CHAPTER -1 UNIT 1 ADDITIONAL PROBLEMS ON PLYING WITH NUMBERS

CHAPTER -1 UNIT 1 ADDITIONAL PROBLEMS ON PLYING WITH NUMBERS I. Choose the correct option 1. The general from of 456 is a) ( 4 x 100) + ( 5 x 10 ) + ( 6 x 1 ) b) ( 4 x 100) + ( 6 x 10 ) + ( 5 x 1 ) c) (5 x 100 ) + ( 4 x 10 ) + (6 x 1 ) d) ( 6 x 100 ) + ( 5 x 10 ) + ( 4 x 1) [a] 2. Computers use a) Decimal system b) binary system c)base 5 system d) base 6 system [b] 3. If abc is 3 digit number, then the number n = abc + acb + bac + cab + cba is always divisible by a) 8 b) 7 c) 6 d) 5 [c] 4. If abc is 3 digit number, then n = abc - acb - bac - bca - cab - cba is always divisible by a) 12 b) 15 c) 18 d) 21 [c] 5. If 1 K x K1 = K 2 K, the letter K stands for the digit a) 1 b) 2 c) 3 d) 4 [a] 6. The number 345111 is divisible by a) 15 b) 12 c) 9 d) 3 [d] 7. The number of integer of the form 3AB4, where A, B denote some digits, which divisible by 11 is

a) 0 b) 4 c) 7 d) 9 [d] 2.What is the smallest 5 digit number divisible by 11 and containing each of the digits 2, 3, 4, 5, 6? The smallest 5- digit number divisible by 11 which contain the digits 2, 3, 4, 5, 6 is 24365 [ 2 43 6 + 5 = 0 ] 3.How many 5 digit number divisible by 11 are there containing each of the digits 2,3, 4, 5, 6? 24365, 26245, 24563, 26543 36542, 34562, 34265, 36245 54263, 56243, 54362, 56342 There are 12 5 digit numbers 4.If 49A and A49 where A > 0, have a common factor, find all possible values of A. If A = 2, 5, 7, 8 then the numbers formed by 49 A and have common factors. Like 1. 492 and 249 common factor 3 2. 495 and 549 common factor 3 3. 497 and 749 common factor 7 4. 498 and 849 common factor 3 But if A = 1, 3, 4, 6, 9 donot have common factor. 5. Write 1 to 10 using 3 and 5, eah al last once, and using addition and subtraction, (For example, 7 = 5 + 5 3 ) 1 = 3 + 3 5 2 = 5 3 3 = 5 + 3 5 4 = 3 + 3 + 3 5 5 = 3-3 + 5 6 = 3 + 3 + 5 5 7 = 5 + 5 3

8 = 5 + 3 9 = 5 + 5 + 5 3 3 10 = 3 3 + 5 + 5 6. Find all 2 digit numbers each of which is divisible by the sum of digits. 2 digit numbers which are divisible by the sum of its digits are 10, 12, 18, 20, 21, 24, 27, 30, 36, 40, 42, 45, 48, 50, 54, 60, 63, 70, 72, 80, 81, 84, 90. 7. The page numbers of a book written in a row gives a 216 digit number. How many pages are there in the book. From Page No 1 to 9 there are 9 digits From page No 10 to 99 there are 180 digits From Page No 100 to 108 there are 27 digits 216 Total No of pages 108 in a book when has 216 digit numbers. 8. Look at the following patters: 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 6 1 1 7 21 35 35 21 7 1 1 8 20 56 70 56 28 8 1 9. This is called Pascal s triangle. What is the middle number in the 9-th row? The Middle number is the 9 th Row is 70. 10. Complete the adjoining magic square. [ Hint: in a 3 x 3 magic square,., Magic some is three times the central number? Ans: The Number is 108 [ 108 = 1 + 0 + 8 = 9] 9 x 12 = 108 11. Find all 3- digit natural numbers which are 12 times as large as the. m of their digits. Find all digits x, y such that 34 x 5y is divisible by 36. The no.s are divisible by 36 only when they are divisible by 4 and 9. According to the divisibility condition of 4 and 9, then the numer is divisible by 4. If the sum of all the digits of a number is divisible 9, then the number s divisible by 9.

These condition one satisfied only When x = 4, y = 2 Or x = 0, y = 6 Or x = 9, y = 6 12.Can you divide the numbers 1,2,3,4,5,6,7,8,9,10 into two groups such that the product of numbers in one group divides the product of numbers in the other group and the quotient is minimum? Group 1 1 x 2 x 3 x 4 x 5 x 6 7 = 5040 Group 2 8 x 9 x 10 = 720 5040 = 7 720 Minimum quotient i s 7 13.Find all 8 digit numbers 273A49B5 which are divisible by 11 as well as 25. The number is divisible by 25 When the number ends with 00, 25, 50 or 75 In the number B must be either 2 to 7. The number is divisible by 11 if and only if ( Sum of odd placed digits) ( Sum of even placed digits) is divisible by 11. It is possible only when i) If A = 1 and B = 2 Then the number is = 27314925

Or If A = 6 and B = 7 Then the number is 27364975 14.Suppose a, b are integers such that 2 + a and 25 b are divisible by 11. Prove that a + b is divisible by 11. It two numbers are divisible by 11 then their sum and difference is also divisible by 11 Let us take the difference of the no. s Then ( 2 + a ) ( 35 b ) is divisible by = 2 + a 35 + b = 2 35 + a + b = - 33 + a + b divisible by 11 Since 33 is divisible by 11 a + b also must be divisible by 11 15.A list of numbers and corresponding codes are given in the adjacent table. Find the Number L. As per the given of numbers and their codes 2 corresponds to 8 0 corresponds to 0 1 corresponds to 6 the number L = 8066 2011

16. In the multiplication table A 8 x 3B = 2730, A and B represent distinct digits different from 0. Find A + B. In A8 x 3B = 2740 To get 0 in unit place 8 is to be multiplied by 5 Ie is 8 x 5 = 40 A8 = 2730 = 78 35 AB = 78 A = 7 17. Find the least natural number which leaves the remainders 6 and when divided by 7 and 9 respectively. LCM of 7 and 9 is 63 Least number which leaves the remainder 6 and 8 Go on finding the remainder by subtract in natural no. from 63 Ie 63 1 = 7 = 8 7) 62 56 6 II ly 63 1 = 6 9) 62 54 08 There is no other no. which gives 6 and 8 when divided by 7 and 9 Least number is 62.

18.Prove that the sum of cubes of three consecutive natural numbers always divisible by 3. Let the 3 consecutive nature no.s be X, x + 1 and x + 2 x³ + ( x + 1) ³ + ( x + 2 ) ³ x ³ + x ³+ 1 + 3x ( X + 1) + x ³ + 6 + 6x ( x + 2) x³ + y³ + 1+ 3x³ + 3x + x³ + 6 + 6x + 12x 3x³ + 9x² + 15x + 9 3 ( x² + 3x + 5x + 3) Sum of cubes of three consecutive natural numbers is always divisible by 3. CHAPTER 2 UNIT 1 ADDITIONAL PROBLES ON ALGEBRIC EPXPRESSIONS I. Choose the correct answer: a. Terms having the same literal factors with same exponents are called a) Exponents b) like terms c)factors d) unlike terms [b] b. The coefficient of ab in 2ab is: a)ab b) 2 c) 2a d)2b [b] c. The exponential from of a x a x a is : a) 3a b) 3+a c) a³ d) 3-a [c]

d. Sum of two negative integer is: a) Negative b) positive C)zero d) infinite [a] e. What should be added to a² + 2ab to make it a complete square? a)b² b) 2ab c) ab d)2a [a] f. What is the product (x + 2) ( x 3 )? a) 2x 6 b) 3x 2 C)x² -x 6 d) x² - 6x [c] g. The value of (7.2)² is (use an identity to expand): a) 49.4 b) 14.4 c) 51.84 d)49.04 [c] h. The expansion of (2x 3y)² is: a) 2x² + 3y² + 6xy b)4x² + 9y² - 12 xy c)2x² + 3y² - 6xy d) 4x² + 9y² + 12xy [b] i. The product 57 x 62 is (use an identity): a) 4596 b) 2596 c) 3596 d) 6596 [c] 2.Take away 8x 8p + 10q from 10x + 10y 7p + 9q. 10x + 10y - 7p + 9 q 8x - 7 y 8 p + 10 q ( - ) (+) (+ ) ( - ) +2x + 17 y + p - q

3.Expand: (i) (4x + 3)² (iii) ( x + 1/x)² ii) (x + 2y)² iv) ( x 1 / x)² i) (4x + 3) ² this is in the form ( a + b )² = a² + 2ab + b² Here a = 4x (4x + 3) ² = (4x)² + 2 (4x) (3) + 3² b= 3 = 16x² + 24x + 9 ii) (x + 2y )² this is in the form (a + b) ² = a² + 2ab + b² Here a= x = x² + 2 (x) (2y) + ( 2y)² b = 2y = x ²+ 4xy + 4y² iii) ( x + 1 )² This is in the form x ( a+ b) = a + 2ab + b Here a = x = x² + 2. x. 1 + ( 1 )² b = 1 X x x = x² + 2 + 1 x²

iv) ( x- 1 ) ² This is in the form X ( a+ b) ² = a² 2ab + b ² Here a = x = x² + 2, x, 1 + (1 ) ² b = 1 x x x = x² + 2 + 1 x² 3. Expand: i) (2t + 5) ( 2t 5) ii) ( xy + 8 ) ( xy 8 ) iii)(2x + 3y ) ( 2x 3y) i) ( 2t + 5) ( 2t 5) This is in the form = ( a+ b) ( a- b) = a² b² = ( 2t + 5 ) ( 2t 5) = (2t) ² 5² = 4t² - 25 ii) (xy + 8 ) ( cy 8) This is in the form of = ( a + b) ( a- b) = a² b² = ( xy + 8 ) ( xy 8) = (xy) ² 9² = x² y² - 64 iii) ( 2x + 3y ) ( 2x 3y ) This is in the form of = (a + b) ( a- b) = a² b² = ( 2x + 3y ) ( 2x 3y ) = (2x) ² - (3y) ² = 4x² 9y²

5.Expand i) ( n-1 ) ( n + 1) ( n² + 1) ii) ( n 1/n) ( n + 1 /n) ( n² +1/n²) iii)(x -1 ) ( x + 1) ( x² + 1) ( x⁴ + 1) iv)(2x y ) ( 2x + y) ( 4x² + y²) i) ( n 1) ( n²+ 1) ( n -1) ( n + 1 ) = n² -1² = ( n²- 1² ) ( n²+ 1²) This is in the form = ( a- b) ( a + b) = a² - b² = ( n² 1²) ( n² + 1²) = (n²)² ( 1²)² = n⁴ - 1 ii) ( n 1 ) ( n + 1 ) ( n² + 1 ) n n n² = ( n² 1 ) ( n² + 1 ) n² n² = n⁴ 1 n⁴ iii) ( x 1) ( x + 1) ( x² + 1) ( x⁴ + 1) = (x² 1² ) ( x² + 1²) ( x⁴ + 1) =[ ( x² ) ² ) ( 1²)²) ] [ x⁴ + 1⁴ )] =(x⁴ 1⁴ ) ( x⁴ + 1⁴ ) =( x⁴ )² ( 1⁴ )² = x ⁸- 1 iv) ( 2x y) (2x + y) ( 4x² + y² ) =[(2x) ² (y) ² ] ( 4x² + y² )

= ( 4x² + y² ) ( 4x² + y²) = (4x²)² ( y²)² =16x⁴ - y⁴ 6. Use appropriate formula and compute: i) ( 103) ² ii) (96) ² iii) 107 x 93 iv)1008 x 992 v) 185² - 115² i) ( 103) ² = (100 + 3) ² This is in the form ( a + b) ² = a² + 2ab + b² ( 100 + 3 ) ² = ( 100 ) ² + 2 ( 100) (3) + 3² = 10000 + 600 + 9 ( 103 ) ² = 10609 ii) (96) ² = ( 100 4) ² This is in the form ( a b) ² = a ² 2ab + b² (100 4) ² = ( 100) ² + 2 ( 100) (4) + 4² = 10000 800 + 16 = 10016 800 ( 96) ² = 9216 iii) 107 x 93 = ( 100 + 7) ( 100 7) This is the form of ( a + b) ( a- b) = a² b ²

( 100 + 7) ( 100 7) = ( 100) ² 7² = 1000 49 = 9951 iv) 1008 x 992 ( 1000 + 8 ) ( 1000 8) This is in the form ( a+ b) ( a- b) = a² b² ( 1000 + 8) ( 1000 8) = ( 1000) ² 8² = 100000 64 = 999936 v) 185² 115² This is in the form a² b² = ( a + b) ( a- b) ( 185) ² (115) ² = ( 185 + 115) ( 185 115) = ( 300) (70) (185) ² (115) ² = (300) (70) (185) ² ( 115) ² = 21000 7.If x + y and xy = 12 find x² + y². ( x + y ) ² = x² + y² + 2 xy ( x + y ) ² = ( x + y ) ² = 2xy = x² + y ² x² + y² 2xy 7² = 2 ( 12) = x² + y² x² + y² = 49 24 = x ² + y² ( x + y) 2xy 25 = x² + y²

X² + y² = 25 8. If x + y = 12 and xy = 32, find x² + y². ( x + y) ² = x² + y² + 2xy ( x + y ) ² 2 xy = x² + y² (12) ² 2 ( 32)= x² + y² 144 64 = x² + y² X² + y² = 144 64 X² + y² = 80 9. If 4x² + y² = 40 and xy = 10, find x ² + y². 4x² + y² = 40 = 2² x² + y² = 40 = ( 2x) ² + y² = 40 ( 2x + y) ² = ( 2x )+ y² + 2 (2x) (y) (2x + y) ² = 4x² + y² + 4 ( xy) = 40 + 4 (6) = 40 + 24 ( 2x + y) ² = 64 2x + y = 64 2x + y =8 10. if xy = 3 and xy = 10, find x² + y² ( x-y) ² = x² + y² 2xy 3² = x² + y² 2 (10)

9 = x² + y² 20 X² + y² = 9 + 20 X² + y² = 29 11. If x =( 1 )= 3, find x² + (1 ) and x³ + ( 1 ) x x² x³ case 1 ( x + 1 ) = 3 x Squaring on both side ( x + 1 )² = 3² X X² + 1 + 2. x. 1 = 9 X ² x X² + 1 = 9 2 x² x² + 1 = 7 x ² Case 2 x + 1 = 3 x Cubing on both side (x + 1)³ = 3³ x x³ + 1 + 3 x. 1 ( x + 1 ) = 27

x ³ x x x³ + 1 + 3 ( 3) = 27 x ³ x³ + 1 + 9 = 27 x ³ x³ + 1 = 27 x ³ x³ + 1 = 18 x ³ 12.If x + 1 = 6, find x ²+ 1 and x⁴ + 1 X x ² x⁴ Case 1 x + 1 = 6 x Squaring on both side ( x + 1 )² = 6² X X² + 1 + 2. x. 1 = 36 X ² x X² + 1 + 2 = 36-2 X ² X² + 1 = 34 X ²

Case 2 Consider X² + 1 = 34 X ² Squaring on both side ( x + 1 )³ = 3³ X ( x² + 1 )² = ( 34)² X² X⁴ + 1 + 2. x² 1 = 1156 34² = 1156 X ⁴ x² X⁴ + 1 + 2 = 1156 X⁴ X⁴ + 1 = 1156 2 x⁴ X⁴ + 1 = 1154 x⁴ 13.Simplify : (i) ( x + y)² + ( x y)² ii) ( x + y )² x ( x y )² i) ( x + y ) ² + ( x y) ²

= [x² + y² + 2 xy ] + [x² + y ² 2xy] =x² + y ²+ 2xy + x² + y² 2xy = 2x² + 2y ² = 2 ( x² + y² ) ii) (x + y) ² x ( x- y) ² = [ ( x + y ) ( x y) ] ² ( a + b) ( a- b) = a² - b² = [ x² y ²]² This is in the form = ( a b) ² = a² + b² 2ab ( a- b) ² = a² + b² 2ab = ( x² y² ) ² = x⁴ + y⁴ 2 x² y² ( or) (x² y² ) = x⁴ 2x²y ² + y⁴ 14.Express the following as difference of two squares: i) ( x + 2z) ( 2x + z) ii) 4 (x + 2y) ( 2x + y) iii) (x + 98) ( x + 102) iv) 505 x 495 [Hint: 4ab = (a + b) ² ( a- b) ² i) ( x + 2z ) ( 2x + z) = 2x² + xz + 4 xz + 2 z² = 2x² + 2z ²+ 4 xz + xz

= 2( x² + z² + 2xz ) + xz = 2 ( x + z ) ² + xz Multiply and divide by 4 = 8 ( x + z ) ² + 4xz ( x + z ) ² (x z) ² 4 4 = 8 ( x + z ) ² + 4 xz [x² + z² + 2xy ] [x² + z² 2zy] 4 4 = 8 ( x² + z ) ² + ( x + z) ² ( x- z) ² 4 = 9 ( x + z ) ² - ( x - z ) ² 4 4 3(x z) ² - x z ² 4 2 ii) 4 (x + 2y) ( 2x + y) 4 [2x² + xy + 4xy + 2y² ] 4[2x² + 2y² + 4xy + xy ] 4[ 2 ( x² +y² + 2xy ) + xy] 4[2 (x +y) ² + xy] 8(x+ y ) 2 + 4xy 8 (x + y ) ² + ( x + y) ² (x y) ² =9 ( x + y) ² ( x- y) ² =[3 (x + y)] ² ( x- y ) ²

iii) ( X + 98) ( x + 102) = x² + 102 x + 98 x + 9996 = x ²+ 200x + ( 10000 4) = x ²+ 200x + 10000 4 = x² + 200 x + (100) ² 2² =(x + 100) ² - 2² iv) 505 x 495 = ( 500 + 5) ( 500 5) = ( 500) ² 5² 15. If a = 3x 5y, b = 6x + 3y and c= 2y 4x, find i) a + b c ii) 2a 3b + 4c i) a + b + c = ( 3x 5y) + ( 6x + 3y) (2 y- 4x) = 3x 5y + 6x + 3y 2y + 4x = 13x 4y ii) 2a 3b + 4c =2(3x 5y) 3 ( 6x + 3y) + 4 ( 2y 4x) =6x 10y -18x 9y +8y -16 x = 6x 13x 16x -10y- 9y + 8y = 28x 11y 16. The perimeter of a triangle is 15x² 23 x + 9 and two of its side

are 8x² + 8x 1 and 6x² 9x + 4. Find the third side. Perimeter of a triangle = sum of all the three sides of a triangle Let the sides of the triangle be a, b and c then P = a + b + c Let a = 5x² + 8x 1 b = 6x² - 9x + 4, and P = 15x² 23 x + 9 C =? C = P ( a + b) = ( 15x² 23 + 9 ) [( 5x² + 8x 1 ) + (6x² 9x + 4) = ( 15x² + 23x + 9 ) [11x² x + 3] = (15x² 23x + 9)- 11x² + x -3 = 4x² 22 x 6 17. The two adjacent sides of a rectangle are 2x² 5xy + 3x² and 4xy x² x². Find its perimeter. Let a and b are the sides of the Rectangle l = 2x² - 5xy + 3x² b = 4xy - x² - x² Then its perimeter P = 2 ( l + b) P = 2 [2x² - 5xy + 3z² + 4xy - x² - z² ] P= 2 [ x² - xy + 2z²]

P = 2x² - 52 xy + 4z² 18.The base and altitude of a triangle are ( 3x 4y) and ( 6x + 5y) respectively. Find its are. Area of a = 1 b x h 2 Let b and h are the base and altitude of the triangle Then Area of triangle = 1 x b x h 2 A = 1 ( 3x 4y ) ( 6x + 5y) 2 = 1 [ 18x + 15 xy - 2 xy 20 y² ] 2 = 1 [ 18x² - 9xy 20 y² ] 2 19.The sides of a rectangle are 2x + 3y and 3x + 2y. From this a square of side length x + y is removed. What is the area of the remaining region? ( 2x + 3y) X + y X + y ( 3x + 2y) Area of the remaining region =

(Area of the Rectangle ) (Area of the square) (l x b ) - ( side) ² =[(2x + 3y) (3x + 2y)] ( x + y) ² = [ 16² + 4xy + 9xy + 6y² ] ( x² + y² + 2xy) = 6x² + 13xy + 6y² - x² - 2xy - y² = 5x² + 11xy 5y² 20. a, b, c are rational numbers such that a² + b² + c² -bc ca = 0. Prove that a = b = c. a² + b² + c² - bc ca = 0 Multiply both side by 2 2(a² + b² + c² - ab bc ca ) = 2 x 0 2a² + 2b² + 2c² - 2ab 2bc 2ca = 0 a² + a² + b² + c² + c² - 2ab 2bc 2ca = 0 a² + b² - 2ab + b² + c² - 2bc + c² +a² - 2ca = 0 ( a- b) ² + ( b c) ² + ( c- a) ² = 0 If the sum of squares of three terms is 0 Then each term is equal to 0 ( a- b) ² = 0 ( b c ) ² = 0 ( c- a) ² = 0 a- b= 0 b c = 0 c a = 0

a=b. ( 1) b = c.(2) c = a.(3) from equation ( 1) ( 2) and ( 3) a = b = c CHAPTER 3 UNIT 1 ADDITIONAL PROBLEMS ON AXIOMS, POSTULATES 1. Choose the correct option: i) a= 60 and b = a, then b= 60 by. a. Axiom 1 b. Axiom 2 c.axiom 3 d.axiom 4 [..]

ii) Given a point on the plane, one can draw. a. Unique b. two c.finite number of d.infinitely many through that point [d] iii) Even two points in a plane, the number of lines which can be drawn to pass through these two points is. a. Zero b. exactly one c.at most one d. more than on [b] iv) Two angles are supplementary, then their sum is.. a. 90 b. 180 c. 270 d. 360 [b] v)the measure of an angle which is 5 times its supplement is a. 30 b. 60 c.120 d.150 [d] 2. What is the difference between a pair of supplementary angles and a pair of complementary angle? Supplementary angles : If the sum of two angles is 180 then they are supplementary angles. Complementary angles: If the sum of two angles is 90 then they are complementary angles. 3. What is the least number of on collinear points required to determine a plane? 3 non collinerar points.

4. When do you say two angles are adjacent? The angles are said to be adjacent if they have a common side and a common and point. A C D B 5. Let AB be a segment with C and D between them such that the order of points on the segment is A, C, D, B, Suppose AD = BC. Prove that AC= DB. AB is s line segment C and D are points on it such that AD = BC AC + CD = BD + CD AC = BD [ Axioms 3 ] 6. Let AB and CD be two straight lines intersecting at 0. Let 0X be the dissector of BOD. Draw 0Y between 0D and 0A such that 0Y 0X. Prove that 0Y bisects DOA. Y A D Q P X 0 X X

C B AB and CD intersecting at 0 OX bisects BOD Let = BOX = DOX = X OY OX Let POY = P and YOA = Q BOX + XOD + DOY + YOA = 180 x + x + p + q = 180.. ( 1) x + p = 90.. ( 2) given x + q = 90 (3) ( Remaining angle of 180 ) From (2) + ( 3) X + p = x + q = 90 P = q = 90 OY bisect DOA 7.Let AB and CD be two parallel lines and PQ be a transversal. Let PQ intersect AB in L. Suppose the bisector of ALP intersect CD in R and the bisector of PLP intersect CD in S. Prove that LRS + RSL = 90. M M M X Y

A X Y B L C S R P Proof: Let ALN = NLP = x and BLM = MLP = y ALB = 180 ALN + NLP + PLM + MLB = 180 X + x + y + y = 180 2x + 2y = 180 X + y = 180 = 90 2 NLM = 90 But NLM = RLS = 90 * V. O. A+ In the LRS

LRS + RSL + LRS = 180 [Sum of all the angles of le] LRS + RSL + 90 = 180 LRS + RSL = 180-90 LRS + RSL = 90 8 In the adjoining figure. AB and CD are parallel lines. The transversals PQ and RS interest at U on the line AB. Given that DWU = 110 and CVP = 70, find the measure of QUS. S Q U X A B 110

C ( ( ) D 70 V W P R Ans: Given - DWU = 110, CVP = 70 Proof- UVP = KVP = 70 ( V.O. A) VWU + UWD = 180 ( Adjacent Angles) V + W + U = 180 ( Sum of angles of triangle is 180 ) U = 180-70 - 70 = 40 X = U = 40 ( V. o. A) QUS = 40 9. What is the angle between the hour s hand and minute s hand Of a clock at i) 1.40 hours, ii) 2.15 hours? ( Use 1 = 60 minute) 10. How much would hour s hand moved from its position to 12 noon when the time is 4.24 p.m? 11.Let AB be a line segment and let C be the midpoint of AB. Extend AB to D such that B lines between A and D. Prove that AD + BD = 2CD.

12.Let AB and CD be two lines intersecting BOD. Prove that the extension of 0X to the left of 0 bisects AOC. 13.Let 0X be a ray and let 0A and 0B be two rays on the same side of 0X and 0B. Let 0C be the bisector of AOB. Prove that AOX + XOB = 2 XOC. 14. Let 0A and 0B be two rays and let OX be a ray between 0A and 0B such that AOX > XOB. Let 0C be the bisector of AOB. Prove that AOX - XOB = 2 Cox. 15. Let 0A, 0B, 0C be the three rays such that 0C lies between 0A and 0B. Suppose the bisectors of AOC and COB are perpendicular to each other. Prove that B, O, A are collinear. 16. In the adjoining figure, AB DB. Prove that ABC + DCB + CDE = 180. A B D E C

17. Consider two parallel lines and a transversal. Among the measurement of transversal. Among the measurement of triangles formed, how many distinct numbers are there? (Answer will be given in next issue) CHAPTER - 1 UNIT1 2 EXERCISE 1. 2. 2 1. Express the following statement mathematically. i) Square of 4 is 16 Ans: 4² = 16 ii) Square of 8 is 64 Ans: 8² = 64 iii) Square of 15 is 225 Ans : 15² = 225 2. Identify the perfect squares among the following numbers: 2, 3, 8, 36, 49, 65, 67, 71, 81, 169, 625, 125, 900,100, 1000, 100000. Ans: Perfect squares : 6² = 36, 7² = 49, 9² = 81, 13² = 169, 25² = 625,

30² = 900, 10² = 100 3. Make a list of all perfect squares from 1 to 500. Ans: Perfect squares: 1,4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484. 4. Write 3 digit number ending with 0, 1, 4, 5, 6,9, one for each digit, but none of them is a perfect square. Ans: 110, 111, 114, 115, 116, 119 ( ending with 0, 1, 3, 5, 6, 9, but not a perfect square). 5. Find numbers from 100 to 400 that and with 0, 1, 4, 5, 6, or 9 which are perfect squares. Ans: Perfect squares from: 100 to 400 100 ( 10²) 144(12²), 169 ( 13²), 196 (14²), 225 (15²), 289 (17²), 324 (19²), 400(20²). EXERCISE 1. 2. 3

1. Find the sum 1 + 3 + 5 + + 51 ( the sum of all odd numbers from 1 to 51) without actually adding them. Ans: 1 + 3 + 5 + +51 = 51 + 1 = 26 26² = 676 2 Sum of n odd numbers = n² Sum of 26 odd numbers = 26² = 676 2. Express 144 as a sum of 12 odd numbers. Ans: 144 = 12² Sum of n odd numbers = n² Sum of 12 odd numbers = 12² 1+ 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 + 21 + 23 12² = 144 3. Find the 15 th and 16 th triangular numbers and find their sum. Verify the statement 8 for this sum. Ans: 1 +2 3 + 3 6 + 4 10 + 5 15 +6 I II III IV V 21 + 7 28 +8 36 +9 45 +10 55 +11 VI VII VIII IX X 66 +12 78 +13 91 +14 105 +15 XI XII XIII XIV 120 +16 136 +17

XV XVI n + ( n + 1) = ( n + 1 ) 2 15 th no + 16 th no = (16) 2 120 + 136 = 256 4. What are the remainders of a perfect square when divided by 5. Ex. Problem Write 5 th and 6 th triangular numbers and verify the formula of statement 8. Ans: 1 +2 3 +3 6 +4 10 +5 I II III IV 15 +6 21 + 7 V VI n+ ( n+1) = ( n + 1) 2 5 th no + 6 th no = ( 6) 2 15 + 21 = 36 Statement 8 is true Methods for squaring a number: 1842 = (180 + 4) 2 *( Using identity) =180 2 + 4 2 + 2 x 180 x 4 = 32,400 + 16 + 1,440 = 33,856 *No. ending with 5

1.9 x 10 5 2 2. 7 x8 5 2 3. 5 x6 5 2 9025 5625 3025 EERSICE 1.2.4 1. Find the squares of : i) 31 = (30 + 1) 2 (a + b) 2 = a 2 + b 2 + 2ab (30+1) 2 = ( 30) 2 + (1) 2 + 2 x 30 x 1 = 900 + 1 + 60 = 961 ii) (72) 2 = (70 + 2) 2 (70 + 2 ) 2 = (70) 2 + (2) 2 + 2 x 70 x 2 = 4,900 + 4 + 280 = 5184 iii) 37 = (30 + 7) 2 (30 + 7) 2 = (30) 2 + (7) 2 + 2 x 30 x 7 = 900 + 49 + 420 = 1369 iv) 166 = ( 160 + 6) 2 (160 + 6) 2 = ( 160) 2 + (6) 2 + 2 x 60 x 6 = 25,600 + 36 + 1920 = 27,556

2. Find the squares of : i) 8 x 9 5 = 7225 ii) 11 x12 5 = 13225 iii) 16 x 17 5 = 27225 3. Find the square of 1468 by writing this as 1465 + 3 Ans: 1465 + 3 ( a + b) 2 = a 2 + b 2 + 2ab (1465 + 3 ) 2 = ( 1465) 2 + (3) 2 + 2 x 1465 x 3 = 21,46,225 + 9 + 8790 = 2146234 + 8790 = 21,55,024 EXERCISE 1.2.5 1. Find the squares root of the following numbers by factorization: i) 196 2 196 2 98 7 49

7 7 196 = 2 2 x 7 2 = 2 x 7 1 196 = 14 ii) 256 2 256 2 128 2 64 2 32 2 16 2 8 2 4 2 2 1 256 = 2 2 x 2 2 x 2 2 x 2 2 = 256 = 16 iii) 10404 2 10404 2 5202 3 2607 3 867 17 287 17 17 1 2x 2x2x2

10404 = 2 2 x 3 2 x 17 2 = 2 x 3 17 = 102 iv) 1156 2 1156 2 578 17 289 17 17 1 1156 = 2 2 x 17 2 = 2 x 17 1156 = 34 v) 13225 5 13225 5 2645 23 529 23 23 1 13225 = 5 2 x 23 2 = 5 x 23

13225 = 115 2. Find the squares of : i) 100 + 36 = 10 + 6 = 16 2 100 2 50 5 25 5 5 1 100 = 2 2 x 5 2 = 2 x 5 100 = 10 3 36 3 12 2 4 2 2 36 = 3 2 x 2 2 = 3 x 2 = 6 1 ii) 1360 + 9

1369 = 37 37 1369 1369 = 37 2 = 37 37 37 1 iii) 2704 + 144 + 289 52 + 12 + 17 64 + 17 = 81 2 2704 2 1352 2 676 2 338 2 169 13 13 1 2704 = 2 2 x 3 2 x 13 2 = 2 x 2 x 13 = 52 2 144 2 72 2 36 2 18 3 9 3 3 1

144 = 2 2 x 2 2 x 3 2 = 2 x 2 x 3 = 12 17 289 17 17 289 = 17 2 = 17 1 iv) 225-25 = 15 5 = 10 5 225 4 45 3 9 3 3 225 = 5 2 x 3 2 = 2 x 2 = 15 1 5 25 5 5 25 = 5 2 = 5 1 v) 1764-1444 = 42-38 = 4 2 1764 2 882 3 441 3 147 7 49 7 7 1

1764 = 2 2 x 3 2 x 7 2 = 2 x 3 x 7 1764 = 42 2 1444 2 722 19 361 19 19 1444= 2 2 x 19 2 = 2 x 9 1 1444 = 38 vi) 169 x 361 2 13 x 19 117 13 x 247 13 169 13 13 169 = 13 2 = 13 1

19 361 19 361 1 361 = 19 2 19 3. Square yard has area 1764m 2. From a corner of this yard, a square part of area 784 m 2 is taken out for public utility. The remaining portion is divided in to 5 equal square parts. what is the parameter of these equal parts? Ans. Area of square = a 2 Area of given square yard = 1764 m 2 Area used for public utility = 784 m 2 Area of remaining portion = 1764 784 980 m 2 If the area is divided into 5 parts, then the area of each Square = 980 = 196 m 2 5 Length of each side = 196 a- = 14m Perimeter of square = 4a = 4 x 14 = 56m

4. Find the smallest positive interger with which one has to multiply each of the following numbers to get a perfect square. Ans i) 847 11 847 11 77 7 7 847 = 11 2 x 7 1 7 is the smallest positive number to be added to 847 to get a perfect square. ii)450 2 450 5 225 5 45 3 9 450 = 5 2 x 3 2 x 2 3 3 1 2 is the smallest positive interger to be multiplied with 450 to get a perfect square. iii) 1445 5 1445 17 289 1445 = 17 2 x 5 17 17 1

5 is the smallest positive interger to be multiplied with 1445 to get a perfect square. iv) 1352 2 1352 2 676 2 338 13 169 1352 = 2 2 x 13 2 x 2 13 13 1 2 is the smallest positive interger to be multiplied with 1352 to get a perfect square. 5. Find the largest perfect square factor of each of the following numbers: i) 48 2 48 2 24 2 12 2 6 3 3 1 2 x 2 x 2 x 2 x 3

4 x 4 x 3 16 x 3 16 is the largest perfect square factor. ii) 11280 2 11280 2 5640 2 2820 2 1410 5 705 3 141 47 47 2 x 2 x 2 x 2 x 5 x 3 x 47 1 4 x 4 x 5 x 3x 47 16 x 5 x 3 x 47 16 is the largest perfect square factor. iii) 729 3 729 3 243 3 81 3 27 3 9 3 3 1

3 x 3 x 3 x 3 x 3 9 x 9 x 9 81 x 9 729 729 is the largest perfect square factor. iv) 1352 2 1352 2 676 2 338 13 169 13 13 1 2 x 2 x 2 x 13 x 13 4 x 2 x 169 676 x 2 676 is the largest perfect square factor. 6. Find the proper positive factor of 48 and a proper positive multiple of 48 which add up to a perfect square. Can you prove that there are infinitely many such pairs? Ans: The factor of 48 are 1, 2, 3, 4, 6, 8, 12, 16, 24, 48. Multiples of 48 are = 48, 96, 144, 192, 240, 280. 1) 48 + 1 2 ) 192 + 4 49 = 7 2 192 = 14 2

3)288 + 1 4 ) 96 + 4 289 = 11 2 100 = 10 2 5)96 + 48 6) 48 + 16 144 = 12 2 64 = 8 2 7)240 + 16 256 = 16 2 Yes, we can prove that there are infinitely many such fairs. EXERCISE 1.2.6 1. Find the nearest integer to the square of the following numbers. Ans: i) 232 225 < 232 < 256 15 2 16 2 Square root of 232 is nearest to 15 ii)600 576 < 600 < 625 24 2 25 2 Square root of 600 s nearest to 24. iii)728 678 < 728 < 729 26 2 27 2 Square root of 728 is nearest to 27.

iv)824 784 < 824 < 841 28 2 29 2 Square root of 824 is nearest to 27 v)1729 1684 < 1729 < 1764 41 2 42 2 Square root of 1729 is nearest to 42. 2. A piece of land is in the shape of a square and its area is 1000m 2. This has to be fenced using barbed wire. The barbed wire is available only in integral lengths. What is the minimum length of the barbed wire that has to be bought for this purpose. Ans: Area of the land = 1000m 2 Area = a 2 = 1000 Perimeter = 4a Squaring, (Perimeter) 2 = ( 4a) 2 = 16a 2 = 16 x 1000 P 2 = 16,000 P = 16000

15876 < 16000 < 16129 126 127 Nearest number is 126 But this is not enough to cover the hard. Length of barbed wire required = 127 m. 3. A student was asked to find 961. He read it wrongly and found 691 to the nearest integer. How much small was his number form the correct answer? Ans. 961 = 31 691 676 < 691 < 729 26 27 26 is nearest number to 691 difference = 31 26 = 5 EXERCISE 1.2.7 1. Looking at the pattern, fill in the gaps in the following: 2 3 4-5 - 8-2 3 =8 3 3 = - - = 64 - = - 6 3 = - - = - - = - 729 Ans: 3 3 = 27, 4 3 = 64 (-5) 3 = -125, 6, 6 3 = 216 8 3 = 512-9, (-9) 3 = -729.

2. Find the cubed of the first five odd natural numbers and the cubes of the first five even natural numbers. What can you say about parity of the odd cubes and even cubes? Ans: Odd Cubes 1 3 = 1 3 3 = 27 5 3 = 125 7 3 = 343 9 3 = 729 Even Cubes 2 3 = 8 4 3 = 64 6 3 = 216 8 3 = 512 10 3 = 1000 Parity Cubes of odd numbers is always odd. Cubes of even numbers is always even. 3. How many perfect cubes you can find from 1 to 100? How many from 100to 100? Ans: 1 1to 100-4 perfect cubes -100 to 100-4 perfect cubes 4. How many perfect cubes are there from 1 to 500? How many are perfect square among cubes?

Ans: There are 7 perfect cubes from 1 to 500. Only 1 3 and 4 3 are the squares among these cubes. 5. Find the cubes of 10, 30, 100, 1000. What can you say about the zeros at the end? Ans: 10 3 = 1000 30 3 = 27000 100 3 = 1000000 1000 3 = 10000000000 The number of zero of a cube are 3 times, the no. of zero of numbers. 6. What are the digits in the unit s place of the cubes 1, 2, 3, 4, 5, 6, 7, 8, 9, 10? Is it possible to say that a number is not a perfect cube by looking at the digit in unit s place of the given number, just like you did for squares? Ans: Numbers The digit in the unit s place

1 2 3 4 5 6 7 8 9 10 1 8 7 4 5 6 3 2 9 0 EXCERSISE 1.2.8 1. Find the cube root by prime factorization. i) 10648 2 10648 2 5324 2 2662 11 1331 11 11 1 3 10648 = 3 2 2 x 11 2 = ii) 46656 2 46656 2 23328 2x11 = 22

2 11664 2 5832 2 2916 2 1458 9 729 9 81 9 9 1 3 15625 = 3 5 3 x 5 3 = 5x5 = 25 2.Find the cube root of the following by looking at the last digit and using estimation. i) 9112 5 4 3 < 91 < 5 3 40 3 < 91125 < 50 3 45 ii) 16637 5 5 3 < 166 < 6 3 50 3 < 1166375 < 60 3 55 iii) 70496 9 8 3 < 704 < 9 3 80 3 < 704969 < 90 3

89 3.Find the nearest integer to the cube root of each of the following. i) 331776 216000 < 331776 < 343000 60 3 40 3 3 331776 lies between 60 and 70. ii) 46656 27000 < 46656 < 64000 30 3 40 3 3 466656 lies between 30 and 40. iii) 373248 216000 < 373248 < 343000 3 373248 lies between 60 and 70. UNIT 2 CHAPTER 2 EXERCISE 2.2.2

1. Resolve into factors: i) x 2 + xy ii)3x 2 6x x(x+y) 3x(x-2) iii)(1.6)a 2 (0.8)a iv) 5-10m 20m 0.8a (2a 1) 5(1-2m 4n) II. Factories. i) a 2 + ax + ab + bx a(a + x) (a + b) (a + x) ( a + b) ii) 3ac + 7bc 3ad - 7bd 3ac + 7bc 3ad 7bd c(3a + 7b) d(3a + 7b) (3a + 7b) (c d) iii) 3xy 6yz 3xt + 6zt 3y ( x-2z) 3t(x-2z) (x-2z) (3y- 3t) iv) y 3-3y 3 + 2y 6 xy + 3x y 2 (y-3) + 2(y -3) x(y-3) (y-3) ( y 2 + 2 x) 3.Factorise i) 4a2 25

(2a)2 ( 5)2 (2a + 5) (2a 5) ii) x 2 9 16 (x) 2 - ( 3 ) 2 4 ( x + 3 ) ( x 3) 4 4 iii) x 4 y 4 (x 2 ) 2 - (y 2 ) 2 (x 2 + y 2 ) ( x 2 y 2 ) iv) 7 3 2-2 1 2 10 10 7 3 + 2 1 7 3 2 1 10 10 10 10 = 94 52 10 10 = 1222 25 v) (0.7) 2 (0.3) 2 (0.7 + 0.3) (0.7 0.3) (1) (0.4) = 0.4

vi) (5a 2b) 2 (2a b) 2 [5a 2b + ( 2a b )] [ 5a 2b ( 2a b)] (5a 2b + 2a b ) ( 5a 2b 2a + b) (7a 3b) ( 3a b) EXERCISE 2.2.3 1. In the following you are given the product pq and the sum p+ q. Determine p and q. i) pq= 18 and p + q = 11 +9 Product = 18 p = 9, q = 2 +2 Sum = 11 ii) pq= 32 and p + q = - 12-8

Product = 32-4 Sum= -12 p = - 8, q = -4 iii) pq = - 24 and p + q = 2 +6 Product = 24-4 Sum = 2 iv)pq = -12 and p + q = 11 +12 Product = - 12-1 Sum = 11 p = + 12, q = -1 v)pq = -6 and p + q = -5-6

Product = - 6 p= - 11, q = + 4 +1 Sum= - 7 II. Factorise i) x 2 + 6x + 8 +4x x 2 + 4x + 2x + 8 x(x + 4) +2 (x + 4) Product = 8x 2 x(x+4) +2 (x + 4) (x + 4) ( x + 2) +2x Sum = 6x ii) x 2 +4x + 3 x 2 + 3x + 1x + 3 +3x x( x + 3) + 1 ( x + 3) Product = 3x 2 (x + 3) ( x + 1) + 1x Sum = + 4x iii) a 2 + 5a + 6 +3x a 2 + 3a + 2a + 6 a(a + 3) + 1 ( a + 3) Product =6a 2

(a + 3) (a + 2) Sum = +5a +2a iv) a 2 5a + 6-2a a 2 2a 3a + 6 a(a 2) 3( a- 2) Product = 6a 2 (a 2) ( a 3) -3a Sum = - 5a v) a 2 3a 40-8a a 2 8a + 5a 40 a(a 8) + 5(a -8) Product = -40a 2 (a 8) ( a+ 5) +5a Sum = - 3a vi) x 2 x 72-9x x 2 8x + 5x 40 x(x 9) + 8 ( x 9) Product = - 72x 2 ( x 9) ( x + 8 ) +8x Sum= - x ADDITIONAL PROBLEM ON FACTORISATION

1. Choose the correct answer : a. 4a + 12 b is equal to a. 4a b. 12b c. 4(a + 3b) d. 3a (c ) b. The perfect of two numbers is positive and their sum negative only when a. Both are positive b. both are negative c. one positive the other negative d. one of them equal to zero ( b) c. F a c t 2 + 6x + 8, we get a. ( x + 1) ( x + 8) b. ( x + 6) ( x + 2) c. ( x + 10 ) ( x + 2) d. ( x + 4) ( x + 2) (d) d. The denominator of an algebraic fraction should not be a. 1 b. 0 c. 4 d.7 (b) e. If the sum of two integers is 2 d their product is 24 the number are a. 6 and 4 b. 6 and 4 c. 6 and 4 d. 6 and 4 ( b)

f. The difference ( 0.7) 2 ( 0.3)2 simplifies to a. 0.4 b. 0.04 c. 0.49 d. 0.56 (a) 2. Factorise the following: i) x 2 + 6x + 9 x 2 + 3x + 3x + 9 x( x+ 3) + 3 ( x + 3) (x + 3) ( x + 3) = ( x + 3) 2 ii) 1-8x + 16 x 2 16x 2 8x + 1 16x 2 4x 4x +1 4x (4x 1) 1(4x 1)) =(4x 1) 4x 1) iii) 4x 2 81y 2 (2x) 2 ( 9y) 2 (2x + 9y) ( 2x 9y) iv) 4a 2 + 4ab + b 2 4a 2 + 2ab + 2ab + b 2 2a(2a + b) + b(2a + b) (2a + b) ( 2a + b) v) A 2 b 2 + c 2 d 2 a 2 c 2 b 2 d 2 A 2 b 2 a 2 c 2 + c 2 d 2 b 2 d 2

A 2 (b 2 c 2 ) + a 2 (c 2 b 2 ) A 2 ( b 2 c 2 ) d 2 (b 2 c 2 ) =( b 2 c 2 ) ( a 2 d 2 ) 3. Factories the following: i) x 2 + 7x + 12 +4x x 2 + 4x + 3x + 12 x( x + 4 ) + 3( x + 4) P=12 x 2 (x + 4) ( x + 3) +3x Sum = 7x ii)x 2 + x 12 +4x x 2 3x + 4x 12 Product =12x x(x-3) + 4(x 3) - 3 ( x 3 ) ( x + 4 ) Sum = +x iii)x 2 3x 18-6x x 2 6x + 3x - 18 Product = 18x2 x( x 6 )+ 3 (x 6) +3x ( x 3 ) ( x 7) Sum = -3x iv)x 2 + 4x 192 +7x x 2 3x + 7x 21 P= +21x2

x( x 3) + 7 ( x 3) -3x ( x 3 ) ( x 7) Sum = +4x v)x 2 4x 192 +12x x 2 16x + 12x 192 P = + 192 x x( x 16 ) + 12 ( x 16) -16x ( x 16) ( x + 12 ) Sum = -4x vi)x 4-5x 2 + 4 x 4 4x 2 1x 2 + 4-4x 2 x 2 ( x 2 4) 1 ( x 2 4) P = + 4x4 (x 2 4) ( x 2 1) -1 x 2 = [x 2 (2) 2 ] [ x 2 ( 1) 2 ] Sum = - 5x 2 (x + 2) ( x 2) ( x + 1) ( x -1) vii)x 4 13x 2 y 2 + 36y 4-9x 2 y 2 x 4 9x 2 y 2 4x 2 y 2 + 36y 2 P=+36x 4 y 4 x 2 (x 2 9y) 4y 2 (x 2 9y 2 ) -4x 2 y 2 (x 2 9y 2 ) (x 2 4y 2 ) [x 2 (3y) 2 ] [x 2 (2y) 2 ] ( x + 3y) ( x 3y) ( x + 2y) ( x 2y) 4. Factorise the following:

i) 2x 2 + 7x +6 24x 2x 2 + 4x + 3x + 6 Product = 12x 2 2x( x+ 2) + 3 ( x + 2) +3x ( x + 2) ( 2x + 3) Sum = 7x ii) 3x2 17x + 20-22x 3x2 12 x + 5x + 20 P = + 60x2 3x ( x 4) 5(x 4) + 5x ( x 4) ( 3x 5) Sum = - 17x iii) 6x 2 5x 14-12xy 6x 2 12x + 7x -14 P= x 2 y 2 6x (x-2) + 7 ( x-2) + 7x (x 2) (6x 7) Sum = -5x iv) 4x2 + 12xy + 5y2 4x2 + 2xy + 10xy + 5y2 +2xy 2x ( 2x+y) 5y (2x + y) P=20x2y2 (2x + y) (2x + 5y) +10xy Sum= 12xy v) 4x 4 5x 2 +1 4x 4 4x 2 1x 2 + 1-4x 2 X 2 (x 2 1) -1 ( +1x 2-1 ) P=4x 4 (x 2 1) (4x 4 1) -1x 2 [(x 2 ) (1) 2 ] [(2x) 2 (1) 2 ] Sum= -5x 2 (x+1) (x -1)(2x+1) (2x-1)

CHAPTER 3 UNIT 2 EXERCISE 3.2.1 1. Match the following i) (a) Equilateral triangle ii) (b) Acute angled triangle iii) ( c) Right angled triangle

iv) (c) Obtuse angled triangle Ans: i. c ii. d. iii. a iv. b. 2. Based on the sides, classify the following triangles ( figures not drawn to the scales) i) ii) 3cm 5cm 7cm 5cm 4cm Scalene Triangle 4cm Scalene Triangle iii). iv) 6.5 cm 3.5 cm 6.5cm 6.5cm 6cm Scalene Triangle 6cm Isosceles Triangle

v). vi) 3cm 5.6cm 2.5cm 4.1cm 4.3 cm 3.2 cm Scalene Triangle Scalene Triangle vii). viii). 3cm 9cm 5cm 3cm 3cm 6cm Scalene Triangle Equilatereal Triangle ix). x). 6cm 6cm 5cm 5cm 8cm 3.5cm Isosceles Triangle Isosceles Triangle

EXERCISE 3.2.2 1. In a triangle ABC, if A = 55 and B = 40, find C Ans: In ABC A + B + C = 180 ( Theorem 3) 55 + 40 + C = 180 C = 180 - ( 55 + 40 ) C = 85 2. In a right angled triangle, if one of the other two angle is 35, find the remaining angle. Ans: A = 35, B = 90, ( In a right angled triangle, one angle should be 90 ), in a right angled triangle A + B + C = 180 ( Theorem ) 35 + 90 + C = 180 C = 180 - ( 35 + 90 ) C = 180 125 C = 55

3. If the vertex angle of an isosceles triangle is 50, find the other angles. Ans: A = 50, Let B and C be x ( Because in isosceles two angles are equal) A + B + C = 180 ( Theorem) 50 + x + x = 180 2x = 180 50 X = 130 2 X = 65 A = 50, B = 65 and C = 65 4. The angles of triangle are in the ratio 1:2:3 Determine the three angles. Ans: 1:2:3 Let the angles be 1x, 2x and 3x 1x + 2 x + 3x = 180 ( Theorem 3) 6x = 180 x = 180 6 x = 30, 2x = 2 x 30 = 60 3x = 3 x 30 = 90

5. In the adjacent triangle ABC, find the value of x and calculate the measure of all the angles of the triangle. Ans: In ABC A + B + C = 180 ( Theorem 3) x+ 15 + x + 15 + x + 30 = 180 30x + 30 = 180 A = x + 15 + 50 + 15 = 65 B = x 15 = 50 15 = 35 C = x + 30 = 50 + 30 = 80 3x = 180-30 x = 150 3 x = 50 6. The angles of triangle are arranged in ascending order of their magnitude. If the difference between two consecutive angles is 10, find the three angles. Ans: Let the angles be x, x + 10 and x + 20 x + x +10 + 20 = 180 ( Theorem 3) 3x + 30 = 180 3x = 180 30 x = 150 3 x = 50

x + 10 = 50 + 10 = 60 x + 20 = 50 + 20 = 70 The angles are 50, 60 and 70 EXERCISE 3.2.3 1. The exterior angles obtained on producing the base of a triangle both ways are 104 and 136. Find the angels of the triangle. Ans: A 136 104 B C T M Given ACT is triangle with the lines BC and TM extending both ways. BCA = 136, ATM = 104 Proof : BCA + ACT = 180 ACT = 180 136 = 44 CTA + ATM = 180 ( Pro.1) CTA = 180 104

CTA = 76 In ACT, A + C + T = 180 ( Interior angle theorem) A + 76 = 180 A + 180 ( 44 + 76 ) A + 180 120 A = 60 The Angles in the ACT are, A = 60, C = 44, T = 76 2. Sides BC, CA and AB of a triangle ABC are produced in an order, forming exterior angles ACD, DAE and CBF. Show that ACD + DAE + CBF = 360. Ans: E E A B C D F

ACB + ACD = 180 ( Proposition 1) BAC + CAE = 180 ( Proposition 1) FBC + ABC = 180 ( Proposition 1) Adding above equations. ACB + ACD + BAC + CAE + FBC + ABC = 360 + 160 ACD + CAE + CBF + { ACB + BAC + ABC } = 360 + 180 ACD + BAE + CBF + 180 ( Interior angle theorem ) = 360 + 180 ACD + BAE + CBF = 360 ( Axiom 3) 3. Compute the value of x in each of the following figures: Ans: i) Given ABC is an isosceles single and CD is extended ABC = 50 A To prove - ACD Proof : In ABC 50 x C = B = 50 ( AB = AC ) B C D BCA + ACD = 180 ( Proposition 1) P ACD = 180-50 T 180 ACD = 130 x = 130 160 x S Q R

ii) Given PTR = 130, SQT = 106 To prove QRT Proof : SQT + TQR = 180 ( Prop. 1) TQR = = 180-106 TQR = 74 PTR + PTQ = 180 ( Prop.1 ) PTR = 180-130 PTR = 50 In PTQ, R + T + Q = 180 ( Interior angle theorem) R = 180 - ( 50 + 74 ) R = 56 QRT = 56, x = 56 iii) Given XRY = 65, RST = 100 To Prove RZS Proof : ZRS = XRY = 65 (Vertically opposite angle) ZSR + RST = 180 ( Prop. 1) ZSR = 180 100 ZSR = 80 In RZS, R + Z + S = 180 ( Interior angle theorem)

Z =180 - ( 65 + 80 ) x y Z = 180 145 Z = 35 65 R RZS = 35 x 100 X = 35 z s T iv) Given ABE = 120, BED = 112 To prove BEC A Proof : ABE + EBC = 180 ( Proo.1) 120 B EBC = 180-120 EBC = 60 EGB + BCD = 180 ( Prop.1) x 112 ECB = 180 120 E C D ECB = 68 In EBC, E + B + C = 180 ( Interior angle theorem) E = 180 - ( 60 + 68 ) E = 52 BEC = 52 x = 52 v) Given ACT is an isosceler triangle. ATC = 20 To prove - ACN Proof : In ACT, A = T = 20 ( AC = CT )

In ACT, A + C + T = 180 ( Interior angle theorem) ACN = 180 - ( 20 + 20 ) A ACN = 140 ACN = 40 x = 40 X 20 N C T 4. In the figure, QT PR, TQR = 40 and SPR = 30, Find TRS and PSQ Ans: In TRQ, QTR = 90 ( data) TQR = 40 (data) TQR = 180 - ( 90 + 40 ) (Remaining angle) = 180-130 TRQ = 50 In PSR, Ext PSQ = SPR + PRS ( Exterior angle theorem) = 30 + 50 Ext PSQ = 80 5. An exterior angle of triangle is 120 and one of the interior opposite angles is 30. Find the other angles of the triangles. Ans: ACB + ACD = 180 ( Proposition 1)

ACB = 180-120 ACB = 60 ( Third angle) In ABC, A A + B + C = 180 ( Interior angle theorem) A = 180 ( 30 + 60 ) 30 120 A = 180-90 B C D A = 90 ( Other interior opposite angle) 6. Find the sum of all the angles at the five vertices of the adjoining star. Ans: 2 + 4 + 6 = 180 (Interior angle theorem) 3 1 + 8 + 10 = 180 ( Th.3) 2 8 9 4 2 + 9 + 5 = 180 ( Th.3) 7 10 1 + 8 + 4 = 180 ( Th.3) 6 3 + 7 + 5 = 180 ( Th.3) 1 5 Adding above equations, 2 1 + 2 2 + 2 3 + 2 4 + 2 5 + 6 + 7 + 8 + 9 + 10 = 5 x 180

2 ( 1 + 2 + 3 + 4 + 5 ) + 6 + 7 + 8 + 9 + 10 = 900 2 ( 1 + 2 + 3 + 4 + 5 ) + 540 ( Sum of angles of a pertragon is 540 ) = 900 2 ( 1 + 2 + 3 + 4 + 5 ) = 900-540 1 + 2 + 3 + 4 + 5 = 360 2 1 + 2 + 3 + 4 + 5 = 180 ADDITIONAL PROBLEM ON THEOREMS 1. Fill in the blanks to make the following statements true. a. The sum of the angles of a triangle is ( 180 ) b. An exterior angle of triangle is equal to the sum of opposite angles. ( Interior ) c. An exterior angle of a triangle is always than either of the interior opposite angle. ( Larger) d. A triangle cannot have more than right angle. (1) 2. Choose the correct answer from the given alternatives: 1. In a triangle, ABC, A = 80 and AB = AC, then B is a. 50 b. 60 c. 40 d.70 (a) 2. In right angled triangle, A is right angle and B = 35, then C is a. 65 b. 55 c. 75 d.45 (d)

3. In a triangle ABC, A = C = 45, then the triangle is a. right angle b. Acute angled triangle c. Obtuse angle triangle d. Equilateral triangle (a) 4. In an equilateral triangle, each exterior angle is a. 60 b. 90 c.120 d. 150 ( c ) 5. Sum of three exterior angles of a triangle is a. Two right angles b. Three right angles c. One right angle d. Four right angles (d)