable garden in the shape of a rhombus. A company hired one of Its walls for 6 months How much rent did It pay? 13. 70m and 40m. Find the area of triangle whose side is 6 cm. 10 cm and 15cm. It the difference in lengths of Its hypotenuse and other side is 14 cm. It one of the equal side is 5 cm. The advertisements yield a caning of Rs 3000 per M2 per year. 6 cm and 8 cm. 12. [2] 8. Find area of quadrilateral ABCD in which AB= 5 cm. 7.9 sq cm (C) 402.9 sq cm (D) 410sq cm [1] [1] 5. The length of each side of garden is 35 m And Its diagonal is 42 m long. BC= 6 cm. then the length of the base is (A) 4 cm (B) 5 cm (C) 6 cm (D) 8 cm 4. CD 6 cm. find Its area. [2] 9. Find the area of the triangles. the measure of one of Its equal side being [2] b and the third side a. (a) Ans03.CBSE TEST PAPER-04 CLASS . (a) Ans05. (b) Ans02. S= a+b+b a + 2b units = units 2 2 ∴ Area of triangle = a + 2b a a a × × × units 2 2 2 2 = = Ans07. a + 2b a + 2b a + 2b a + 2b × − a − a − a units 2 2 2 2 a 4 4b − a 2 2 sq units 12 + 12 + 12 cm 2 36 = cm = 18 cm 2 S= ∴ Area of equilateral = s (s-a) (s-b) (s-c) = 18 (18-12) (18-12) (18-12) = 18 × 6 × 6 × 6 = 36 3 sq cm 134 .X Mathematics (Heron’s formula) [ANSWERS] Ans01. S= 168 42 + 56 + 70 m = m or 84 2 2 ∴ Area of ∆ ABC = = s (s-a) (s-b) (s-c) 84 (84-42) (84-56) (84-70) sq m = 42 × 28 sq m = 1176sq m Ans06. (c) Ans04. 4 sq cm = 228.8 sq cm Ans09.6+ 6+8 cm 2 20 = = 10 cm 2 Ans08. Area of ∆ ABD = 26 × 6 × 6 ×14 sq cm = 114. ∴ S= 20 + 20 + 12 2 so.4 sq cm Area of rhombus = 2 × area of ∆ABD = 2 × 114. ∴ AB = BC = CD = DA = 80 = 20 cm 4 Now in ∆ ABD.8 sq cm 135 (iii) . Let AB = y and AC = x and BC = 42 cm ∴ By the given condition. we get 2x = 140 i. x = 70 y = 126-x ∴ y = 126 – 70 = 56 Ans10.y = 14 (i) By Pythagoras theorem.e. x2 – y2 = 1764 (x + y) (x – y) = 1764 14 (x + y) = 1764 using (ii) ∴ 1764 x+y= = 126 ∴ 14 Adding (ii) and (iii). x . S= ∴ Area of isosceles triangle = 80 (10-6) (10-6) (10-8) = 10 × 4 × 4 × 2 sq cm = 17. 9) sq cm = (14. 22m and 120m So.3 sq cm. S= 5+6+7 = 9 cm 2 ∴ Area of ∆ ABC = 9 (9 − 5)(9 − 6)(9 − 7) sq cm 9 × 4 × 3 × 2 sq cm = = 6 6 sq cm In ∆ ACD. Area of one wall = 1 × 120 × 22 2 = 1320 sq m Rent = Rs 3000/sq m per year 3000 × 1320 × 6 ∴ Rent for 6 months = Rs 12 = Rs 198000 136 (i) . 1202 + 222 = 14400 + 484 = (122)2 ∴ Walls are in the form of right triangles.9 sq cm Area of quadrilateral ABCD = ( 6 6 + 18. S= 7+7+6 = 10 cm 2 ∴ Area of ∆ ACD = = 10 (10 − 7)(10 − 7)(10 − 6) sq cm 10 × 3 × 3 × 4 sq cm = 18. Area of quadrilateral ABCD = Area of ∆ ABC + Area of ∆ ACD In ∆ ABC. The lengths of the sides are 122m.9) sq cm = 33.Ans11. Ans12.4 + 18. . .Ans13. . Let ABCD be garden ∴ DC = 35 m . . ∴ DE = 1 1 DB = × 42 2 2 or 21 m Now CE2 = CD2 – DE2 = 352 – 212 = 784 CE = 28 m Now area of ∆ DBC. daily = 1176 sq m 2 = 168 sq m 137 . . (Given) Draw CE ⊥ DB We know that the diagonals of a rhombus bisect each other at right angles. = = 1 × DB × CE 2 1 × 42 × 28 2 = 588 sq m ∴ Area of the garden ABCD = 2 × 588 sq m = 1176 sq m Area of the garden he has to look after. . . . . (Given) DB = 42 m . . [2] 7. Its area is (a) 18 3 sq cm (b) 72 3 sq cm (c) 64 3 sq cm (d) 60 3 sq cm If area of isosceles triangle is 48 sq cm and length of one of its equal sides is 10 [1] [1] m.X Mathematics (Heron’s formula) 1. (a) 16 cm or 12 cm (b) 12 cm or 14 cm (c) 14 cm or 16 cm (d) 16 cm or 18 cm If AB = 14 cm. 6. 4 cm and 10 cm? [2] 8. find area of triangle whose sides are 6 cm. If side of equilateral triangle is 25 m. Its area is (a) 5 2 3 sq cm (b) 5 4 3 sq cm (c) 5 3 sq cm (d) [1] 3 sq cm The perimeter of an equilateral triangle is 48 cm. Find the area of equilateral triangle the length of one of its sides being 24 cm. CD = 17 cm. DA = 8 cm and AC = 15 cm then area of [1] quadrilateral ABCD is 5. Find the cost of leveling the ground in the form of equilateral triangle whose side is 12 m at Rs 5 per square meter. BC = 13 cm. [3] Find the area of triangle? 10. The perimeter of a triangle is 480 meters and its sides are in the ratio of 1:2:3. Find the perimeter and area of a triangle whose sides are 3 cm. 3. 2. 138 [3] . (a) 150 sq cm (b) 144 sq cm (c) 142 sq cm (d) 140 sq cm Find the area of an isosceles triangles. 8 cm and 10 [2] cm? 9. the measure of one of its equal sides [2] being 10 cm and the third side is 6 cm.CBSE TEST PAPER-05 CLASS . Using Heron’s formula. then what is the base? 4. 36 m and 24 [3] respectively. How much paper of each shade has been used in it? ( use 5 = 2.24) 12. Find the area of a triangle whose perimeter is 180 cm and two of its sides are 80 cm and 18 cm. taken in order are 29 m. hence calculate the altitude of the triangle taking the longest side as base. Find its area. The sides of a quadrangular field. The angle contained by the last two sides is a right angle. A kite in the shape of a square with [3] diagonal 32 cm and an isosceles triangle of base 8 cm and side 6 cm each is to be made of three different shades. 13. 139 [3] .11. X Mathematics (Heron’s formula) [ANSWERS] Ans01. a = b = c = 24 cm 24 + 24 + 24 72 ∴ S= cm = cm 2 2 = 36 cm ∴ Area of triangle = 36 (36-24) (36-24) (36-24) sq cm = 246. (a) Ans02.12 sq cm Ans07. Perimeter = 3+4+5 = 12 cm ∴ 12 2 = 6 cm S = semiperimeter = or area of triangle = 6 (6-3) (6-4) (6-5) sq cm = 6 sq cm 140 . (b) 10 + 6 16 = 2 2 = 8 cm Ans05. S= ∴Area if tringle = 8 (8-5) (8-5) (8-6) sq cm = 8 × 3 × 3 × 2 sq cm = 12 sq cm Ans06.CBSE TEST PAPER-05 CLASS . (a) Ans04. (c) Ans03. 12 + 12 + 12 ∴ S= 2 = 18 cm And. ∴ Area of triangle = s (s-a) (s-b) (s-c) sq m = 240 (240-80) (240-160) (240-240) sq m = 0 sq cm ∴ Triangles don’t exist with the ratio 1:2:3 whose perimeter is 480 m.73) = Rs 311. Ans10. ∴ Area of equilateral triangler = s (s-a) (s-b) (s-c) sq m = 18 (18-12) (18-12) (18-12) sq m = 18 × 6 × 6 × 6 sq m = 6 × 3 × 6 × 6 × 6 sq m = 36 3 sq m ∴ Cost of leveling ground = Rs (5 × 36 × 1. 12 m.6 + 8 + 10 24 = 2 2 = 12 cm S= Ans08. 3x Perimeter of the triangle = 480 m x + 2x + 3x = 480 m ∴ 6x = 480 m x = 80 m ∴ The sides are 80 m. 240 m So. 12 m.4 m 141 . Let the sides of the triangle be x. Here. 2x. 160 m. 80 + 160 + 240 480 S= = 2 2 = 240 m And. ∴ Area of triangle = 12 (12-6) (12-8) (12-10)sq cm = 24 sq cm Ans09. sides are 12 m. Ans11.92 sq cm Thus. 256 sq cm and 17. (II) and (III) we get Area of quad ABCD = (84+360) sq m = 444 sq m 142 .92 sq cm. In ∆ ADC. a 4b 2 − a 2 4 8 Area of portion III = = 4 × (6)2 − 8 4 = 17. Let the diagonals intersect each other at O. 1 AO = × 32 cm 2 = 16 cm 1 Area of shaded portion I = × 16 × 32 sq cm 2 = 256 sq cm And. which are right angles? ∴ AC2 = AD 2 + DC 2 = (24) 2 + (7) 2 = 576+49=625 = ∴ AC = 625 = 25 Area of quad ABCD =Area of ∆ ADC+ Area of ∆ ABC → (I) 1 1 AD × CD = × 24 × 7 sq m Area of ∆ ADC= 2 2 = 84 sq m → (II) For ∆ ABC. Let ABCD be the square and ∆ CEF be an isosceles triangles.= 45(45 − 36)(46 − 25)(46 − 29) ∴ S= = 360 sq m → (III) From (I). 36 + 25 + 29 M = 45M 2 ∴ Area of ∆ABC. Ans12. Then. the papers of three shades required are 256 sq cm. 1 ∴ Area of triangle = × 82 × hcm 2 1 720 = × 82 × h cm 2 720 h= cm 41 23 h= 17 cm 41 Hence. And length required altitude is 17 231 cm.Ans13. Area of triangle is 720 sq cm. S= = 90 cm 2 ∴ Area of triangle = 90(90 − 80)(90 − 18)(90 − 82) = 90 × 10 × 72 × 8 sq cm = 720 sq m The longest side of triangle is 82 cm Let h cm be length of altitude corresponding to longest side. perimeter = 180 cm ∴ 180= a+b+c =80+18+c C= 82 cm 180 Now. Let a= 80 cm and b=18cm. 41 143 . 55. from 1 to 5 carries 1 mark each. No. 45. 2. No. Find mean of the following data. 4.50 hrs. The following data has been arranged in ascending order. Xi 5 10 15 20 35 Fi 3 5 7 7 9 144 . If the media of the data is 23.6. 26. 6. 36. 32. 27.x. 3. 20. In triangle ABC.5. 28.7.CBSE MIXED TEST PAPER-01 (Unit Test) CLASS . 5. For what value of x.: 40] General Instructions:- 1.8. No.6. from 10 to 14 carries 3 marks each. 17. from 6 to 9 carries 2 marks each. No. 15.IX MATHEMATICS [Time : 1. the mode of following data is 7. What is the mean of first five prime numbers. 3. M. 36.5 cm. (v) Q.] [M. (iii) Q.7. from 15 to 16 carries 6 marks each Determine the range of the following data. AC = 4. 75. (iv) Q. 80. 25. 90. 7. State congruence Axiom. 12.5 cm. 14. (i) All questions are compulsory. (ii) Q.4. 95. Name the greatest and the smallest angle. AB = 6.7. 22.5. find x. BC = 5cm.5. x. 20 35 60 45 75 90 50 43 80 70 72 37 60 60 40 47 89 93 15 73 51 45 87 68 29 145 . In fig. of Employess 50-59 45 60-69 12 70-19 33 80-89 36 90-99 24 Find a. Show that BF = CE. Class size of 4th class interval 11. OA = OB. triangle AOD ≅ triangle BOC ii. E and F are the mid points of equal sides AB and AC of ABC. 10. AD parallel BC 9. True limits of 2nd class interval c. and OC = OD Show that i. Wages (in Rs) No. Below are the marks obtained by 25 students of class IX in Maths. Lower limit of 3rd class interval b. The following table shows the frequency distribution of daily wages of employees of a company.8. Prepare a grouped frequency distribution table from the above data taking 10 as class size & one of the class intervals be 20-30. 7. angles opposite to equal sides are equal. Prove that.7. 9. 7. 13. 12. Age(in yrs) Frequency 0-5 2 5-10 4 10-15 5 15-20 7 20-25 9 25-30 9 3-35 6 35-40 5 40-45 3 45-50 1 146 . 14. (30 not included) 12. in an isosceles triangle . 13. 12. 7. angle B < angle A and angle C < angle D. prove the following result: In fig. Find median and mode of the following data. Show that AD < BC. Construct a frequency polygon for the following data. in which AB = AC. 18. 12. Using above theorem. Show that (i) AD bisects BC (ii) AD bisects angel A In fig. if AB = AC and DB = DC then show that angle ABD = angle ACD 16. 15. 15. AD is the altitude of an isosceles triangle ABC. 25. 7. 147 . Prove that the sum of the angles of a triangle is 1800. 1.IX MATHEMATICS [Time : 1. Ray OR is perpendicular to line PQ. In figure. (ii) There is no overall choice.50 hrs.CBSE MIXED TEST PAPER-02 (Unit Test) CLASS . Prove that angle ROS = ½ [angle QOS – angle POS] 4 marks 3. State and prove ASA congruence rule. In figure POQ is a line. If the bisector of angle PQR and 5. However internal choice has been provided. Does Euclid’s fifth postulate imply the existence of parallel lines? Explain. the side QR of triangle PQR is produced to a point S. then prove that angel QTR = ½ angle QPR 4 marks.: 40] General Instructions:(i) All questions are compulsory. OS is another ray lying between ray OP and OR. M.] [M. 4 marks. 4 marks 4. (iii) Marks allotted to each question are indicated against it. PRS meet at point T. 4 marks 2. 4 marks 8. 17. 14. 18. 25. AB is a line segment and P is its mid point. Prove that the diagonals of a parallelogram bisect each other. 14.6. 18. 14. is a square. 4 marks 148 . In fig AB and CD are respectively the smallest and the longest sides of a quadrilateral AB CD show that angle A > angle C and angle B > angle D. 23. 4 marks 9. In fig. 22. 28. D and E are pints on same side of AB such that angle BAD = angle ABE and angle EPA = angel DPB show that 4 marks (i) triangle DAP ≅ triangle (ii) AD = BE 7. 18. 4 marks Find the mode of 14. Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles then it 10. 235 can be expressed in p/q form where p and q are integers and q≠0 Q2.621621…. (ii) There is no overall choice. However internal choice has been provided.235353535 = 0.621 can be expressed in p/q form where p and q are integers and q≠0 (b) Show that 0. Find the reminder when x3 + 3x2 + 3 x + 1 is divided by (i) x + 1 (ii) x + Π Q5. (iii) Questions 1 to 10 carry 3 marks..: 100] General Instructions:(i) All questions are compulsory. Solve the equation graphically 2x + 3y = 7 and 3y + 2x = 8.. = 1. Q1. (v) Questions 21 to 26 carry 5 marks. (a) Show that 1.00 hrs.CBSE MIXED TEST PAPER-03 (complete test paper) CLASS . (iv) Questions 11 to 20 carry 4 marks. Rationalize the denominator of 5− 4 2 6 −4 2 Q3. 149 .IX MATHEMATICS [Time : 3. P (1) and P(2) for each of the following polynomials (i)P(y) = y2 – y + 1 (ii) P(t) = -t3 + 2t2 + t+ 2 Q4. M. Find P(0).] [M. Q6. In the fig. bisector of angle B and angle D of quadrilateral ABCD meet CD and AB produced at P and Q respectively prove that: angle P + angle Q = ½ (angle ABC + angle ADC) Q7. In the fig. PR>PQ and PS bisect angle QPR. Prove that angel PSR > angle PSQ Q8. Two parallel lines I and m are intersected by transversal p. Show that the quadrilateral formed by bisector of interior angels is a rectangle. Q9. Prove two triangles on the same base and between same parallel lines are equal in area. Q10. A field is in shape of trapezium those parallel sides are 25 m and 10m. The non-parallel sides are 14 m and 13 m. Find the area of trapezium. Q11. Factorize (a) x3 – 23x2 + 142 x – 120 (b) 8a3 + b3 + 12 a2b + 6ab2 Q12. Evaluate the following using the suitable identities. (a) (998)3 (b) (2a – 3b)3 150 Q13. In the fig . POQ is a line. RAY OR is perpendicular to PQ. OS is another ray lying between ray OP and OR. Prove that angle ROS = ½ (angle QOS – angle POS) Q14. ABCD is quadrilateral in which AD = BC and angle DAB = CBA. Prove that (i) triangle ABD ≅ triangle BAC (ii) BD = AC (iii) angle ABD = BAC Q15. (a) Show that the bisectors of angles of parallelogram form a rectangle. (b) Show that the diagonals of a square are equal and bisect each other at right angles. Q16. (a) Prove that the median of a triangle divides a triangle into two triangles of equal area. (b) In the fig. side QR of triangle PQR is produced to point to S. If the bisector of angle PQR and PRS intersect at point P than prove that angel QTR = ½ angle QPR. Q17. The sides AB of a parallelogram ABCD is produced to any point P, a line through A and parallel to CP meets CB produced at Q and then parallelogram PQBR is completed. Show that ar(ABCD) = ar(PBQR). Q18. (a)In an triangle ABC,E is the mid point of AD. Show that ar(triangle BED) = ¼ ar(triangle ABC) (b) In the fig. D is a point on side BC of triangle ABC. Such that AD = AC. Show that AB > AD. Q20. In a parallelogram ABCD, E and F are the mid point of sides AB and CD respectively. Show that the line segment AF and EC trisect the diagonal BD. 151 Q21. (a) Prove that the um of the angles of a triangle is 1800. (b) Find angle x and angle y in the following figures. Q22. (a) A rhombus shaped field has green grass for 18 cows to graze. If each side of rhombus is 30 m and longer diagonal is 48m. How much area of grass field will each cow be grazing? (b)An umbrella is made by stitching 10 triangular pieces of cloth of two different colors, each piece measuring 20m, 50cm, and 50cm. How much cloth of each color is required for the umbrella? Q23. (a) In the fig. P is a point in the interior of a parallelogram ABCD. Show that (b) (i) ar(triangle APB) + ar(triangle PCD) = ½ ar(ABCD) (ii) ar(triangle APB) + ar(APCD) = ar(triangle APD) + ar(triangle PBC) In the Fig. ar(triangle DRC) = ar(triangle DPC) and ar(triangle BDP) = ar (triangle ARC). Show that both the quadrilaterals ABCD and DCPR are trapeziums. 152 Q24. (a)Prove that line segment joining the mid point of two sides of a triangle is parallel to the third side and is half of it. (b)ABCD is a trapezium in which AB parallel CD and AD = BC. Show that (i)Angle A = Angle B (ii) Angle C = Angle D (iii) Triangle ABC ≅ triangle BAD (iv) Diagonal AC = Diagonal BD. Q25. (a) Triangle ABC and Triangle DBC are two isosceles triangles on the same base BC. Show that triangle ABD ≅ triangle ACD (b) AD is altitude of an isosceles tingle ABC in which AB = AC show that (i) AD bisects BC (ii)AD bisects A Q26. In the following figure the sides AB and AC of a triangle ABC are produced to point E and D respectively. If bisectors BO and CO of angle CBE and angle BCD meet at point O. Prove that angle BOC = 90 – ½ angle BAC . 153 CBSE MIXED TEST PAPER-04 (Unit test paper) CLASS - IX MATHEMATICS [Time : 2.00 hrs.] [M. M.: 40] General Instructions:(i) All questions are compulsory. (ii) There is no overall choice. However internal choice has been provided. (iii) There are three Section – A, B & C in question paper Section A : Q no. 1 to 5 carry 2 marks each. Section B : Q. no. 6 to 11 carry 3 marks each. Section C : Q no. 12 to 14 carry 4 marks each. 3 4 and 5 5 1. Find two rational numbers between 2. Verify whether the following are zeroes of the polynomial: P ( x ) = lx + m, x = − 3. In which quadrant of the point (-2,4), (3,-1) lie? 4. In figure, if AC = BD, then prove that AB = CD. 154 m l 5. In quadrilateral ABCD, AC = AD and AB bisects angle A, show that: triangle ABC ≅ ABC 6. Express 0.6 in the form 7. Rational the denominator of 8. Evaluate using identity 104 x 106. 9. Draw the graph of x + y = 7. p where p and q are integers and q ≠ 0. q 1 . 7− 6 10. Show that in a right angled triangle, the hypotenuse is the longest side. 11. The angles of quadrilateral are in the ratio 3:5:9:13. Find all the angles of quadrilateral. 12. If x + y + z = 0, show that x3 + y3 + z3 = 3xyz. 13. Factories: X3 – 3x2 =-9x- 5 Or In fig. POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that: Angle ROS = ½ (angle QOS – angle POS) 14. Prove sum of the angles of a triangle is 1800. 155 CBSE MIXED TEST PAPER-05 (Unit test paper) CLASS - IX MATHEMATICS [Time : 2.00 hrs.] [M. M.: 40] General Instructions:(i) All questions are compulsory. (ii) There is no overall choice. However internal choice has been provided. (iii) There are three Section – A, B C D in question paper Section Section Section Section A : Q no. 1 to 5 carry 1 marks each. B : Q. no. 6 to 9 carry 2 marks each. C : Q no. 10 to 14 carry 3 marks each. D : Q no. 15 to 16 carry 6 marks each. 1. State the axiom used to prove AB + BC = AC in figure. 2. In train gel ABC and triangle PQR are to be made congruent in figure, Name one additional pair of corresponding part and the congruence criteria used. 3. Find ‘x’ and ‘y’ in figure 156 4. Arrange the data given under and find its range: 8, 7, 3, 9, 6, 2, 1, 1, 4, 4, 3 5. Find the mode of : 1, 2, 3, 4, 2, 3, 2, 4, 5 and 2. 6. The following distribution is arranged in ascending order. If the median of the data is 63. Then find the value of x : 29, 32, 48, 50, x, x + 2, 72 , 78, 84 , 95. Or The mean of the date given under is 9 find ‘x’ Xi 4 6 9 10 fi 5 10 10 x 8 15 7. In fig. x + y = w + z, then show that AOB is a line 8. Prove that every line segment has one and only one mid point. Or Prove that an equilateral triangle can be constructed on any give line segment . 9. In fig. PR > PQ and PS bisects angle PPR . Prove that angle PSR > angle PSQ. 10. Prove that the sum of three angles of a triangle is 1800. 11. Twenty student of class IX were asked about the number of hours they watched TV programmes in the previous week. The result was found as follows: 2 13 11 9 17 4 5 12 2 7 5 14 12 6 10 10 18 5 4 9 Make a grouped frequency table for this data taking class width 4 and one of the intervals as 4-8. 12. A random survey of number of persons of various age group present in a park was found as follows: 157 (iii) true class limit of the class with maximum frequency. 16.Age group: 1-10 11-20 21-30 31-40 41-50 No. If a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel. 15. ABCD is a quadrilateral in which AD = BC and angle DAB = angle CBA. 13. 158 . of persons 12 29 33 19 7 Answer the following (i) class size of the data (ii) class mark of 21-30. then prove that the two lines are parallel. 14. Or In fig. In a city the weekly observation made in a study on the cost of living index are given in table: Cost of living index 140 – 150 150-160 No. of weeks 7 10 160-170 14 170-180 12 180-190 6 190-200 Total 3 52 Draw histogram and frequency polygon of this data. Prove that BD = AC. show AB parallel EF. Prove that the angle opposite to equal sides of isosceles triangle is equal. Use it to show angle ABD = angle ACD when AB = AC and BD = CD Or Prove that: Two triangles are congruent if two angels and the included side of one triangle are equal to two angles and the included side of another triangle. Give an example such that sum of two irrational numbers is a rational number. 1-mark 2. x) Section . Write the degree of polynomial 2 – y2 – y3 + 2y8 1-mark 4. 1-mark 5. Why (x.IX MATHEMATICS [Time : 1. Represent. y) up to three decimal places.] [M. What do you mean by ordered pairs 2-marks 7. Section – A 1.: 40] General Instructions:(i) All questions are compulsory. (ii) There is no overall choice. 159 .B 6. However internal choice has been provided.5 hrs. Write the value of 1-mark 3. 1-mark (y.CBSE MIXED TEST PAPER-06 (Unit test paper) CLASS . Expand (1 + X)3 2-marks If x + y + z = 0 then write the value of x 8. M.25 on the number line. 1. 2-marks x + y3 + z3 – 3xyz. Represent 11. 3-marks P(x) = x2 – 5x + 6 Section . Using graph Check whether x = 1. Write the factors of 12 ky2 2-marks Section . on the number line upto 4 decimal places. 6-marks 2x + 3y = 12 on the same graph. 160 . on the number line. 3-marks can be expressed in the form. Also find zeros of polynomial 14. y = 2 is the solution of x + 2y = 5 State remainder theorem and factor theorem. 6-marks X3 – 23x2 + 142x – 120 Using factor theorem.9. Show that 12.D Draw the graph of X + 2y = 5 and 15. Visualize 13. Factorize 16.C 10. 3-marks 3-marks 3-marks Renationalize the denominators of Define zero of a polynomial. Factorise 3x2 – x – 4 2-marks Or Factorise 27y3 + 125 x3 6.: 40] General Instructions:(i) All questions are compulsory. Evaluate using identities 97 x 105. (ii) There is no overall choice.A 1.IX MATHEMATICS [Time : 1. Insert two rational and two irrational numbers between Represent on same number line. . 2-marks 161 . 2. 2-marks 7.] [M. Write two solutions of 2x + 3y = 4. 3. M. 2-marks 2-marks 2-marks (i) Rationalise (ii) Find 4. However internal choice has been provided. Section .CBSE MIXED TEST PAPER-07 (Unit test paper) CLASS .5 hrs. Find the value of polynomial 5x – 4x2 + 3 at x = -2 2-marks 5. If x-4 is a factor of 5x3 – 7x2 – kx – 28 find value of k.C 13. Section . 12. Q (5. Find remainder. using remainder theorem. Evaluate 4-marks 162 . R (7.6).B 9. Plot points P (0. Give geometrical representation of 2x + 9 = 0 as an equation 3-marks (i) In one variable (ii) In two variable Section .3) S(2.8.3) on cartesian plane and identify the 3-marks figure PQRS.0). Express 2-marks in form . Using factor theorem factorise x3 – 3x2 – 9x – 5 4-marks 14. 3-marks 11. 10. Draw graph of x + 3y = 6 and find points where it meets 4-marks (i) x-axis (ii) y-axis 15. which x3 + 3x2 + 3x + 1 is divided by 3-marks . Write a linear equation in two 1-marks variables for the statement. In figure.x2 – y2 1-marks 7.0 hrs. Explain parallel and perpendicular line with diagram. (ii) There is no overall choice. if AC = BD then prove that AB = CD 163 2-marks .IX MATHEMATICS [M. are two isosceles triangle on the same base BC. Check whether (4. Represent (625)-4 as positive number in lowest form. Find the product of 1-marks 3. 0) is a solution of equation x – 2y = 4 or not? 1-marks 8. Express 5 = 2x in the form of ax + by=c = 0 and indicate the value of a. 5. M. Factorize: . b and c. Find the value of 5x3-4x2+3 at x = 2 1-marks 4. 1-marks 10.: ] [Time : 3. However internal choice has been provided.CBSE MIXED TEST PAPER-08 (Unit test paper) CLASS . show 2-marks and that angle ABC = angle ACD 12. The cost of a note book is twice the cost of a pen. 1-marks 9. 1-marks 6. Check whether is form of p(x) = lx + m or not? 1-marks 11. 1. 1-marks Write the value of 2.] General Instructions:(i) All questions are compulsory. angle BAC = 350. In the given figure. angle CDE = 530 E. 17. Find the value of k if x-1 is a factor of 4x3 + 3x2 – 4x + x. AC = AE.13. 3-marks (2)Evaluate 0. Write all the postulates of Euclid in geometry. 23. 3-marks 22.265 (express in p/q form). AB and CD are respectively the smallest and largest side of quadrilateral ABCD 3-marks than show that <A> angle C and <B> angle D <A> (OR) If the bisectors of angle B and angle C of a triangle ABC meet at o the prove that BOC = 900 + angle BAC 20. other find 2-marks angle DCE = ? 16. Represent or on a number line. 3-marks 19. Give the geometrical representation of y = 3 in one variable and two variable on 3-marks your answer sheet. Triangle ABC is an isosules triangle in which AB = AC side BA is produced to D 164 3-marks . AB parallel DE. Find the value of (-15)3 + (8)3 + (7)3 without actual multiplication 2-marks 15. In figure. 2-marks 14. AB = AD and angle BAD = angle EAC that show that BC = DE 3-marks 21. (1)Find four rational number between 3 and 4. 3-marks Find a and B if 18. angel CNO = 45o 165 5-marks . Factories (i) 3x2 – x – 4 (ii) 4y2 – 4y + 1 3-marks 25. 5-marks 27.y2 + Z2 – 4xy – 2yZ + 4xZ using identify. (i)AT line XY and MN intersect at o and angle POY = 900 with a: b = 2:3 than find the valued a.such that AD = AB then show that angel BCD = 900 24. write a linear equation for it taxing the distance covred as x km and total fare a Rs y. 28. 30. The taxi fare in a city is comprises of the first kilometer fare an is 8 and far the 5-marks subsequent distance it is Rs5 per km. b. and c. Also draw its graph by taking at least four solution of it. 5-marks (ii)Find the remainder when x3 – 9x2 + 6x – a is divided by (x – a) (iii)Factories 4x2 . Evaluate the factors of (i) 27y3 + 125Z3 (ii) 8x3 + y3 + 27Z3 – 18xy2 using 3-marks suitable identify: 26. Factories p(x) = x3 + 13x2 + 32x + 20 completely using the factor theorem. Hence evaluate x in the figure if AB||CD and angle AMO = 50o. 5-marks 29. “The sum of any two side of a triangle is greater than the third side” prove it. “If a traversal intersects two parallel line than each pair of alternate introvert angle is equal” Prove it. 2. 26 to 30 carry 6 marks each. 3.3) and B(2. 1 to 10 carry 1 mark each.CBSE MIXED TEST PAPER-09 (Unit Test) CLASS . Plot the points A(-2. right angle ed. 16 to 25 carry 3 marks and Q. Find two irrational numbers between 2 and 3. Q. At B. If l1 and l2 are parallel.No. Find the degree measure of angle A in right angled isosceles triangle ABC. SECTION A 1. 7 5 −1 2 4. No. No. 9. 166 . Q. Q. Evaluate (99)2 using some identity. 11 to 15 carry two marks.: ] [Time : 3.00 hrs. Simplify: x −2 y 4 6. How many lines can pass through two points? 8. each.No. 7. All questions are compulsory. Simplify: 16 5 5.-3) in Cartesian Coordinate system. M. then find x.IX MATHEMATICS [M. Express 1 into decimal form. 2.] General Instructions:1. 17. Prove. Two parallel lines are intersected by a transversal. Check your answer by actual division. Prove that he diagonals of rhombus are perpendicular to each other. If the altitude from two vertices of a triangle are equal. of students 4 3 2 2 1 167 . Construct a frequency polygon for the following data: Age: (in year) 0-2 2-4 4-6 6-8 8-10 10-12 12-14 14-16 Frequency 2 4 8 10 6 5 3 6 25. 20. SECTION C 16.10. 22. Using factor theorem. 23. What is the relationship between their areas? SECTION B 11. Find the mean of the following distribution: Wt. 18. Prove that the bisector of any pair of alternate interior angles are also parallel. 21. Factories: 64x3 – 27y3 + 8z3 + 72 xyz. Using remainder theorem. Rationalize the denominator of : 1 5−3 2 13. 19. Prove that the sum of any two sides of a triangle is always greater than the third side. find the remainder when x3 – 6x2 + 2x – 4 is divided by x + 1. Express 2. Show that ar (∆APB ) = ar (∆BQC ) 24. 12. A triangle and a gm have common base and lie between same parallel lines. then the triangle is isosceles. Represents 6 on the number line. Prove that any exterior angle of a triangle is equal to the sum of its interior opposite angles. Factories: 2x2 – 3x + 1. 14. find whether x – 2 is a factor of 2x4 – 5x3 + 2x2 x + 2? 15.765 into p/q form of a rational number.(in Kg) 67 70 72 73 75 No. P and Q are any two points lying on the sides DC and AD respectively of gm ABCD. 29. In fig. Prove that the four triangles formed by joining the midpoints of the sides of a triangle are congruent. Factories: (a+b)3 + (b + c)3 (c + a)3 -3(a+b) (b+c) (c +a) Show that ∠TPS = 168 . 28.SECTION D 26. 27. 1 (∠Q − ∠R ) 2 30. If a transversal makes equal intercepts on three parallel lines then prove that they make equal intercepts on any other transversal. Prove that the triangles on the same base and between same parallel lines are equal in area. PS is the bisector of ∠QPR and PT ⊥ QR.. 3. However.No. 7− 6 Find the zero of the polynomial p(x) = 2x + 5. 17 & 18 and another for Q. Section A 1 . 8. 1.: 80] General Instructions:1. 7. 4. M. B. Find the value of x and y in the given figure if AB parallel CD. Cancel the previous question. All questions are compulsory. The question paper consists of thirty questions into 4 sections. b and c. Section C comprises of ten questions of 3 marks each and Section D comprises of five questions of 6 marks each. Section-A comprises of ten questions of 1 mark each. In questions on construction (if any). You have to attempt only one of the alternative in all such questions. drawing should be neat and exactly as per the given measurements. Use two graph paper one for Q. All questions in Section A are to be answered in one word. 6.CBSE MIXED TEST PAPER-10 (Second Unit Test) CLASS . There is no overall choice. 169 . Rationalize the denominator of 2. if attempted again.IX MATHEMATICS [Time : 3 hrs. for any reason. What are the coordinates of the origin? Write 4 = 5x – 3y in the form of ax + by +c = 0 and indicate the value of a. Section B comprises of five questions of 2 marks each. 3. 28 & 30. No. three questions of 3 marks and two questions of 6 marks each. 5. However you may ask for mathematical tables. A. one sentence or as per the exact requirement of the questions.] [M. Use of calculator is not permitted. internal choice has been provided in one question of 2 marks each. 2. 4. 5. C and D. 3. Find the mode of the following marks (out of 10) obtained by 20 students: 4.4.7.9. In figure line XY and MN intersect at O.5.6.10. 13.4. angle ACB = 300 Find angle BDC.6. 9.9.9. 10.10. x3 – ax2 +6x – a is divided by (x-a). In figure. 8. 12.9 Section B 11. 170 .6.7. If angle POY = 900 and a : b = 2 : 3 find c.5.6. Write the statement of SAS congruence rule.7. Write the condition for a quadrilateral to be parallelogram.2. Show that CD bisect AB. If two figures A and B are congruent then write a relation between the area for the figure A and B. angle ABC = 700.3. In figure AD and BC are equal perpendicular to a line segment AB. 7. Represent 9. . 20.14. AE = 8 cm and CF = 10 cm.0). AE perpendicular DC and CF perpendicular AD. Section C 16. Draw the graph of the linear equation in two variables. R are points on a circle with centre O. In figure if AC = BD then prove that AB = CD. In figure. (0. then prove 1 that ∠BOC = 900 − ∠BAC 2 171 . 15.5). X + y = 4 19. find AD. If the bisector BO and CO of angle CBE and angle BCD respectively meet at point O. (-3. In figure. In figure. 18. (-3. Or A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at point on the major arc. the sides AB and AC of triangle ABC are produced to points E and D respectively. Q. if AB = 16 cm.5). -5) and (5. find angle OPR. where P.3 on the number line. angle PQR = 1000. 17.3) in the Cartesian plane. ABCD is a parallelogram. (2.5). Locate the points (5. Or ABC is a triangle right angled at C. ABCD is a quadrilateral in which P. Show that (i) D is the mid point of AC (ii) MD ⊥ AC 1 (iii) CM = MA = AB 2 172 . the hypotenuse is the longest side. if the bisector BO and CO of angle ABC and angle ACB respectively 1 meet at point O. then prove that angle BOC = 900 + 2 21. show that: 1 (i) SR AC and SR = AC 2 (ii) PQ = SR (iii) PQSR is a parallelogram. AC is a diagonal. BC and DA. Q. R and S are mid points of the sides AB. 22.Or Given a triangle ABC. A line through the mid point M of the hypotenuse AB and parallel to BC intersect AC at D. Show that in a right angled triangle. Write a linear equation for this information and draw its graph. of students 4 11 23 8 4 Total 50 Draw a histogram and a frequency polygon for the above data on the same graph. x. q 27. 5 per km. where p and q are integers and q ≠ 0.27 in the form of Marks 0-20 20-40 40-60 60-80 80-100 No. Prove that: The line drawn through the mid-point of one side of a triangle. Factories : x3 + 13x2 + 32x + 20. 31. 26. The following observations have been arranged in ascending order. the fare is Rs. 25. If the median of data is 63. 95. Section D p . 29.23. State and prove that the mid-point theorem. Or Prove that: Chords equidistant from the centre of a circle are equal in length. 32. Find the value of x. 84. In figure E is any point on median AD of a triangle ABC. 48. (i) In one variable (ii) In two variables. 29. y. 72. 50. parallel to another side bisect the third side. The taxi fare in a city is as follows: For the first kilometer. Show that ar (△ ABe) = ar (△ ACE ) 24. Express 1. x+2. Or 30. 78. The following table gives the number of marks obtained by 50 students. 8 and for the subsequent distance it is Rs. 28. Taking the distance covered as x km and total fare as Rs. 173 . Or Give the geometric representation of y = 3 as an equation. Prove that : Equal chords of a circle (or of congruent circles) are equidistant from the centre (centers). : ] General Instructions:1. 8. Find the median of the points scored by the team.47 in form of p q 12. 9. 7. 2. Coordinates of a point P(-3. 7. 3.4). AB = AC . (-8. Section C comprises of 10 questions of 3 mark each. In which quadrant or on which axis do each of the point lie? (-3.(3.7). 5. 18. 4. Section A comprises of 10 questions of 1 mark each.0). ∠BAE = 1260 . 13.4) 14. 3. find ∠A 10. 5. Write the coefficient of x2 in -7x4 + 8x3 -3x2 + x + 1. 17. M. 4.-7) write the value of abscissa and ordinate. Find ∠ABC 174 . All questions are compulsory. Rationalize the denominator of . The question paper consists of 30 questions into 4 sections.CBSE MIXED TEST PAPER-11 (Unit Test) CLASS . Find the value of the polynomial 5x – 4x2 + 3 at x = -1. 6. Section B 11. Show how 5 can be represented on the number line. 7 1 2. (0. 3. 4. 19. Express 0. 3 questions of 3 marks each and 2 questions of 6 marks each. What is the name of horizontal and the vertical lines drawn in the cartesion plane.IX MATHEMATICS [Time : 3 hrs. is produced to E ∠DCA = 1050 . △ ABC . The points scored by a term in a series of matches are as follows. However internal choice has been provided ‘1’ question of 2 marks . Write the name of point where x axis and y axis intersect. ∠b = 600 . 3. Express 3x + 9 = 0 as an equation in two variables. ABC is a triangle in which BC is produced to DCA. Section D comprises of 5 questions of 6 mark each.] [M. Find the value of (125) 3 . There is no over all choice. Section B comprises of 5 questions of 2 mark each. Section A 1 1. p(x)=x3 – 4x2 + x + 6. Evaluate: 81 23. Locate the points (5.14). 15 by arranging in ascending order.0) in the Cartesian plane. Give the equations of three lines passing through (2. By division method find the remainder when x3 + 3x2 + 3x + 1 is divided by 2x + 5. Section C 1 16. (8.0). Show that angle ABD = angle ACD 175 . 13. If the points (3. 4 2) 21. 18. g(x) = x-3 Or Find the value of K. 15. then find the value of a? Or Check which of the following are solutions of the equation x-2y = 4 and which are not (i) (0. 14. Find the mode of 15. if x-1 is a factor of p(x) = kx2 – 3x + k. Rationalize the denominator of 7− 6 17. Use the factor theorem to determine whether g(x) is a factor of p(x).2) (ii) (2. (2. (5. 9.5). 16. 12. 15.Or Calculate the value of x in the following fig. 3. (-3. ABC and DBC are two isosceles triangles on the same base BC.5).2). 0.8).4) lies on the graph of the equation 3y = ax + 7. (0. 20.0) (iii) ( 2. −1 625 4 22. 13. 19. 15. 7. 2. 15. 4.5.15.2.4.14.5.2.6.6.8.7.13 Make a grouped frequency distribution table for this data. E is the mid-point of median AD show that ar △ BED = Section D 26. In triangle ABC.7. In above fig E is any point on median AD of a triangle ABC.12. 1. Show that ar △ ABe = ar ( AEC ) Or 1 arABC 4 25. Thirty children were asked about the number of hours they played in the previous week . The length of 40 leaves of a plant are measured correct to one millimeter and the obtained data is represented in the following table.24.13. 176 .2.8. The results were found as follows.3.5.5 10. taking one of the class intervals as 5-10.9.3.1.13.6 3.8. Prove that a diagonal of a parallelogram divides it into two congruent triangles. diagonal AC bisects angle A. 28. a. the fare is Rs 8 and for the subsequent distance it is Rs 5 per km. b. Or Prove that parallelogram on the same base and between the same parallels are equal in area By sing above using theorem. Show that the quadrilateral formed by joining the mid – points of the consecutive sides of a rectangle is a rhombus. A taxi fare in a city is as follows: For the first kilometer. Prove that it also bisects angle C. In above fig AB and Cd are respectively the smallest and longest sides of a quadrilateral ABCD. 177 .Length (in mm) 118-126 127-135 136-144 145-153 154-162 163-171 172-180 Number of leaves 3 5 9 12 5 4 2 27. Find the area of gm. Taking a distance covered as x km and total free as Rs 4 write a linear equation for this information and draw its graph. The angles opposite to equal sides are equal prove it. Prove that the sum of the three angles of a triangle is 1800. 29. (ii )∠B > ∠D 30. gm ABCD. By using above theorem. If both are on the same base AB and between the same parallels FC AB. Show that (i )∠A > ∠C . ABEF is 50 squcm. What property q must satisfy such that is terminating. Marks are indicated in brackets. All questions are compulsory. 1 mark. AB CD 9. 1mark q 2. Find the angle ∠ AOC in the figure. identify one triangle and a quadrilateral on the same base and between the same paralles. 1mark 6. B . Factorize : x3 + 8y3. M. 11). Why 3 t + t 2 is not a polynomial? 1mark 3.IX MATHEMATICS [Time : 3 hrs. 178 . Write the co-ordinates of a point which lies 4 units above x-axis and lies on the y-axis. 2.: ] General Instructions:1.] [M.CBSE MIXED TEST PAPER-12 (Half Yearly Test) CLASS . C and D. Section A p 1. Give a linear equation passing through (2.1mark 4. In figure. Find angle ABC in the given figure. Write two conditions that a triangle can be formed. 1mark 8. Questions are divided into four sections A. 1mark 7. 1mark 5. AE = 8 cm and CF = 10 cm find AD. 2 marks 15. 2 16. 2 marks 1 2 and x = are zero of polynomial. Find the mean of first four prime numbers. 3 marks 20. OS is another ray lying between rays OP and 1 OR. 2 marks 12. 1 mark Section B 11. ABCD is a parallelogram AE ⊥ DC and CF ⊥ AD.10. In fig. Prove that ∠ROS = (∠QOS − ∠POS ) 3 marks 2 179 . 5) and (-3. Divide: 3x4 – 4x3 – 3x – 1 by x – 1. 2 marks. Find the value of k if X = 2 and y = -1 is a solution of equation 2x + 3y = k. In the given figure. Locate the points : (5. (2. Represent 3 on number line. 0). 3 marks Or Factories: 6x2 + 5x – 6 18. 5). (-3. 1 19. Also write its quadrant. POQ is a lies Ray. prove that ∠QTR = ∠QPR. -5). 3 marks. If the bisector of angle PQR and angle PRS meet at point T. OR is perpendicular to PQ. If AB = 16 cm. In an equilateral triangle prove that median is ⊥ to the opposite side. 2 marks 14. Check whether x = − 3 3 p(x) = 3x2 – 1 13. Simplify: Given QR of triangle PQR is produced to a point S. Section – C 5 2 6 2 + 3 marks 7−3 2 7+3 2 17. x + 2. 3 marks 23. 84. 95. 2 22. 72. Also find its point of intersection. Verify: 1 (x + y + z) x [(x – y)2 + (y-z)2 + (z – x)2]. 3 marks. In triangle ABC. Or State and prove ASA congruence rule. 92.21. 28. 50. Given angle A = angle D = 900. 60. 3 marks. State and prove angle sum property of a triangle. Show that a median of triangle divides it into two triangles of equal areas. 20. 24. Section – D x3 + y3 + z3 – 3xyz = 26. x3 – 23x2 + 142x -120. x. 10. Show that the bisectors of angles of a parallelogram form a rectangle. 70. Show that AD = AE. 48. 36. Find its Median and mode. 6 marks. Draw graph of the linear equations x + y = 5 and 2x + 3y = 7. 6 marks Using above statement find the value of x and y in figure. Or Median of data 29. D and E are points on BC such that BE = CD. Using ASA prove △ AOB ≅△ DOC in fig. State factor theorem. Find x. 50. 6 marks 27. OA = OD 180 . 40. 95 arranged in increasing order is 63. 3 marks 25. 3 marks. Marks obtained by 10 students are given below: 10. Using factor theorem factories. 32. 78. AB = AC. of letters 1-4 4-6 6-8 8-12 12-20 No.29. 100 surname were randomly picked up and information arranged is alphabet as given below in tabular form: No. D. BC and CA respectively. 30. Diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD. of surname 6 30 44 16 4 (i)Draw a histogram and frequency polygon for above information. In triangle ABC. number of surnames lies. E and F and 1 ar (△ DEF ) = ar (△ ABC ) 4 Or In fig. E and F are respectively the mid points of sides AB. (ii) Write the interval in which max. Show that triangle ABC is divided into four congruent triangles by joining D. If AB = CD. 6 marks. 181 . then show that (i )ar (△ DOC ) = ar (△ AOB ) (ii )ar (△ DCB ) = ar (△ ACB )and (iii) ABCD is a parallelogram. Section C: Q 16 to 25 carry 3 marks each. y = 1 is a solution by 2x + 3y = k. 55. Find the value of 16 4 182 . 5.] [M. 8.3. 61. Section B: Q 11 to 15 carry 2 marks each. 35.4. Section A: Q 1 to 10 carry 1 mark each. M. 71. Find AB2. 2. BC= 9cm. Find angle A. 2. Find range of data 70. Write an irrational number between 1. There are four sections A. Write the Zero of polynomial p(x) = 2x + 5. Find value of k if x=2. and 1. All questions are compulsory. 39. B . Section A 1. 4. Write the equation of x-axis Factorize x2 – 81y2 If AB = AC and angle ACD= 1200. 36. 9. 65.IX MATHEMATICS [Time : 3 hrs. 41. Is ( 3 + 4)( 3 − 4) a rational or irrational number? 7. 3.: 80] General Instructions:1. Internal choices have been provided in some questions.CBSE MIXED TEST PAPER-13 (Half Yearly Test) CLASS . Section D: Q 26 to 30 carry 6 marks each. 40. If B lies between A and C and AC = 15cm. 62. 3 10. C and D. 6. Factorize: x 2 − 2 x + 7 16 Or x − 5 3 x + 30 2 12. CF = 10 cm. AE = 8cm. ABCD is a parallelogram. (i) In fig find x 183 . If AB CD . Represent 2. Find AD. If AB = 16 cm. 3and 5 on same number line. AE perpendicular DC and CF perpendicular AD. 17. Find the value of a if (2x-4) is a factor of 2x3 + ax2 + 11 x + a + 2. 13. find value of x and y. 14. 5 1 and 7 11 15. Find two rational numbers between Section C 16.Section B 11. 18. CD and DA respectively.(ii) AB CD. R. P and 17. Find x 11. 24. Show that (i) △ APB ≅△ AQB (ii)B is equidistant from arms of angle A. is 15. 35. x+2. 13. 18. Give geometric representation of y = 4 as an equation (i) in one variable (ii) in two variable. Line l is bisector of angle A and B is any point on i. 19. Or If mean of 10. ABCD is a rectangle. Find value of P. x+4. 30. BP and BQ are perpendicular from B on arms of angle A. Rationalize: 7− 5 23. 39. EF ⊥ CD ∠GED = 1260 Find ∠AGEand ∠GEF . 184 . 15. 20. Show that the quadrilateral PQRS is a rhombus. BC. S are mid points of sides AB. The median of following observation arranged in ascending order is 25. AB and CD are respectively the smallest and longest sides of Quadrilateral ABCD show that (i )∠A > ∠C (ii )∠B > ∠D 21. Q. 46. 13. Calculate the mean and mode of following x : 10 30 50 70 89 f: 7 8 10 15 10 5 22. 19. P. 12. 38 46.28. 27. If lines segment CD is bisected by AB at O. Factorize x3 – 23 x2 + 142x + 120. 33.20.46.52. (i)Construct a frequency distribution for the above data using class interval of 10 seconds. 31.60 40.27.33. AB.26. (ii)Draw a histogram and frequency polygon to represent the frequency distribution 32. 16.46 30. (ii)In fig PQ RS . ABC and ABD are two triangles on same base. Draw the graph of equation 2x + y = 5. Show that ar (△ ABC ) = ar (△ ABD ) Section D 26.37. 28.48. ∠ACS = 1000 Deter min e∠ABC & ∠BAC 185 . (i)Prove that sum of angles of a triangle is 1800.42.49.50. Prove that parallelogram on same base and between same parallel lines are equal in area. From the graph find two more solution from the graph find y.43.59.53.25.58.64. ∠PAB = 700 . 29.30. if x = 3. The time taken in seconds to solve a problem for 25 pupils is as follows.20. Q4.B Contains 05 Questions of 2 marks each. Section -C Contains 10 Questions of 3 marks each. (4). (5). All questions are compulsory. SECTION –A Q1. The abscissa of a point on the y-axis is …………………. find < P Q6. AB=BC=5cm and angle A=55˚. Section. find angle B Q10. If ∆ABC=∆PQR.A Contains 10 Questions of 1 marks each.Sample Paper-I Half Yearly Examination Subject: Mathematics Class: IX Max: Marks: Time Allowed: 2½Hours General Instructions: (1). Write 3/13 in decimal form. Q3. Section. Find two rational numbers between 3 and 4. 186 . Find the remainder when x³+3x²+3x+1 is divided by x+1. (2). Q8. Check whether x=-1 and x=2 are the Zeros of the polynomial p(x) = (x+1) (x-2) Q7. Section. Write a linear equation in two variables for the statement:Cost of a notebook is twice the cost of a pen. Find: (125) ⅓ Q9.D Contains 05 Questions of 6 marks each. Find the product by using suitable identity:(x+4) (x-10) Q5. <B=55˚ and <C= 95˚. (3). Q2. In ∆ABC. Q21. (2. Show that ar (EFGH) = ½ ar(ABCD) Q17. Factorise: 3√3x³+2√2y³ Q19. 5) and (4.-3) in the Cartesian plane by using a graph paper. Show that1.(-2x+3y+2z) ² by using a suitable identity. if x-1 is a factor of 2x²+kx+√2 __ Q13. Q15. 0). Show that the sum of the Angles of a triangle is 180˚.√6 SECTION-C Q16. If E. Rationalize the denominator of : _ 1____ √7.SECTION-B Q11. Factorise:8x³+27y³-z³+18xyz 187 . Draw the graph of following linear equation in two variables.27 can be expressed in the form p/q. y = 2x+4 Q12. Expand :. Plot the points (4. where p and q are integers and q ≠ 0. Find the value of k. Q20. Q14. F. G and H are respectively the mid points of the sides of a parallelogram ABCD. Represent √3 on number line Q18. <RPT=95˚and <TSQ = 75˚. In given figure if lines PQ and RS intersect at point T. P 95˚ 40˚ R T 75˚ S Q Q24. A F E C B Show that i) ∆ABE= ∆ACF ii) AB = AC Q23. find <SQT. ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. such that <PRT =40˚. Divide the polynomial 3x³-4x²-3x-1 by x-1 Q25.Q22. Factorise:27y³+125z³ by using the identity a³+b³= (a+b) (a²-ab+b²) 188 . SECTION-D Q26. Show that the diagonals of a rhombus are perpendicular to each other. Verify that: x³+y³+z³-3xyz = ½(x+y+z) [(x-y)² + (y-z)² + (z-x)²] Q30. Q29. Show that < ABD = <ACD. A B C D Q27. ………………………………………… 189 . ∆ABC and ∆DBC are two isosceles triangles on the same base BC. Factorise: x³+13x²+32x +20 Q28. Prove that two triangles are congruent if any two angles and the included side of one triangle is equal to any two sides and the included side of other triangle. (5). Find the value of ‘k’ if x=2. Section. Determine whether x-3 is a factor of x³-4x²+x+6. then prove that AB = CD. 4) lies on the graph of equation 3y=ax+7. 8½ Q10. Simplify: (√ 5-√2) (√5+√2) Q3. Q5. Find the zero of the polynomial p(x)= 2x+5 Q4. y= 1 is a solution of the equation 2x+3y=k. Section. then find the value of ‘a’ 190 . Find the remainder when x³-ax²+6x-a is divided by x-a Q6.C Contains 10 Questions of 3 marks each.A Contains 10 Questions of 1 marks each.D Contains 05 Questions of 6 marks each.Sample Paper-II Half Yearly Examination Subject: Mathematics Class: IX Time Allowed: 2½Hours General Instructions: (1). If the point (3.B Contains 05 Questions of 2 marks each. Find the product by using suitable property: (3-2x)(3+2x) Q7. Q2. Evaluate by using suitable property: (99)³ Q8. Section. (4). (2). Simplify: 7½. Section. All questions are compulsory. Max: Marks: SECTION –A Q1. D B A C Q9. (3). In the given figure if AC=BD. AE┴ DC and CF ┴ AD. Show that ar( BED)= ¼ ar(ABC) 191 . In a triangle ABC. AE=8cm and CF= 10cm. If AB=16cm.235 can be expressed in the form p/q. Give two different irrational numbers between 5/7 and 9/11. Q14.SECTION-B Q11. Find the value of k. Represent √5 on number line Q17. Factorise:8x³+y³-27z³+18xyz Q19. A diagonal of a parallelogram divides it into two congruent triangles. Draw the graph of following linear equation in two variables. Q21. using successive magnification. Q20. SECTION-C Q16. Find AD A B F D E C Q15. In the given figure ABCD is a parallelogram. y = 2x-4 Q12. Factorise: 64m³ -343n³ Q18. Show that 0. if x-1 is a factor of p(x) = kx²-√2x+1 ___ Q13. E is the mid point of median AD.765 on the number line. where p and q are integers and q ≠ 0. Visualise 3. Q22. D A O C B 192 . Rationalise the denominator of the: ___ 1_____ 7-3 √2 SECTION-D Q26. Q23. If x+y+z=0. then show that (i) ar (DOC)= ar(AOB) (ii) ar (DCB= ar(ACD) (iii) DA║ CB or ABCD is a parallelogram. figure diagonals AC and BD of quadrilateral ABCD intersect at O such that OB = OD if AB = CD. Q24. In given. and represent the solutions on (i) The Number line (ii) The Cartesian plane. Solve the equation 2x+1=x-3. Prove that the angles opposite to equal sides of an isosceles triangle are equal. Expand by using suitable identity: (-2x+5y-3z)² Q27. Show that x ³ + y³+ z³ =3xyz Q25. Prove that angle PSR > angle PSQ. In the figure.Factorise: x³-23x²+142x-120 Q30. Factorise: 2x² + y ²+ 8z²-2 √2 xy + 4√ 2yz – 8xz Q29.Q28. PR >PQ and PS bisects < QPR. P Q R S ************************ 193 . from 6 to 10 two marks each. 22. If 2 is added to every number. from 11 to 15 three marks each. 23. Find mean of firs five prime numbers. 14. 18. 28. then find P. Q5. what will be new mean. Q. In the give x||y.CBSE MIXED TEST PAPER-19 UNIT TEST CLASS . Q2. Find y. The mean of 16 numbers is 8. 25. Find mode of 14.IX MATHEMATICS [Time : 11/2hrs. 18.: ] GENERAL INSTRUCTIONS:1. O. No. 14. 194 . Q3. 18. No. b are collinear Q4. Q. No. The ratio of angles of Quadrilateral are 1:1:1:1 the quadrilateral is ……………… Q6. No. Q1. Q. 14. M. 3.] [M. Q. 17. from 1 to17 five marks each. 2. z ⊥ y. 4. from 1 to 5 one marks each. if A. D is point on side BC of ∆ ABC such that AD = AC show that AB >AD Q10. 46. x + 4. 50. Find x Q9. 83. 59. 195 .Q7. ABC is a triangle in which BE and CF are altitude to sides AC and AB are equal Prove that ∆ ABC ≅ ∆ ACF Q8. 36. x – 2. In the fig BO and CO are bisectors o interior angles ∠ B and ∠ C intersecting at O. 1 Show that ∠ BOC = 90o + ∠ BAC 2 Q12. ∠ PQR = 110o and ∠ RST = 130o Find ∠ QRS Q11. If PQ||ST. 80. 20. 70. 88 If median is 57. The following data is arrangement in ascending order 19. In a parallelogram ABCD E and F are mid-points of sides AB and CD respectively show that line segment AF and EC trisect diagonal BD. ∠ X = 62o. Find ∠ OZY and ∠ YOZ Q15. Find value of P in the mean of following data 40.4 Variable (x) Frequency (f) 10 3 20 8 30 12 40 5 50 P 60 7 70 5 Q14. Prove the sum of angles of a triangle is 180o.Q13. The distribution of weight (in kg) of 100 people given below: Weight (in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75 Frequency 13 25 28 15 12 5 2 100 Construct a histrogram and frequency polygon for the above data Q17. if YO and ZO are bisectors of ∠ XYZ and ∠ XZY. 196 . Prove that two triangles are congruent if two angles and an included side of one triangle are equal to two angles and an included side of other triangle. ∠ XYZ = 54o. Q16. 7+3 2 [10] m where m is a natural number. Factories: 12x2 – 7x + 1 [3] Q6. 1 Q2. [3] Q3. In fig. p(1) and p(2) for the polynomial: p(t) = 2 + t + 2t2 – t3 [3] Q5. [3] Q7.IX MATHEMATICS [Time : 3hrs. Show that the sum of the angles of a quadrilateral is 360o. Write True of False: (a) Every natural is whole number. (j) A kite is a parallelogram. (b) Every integer is a whole number.27 [3] Q4. Find all the angles. lines PQ and RS intersect each other at point O if ∠ POR: ∠ ROQ o= 5:7. (i) A parallelogram is a trapezium.: 100] Q1. Rationalise the denominator of . (g) Every integer is a rational number.CBSE MIXED TEST PAPER-20 HALF YEARLY EXAMINATION CLASS . [3] 197 . (h) Every rational number is an integer. (d) Every irrational number is a real number. Express as a rational number: 1. (e) Every point on the number line is the form (f) Every whole number is natural number. (c) Every rational number is whole number. Find the remainder when x3 + 3x2 + 3x + 1 is divided by x – ½. M. Find p(0). [3] Q8.] [M. (h) A quadrilateral is a parallelogram if a pair of ……………… is equal and……………. + (b) x3 + y3 + z3 – 3xyz = (……….-…………. 84. the angle opposite to the longer side is …………….. Fill in the blanks: [10] 2 (a) (x + y + z) = ………. (i) In any triangle the side opposite to the larger angle is ……………. (d) If two sides of a triangle are unequal.. then it is a …………………. 32.. If the median of the data is 63... +…………. In fig. x. then prove that ∠ YQZ = ½ ∠ YXZ. 48. the side YZ of ∆XYZ is produced to a point P.) (c) The sides opposite to equal angles of a triangle are ………………….+…………….-………. [4] Q12. 72.. find the value of x. PR>PQ and PS bisects ∠ QPR prove that ∠ PSR> ∠ PSQ.. 95 Q10. The following observations have been arranged in ascending order. To the …………… (g) Parallelogram on the same bases and between the same parallels are ………………. 198 . 50.+…………+……………. (e) If each pair of opposite sides of a quadrilateral is equal.. 78. x+2. if the bisectors of ∠ XYZ and ∠ XZP meet at point Q.Q9. (j) Lines which are parallel to a given line are…………… to ……………… Q11. In fig.. (f) The line segment joining ht mid points of two sides of a triangle is……….+…………+…………)(………+…………-…………. [3] 29. Q17.Q13. (b) Write the class interval in which maximum numbers of surnames lie. Q14. and then prove that two liens PQ and RS are parallel. Show that a median of a triangle divides its who triangles of equal area. In fig. 100 surnames were randomly picked up from a local telephone directory and a frequency distribution of the numbers of letter in the English alphabet in the surnames was found as follows: Number of letters 1-4 4-6 6-8 8-12 12-20 Number of surnames 4 30 44 16 4 (a) Draw a histogram to depict the given information. Prove that parallelogram on the same base and between the same parallels are equal in area. if a transversal intersects two lines such that the bisectors of a pair of corresponding angles are parallel. 199 [6] . Q15. The following table gives the distribution of students of two sections according to the marks obtained by them: [6] SECTION – A MARKS FREQUENCY 0-10 3 10-20 9 20-30 17 30-40 12 40-50 9 SECTION – B MARKS FREQUENCY 0-10 5 10-20 19 20-30 15 30-40 10 40-50 1 Q16. B and C are joined to vertices D. D and E are points on BC such that BE = CD Show that AD = AE 200 . In ∆ABC and ∆DEF AB = DE. In fig. In fig. ABC is an isosceles triangle with AB = AC. BC = EF and BC//EF vertices A. E and F respectively show that “ (1) Quadrilateral ABCD is a parallelogram (2) Quadrilateral BEFC is a parallelogram (3) AD//CF and AD = CF.Q18. AP//BQ//CR Prove that ar(ACQ) = ar(PBR) Q19. (4) Quadrilateral ACFD is parallelogram (5) AC = DF (6) ∆AB≅C ∆DEF Q20. AB//DE.