SETION HPTER REL NUMERS TOPI- Rational Numbers.. 58 0.058 (Decimal point is shifted three places 000 to the left) 5 6 S O L U T I O N S.0.... It is non-repeating, so it is not a rational number. 0.60.... It is non-repeating, so it is not a rational number.. 7.77.... s, 7 is repeating continously, so it is a rational number.. 6.9.... It is non-terminating, non-repeating So it is not a rational number. So,. 7 is a rational number.. Let, 0.999... 0 9.999... 0 (9.999...) ( 0.999...) 9 9. 98 98 9 7 So, it is a rational number. 5. Yes, zero is a rational number. Zero can be epressed as 0, 0, 0 etc, 5 6 00 which are in the form of, p q where p and q are integers and q 0. 6. Let, 0. 6 0.6666...(i) multiplying 0 on both the sides, we get, 0 6.6666...(ii) From (ii) (i), we get 9 6.0 6 9. 7. 7 0.857 7 0 7 0 8 0 60 56 0 5 50 9 WORKSHEET- \ 0. 857 [SE Marking Scheme, 0] 8. Since LM of 7 and is 77, 9 \ 9 7 6 7 77 59 and 5 7 5 7 55 77 85 59 Hence, three rational numbers between 5 7 and 9 are : 86 88 59 59 59 9. ny eample & verification of eample : Let m /5, n 9/ 9 5 Sum + (Rational Number) 5 0 product 6 (Rational Number) 0 9 7 diffrence (Rational Number) 5 0 division 9 5 5 8 (Rational Number) P-
. 77 59 59 7 7.. 7 8 0.875... 0.000..... n. 5. 000 7. 7 7 [SE Marking Scheme, 0] 999 6. lternative Method : Let 0. 7 0.7777... 000 7.77... 000 (7.77...) (0.77...) 999 7 \ 7 999. 57 65.5 Terminating [SE Marking Scheme, 0] lternative Method :.5 65) 57 875 80 500 00 5 750 65 50 50 57.5 (Terminating) 65 7. lternative Method : LM of 5 and 7 is 5 \ and so, 5 5 5 5 7 7 5 7 7 5 5 5 5 5 < 5 < 5 < 5 < 5 5 WORKSHEET- The required three rational numbers are 5, 5 and 5. [SE Marking Scheme, 0] 8. Let a & b Here, we find si rational numbers, i.e., n 6 So d b a n+ 6 + 7 st rational number a + d + 7 7 nd rational number a + d + 7 7 rd rational number a + d + 7 7 th rational number a + d + 5 7 7 5 th rational number a + 5d + 5 6 7 7 6 th rational number a + 6d + 6 7 7 7 So, si rational numbers are 7, 7, 7, 5 7, 6 7 & 7. 7 9. LM of,, 6, and is 6 5 5 5 5 6 7 6 7 7 9 Descending order is 5, 9, 6, 6 i.e., 5, 7,,. [SE Marking Scheme, 0] lternative Method : Since, LM of,, 6, is \ 6 5 5 5 5 5 6 6 7 7 7 7 9 P- M T H E M T I S - I X T E R M -
In Descending order i.e., 5, 9, 6, 6 5, 7,,. 0.. 9 90 0. 5 5 99. + 0. 5 659 990 [SE Marking Scheme, 0] lternative Method : Let,.... 0... 00... 00 0 (...) (...) TOPI- Irrational Numbers 90 9.00 9 90 gain, let y 0. 5 0.555... 00y 5.55... 00y y (5.55...) (0.55...) 99y 5 y 5 99 \. + 0. 5 + y 9 90 + 5 99 9 + 5 0 990 09+50 990 659 990. 0. is a terminating number. So, it is not an irrational number. 0. 5 0.55..., 5 is repeating continuously, so it is not an irrational number. 0.5 0.55..., 5 is repeating continuously, so it is not an irrational number. 0.000000..., non-terminating and nonrecurring decimal. Hence, it is an irrational number. So, 0.000000 is an irrational number.. No, it may be rational or irrational.. Let, 0.777... 0 7.777... 0 (7.777...) (0.777...) 9 7 7 9.. Sum of 5 and 7 5 + 7. 5. Required two irrational number are : 0. 0.0 (i).0000000000... (ii).000000000... 6. 0.50000000000... and 0.50000000000.... [SE Marking Scheme, 0] 7. D.5.5 O E.5 WORKSHEET- Mark the distance.5 units from a fied point on a given line to obtain a point such that.5 units. From, mark a distance of unit and mark the new point as. Find the mid-point of and mark that point as O. Draw a semi-circle with centre O and radius O. Draw a line perpendicular to passing through and intersecting the semi-circle at D. Then, D.5 To represent.5 on the number line, let us treat the line as the number line, with as zero, as, and so on. Draw an arc with centre and radius D, which intersects the number line at E. \E represents.5 8. Let two irrational numbers are : 6 and, (i) 6 Difference is an irrational number. S O L U T I O N S P-
(ii) 6 + sum is an irrational number. (iii) 6 8. Since, 5.6 Hence, the irrational number between and.5 is 5.. 0. + 0. (0...) + (0...) 0.777... Let, 0.777... 0 7.777... 0 (7.777...) (0.777...) 9 7.0 7 9.. is non-terminating, non-recurring.. ( + 5 ) + 5 + 0. ( ) + ( 5) + 5 7 + 0 (Irrational Number). 5. Let.8.8888... 0.8888... 000 8.888... 000 0 (8.888...) (.888...) 990 96.00 96 990 55 6. Given, 55 [SE Marking Scheme, 0] 7 7 0.857857... 0.857857... Hence, required number can be 0.606006000... [SE Marking Scheme, 0] 7. D O P Q product is an irrational number. (iv) 6/ division is an irrational number. Let WORKSHEET- unit length Using Pythagoras theorem, we see that O + onstruct D unit length perpendicular to O, then using Pythagoras theorem, we see that OD ( ) + Using a compass with centre O and radius OD, draw an arc which intersects the number line at the 8. point Q, then Q corresponds to. D O E 9. 9. 9. Mark the distance 9. units from a fied point on a given line to obtain a point such that 9. units. From, mark a distance of unit and mark the new point as. Find the mid-point of and mark that point as O. Draw a semi-circle with centre O and radius O. Draw a line perpendicular to passing through and intersecting the semi-circle at D. Then, D 9. To represent 9. on the number line,let us treat the line as the number line, with as zero, as, and so on. Draw an arc with centre and radius D, which intersects the number line at E. \ E represents 9. [SE Marking Scheme, 0] 9. 0.7878...() 0000 78.7878...() Subtracting () from e.q. () 9990 (78.78....78...) 75 65 9990 998 P- M T H E M T I S - I X T E R M -
TOPI- n th Root of a Real Number. ( 5+ 5 ) ( 5 5 ) 5 ( 5 ) S O L U T I O N S { } {5 5} 0 [SE Marking Scheme, 0]. 8 6.. 8 7 7 7. 8 + (8 + ) 5. 50 0 8 8 0 880 6. + ( ) + + + + 7. ( + ) 6 7 7 7 7 5 5 96. 5 p 7 5 5 7 p 7 WORKSHEET-5 \ 6 6 + 8 5 96 6 7 + 8 + 5 56 56 0. 8. ( + ) ( ) ( ) ( ) 8 8 0. [SE Marking Scheme, 0, 0] 9. 5 5 00 50 0. (i) 9 55 5+0 5 5 5 ( ) ( + ) ( ) ( + ) is an irrational number ( ) ( ( + ) ( ) ( ) ( ) which is an irrational number. Let, there is a number such that is an irrational number but 5 is a rational number. Let, 5 7 be the number ( ) 5 7 5 (7) is an irrational number ut, 5 ( ) 5 5 7 (7) 5/5 7 7, is a rational number (ii) ccepting own mistakes gracefully, co-operative learning among the classmates. p 5 7 7 WORKSHEET-6 5 7.. b a a b. P-5
. ( a+ b) ( a b ) (a) ( ) b a b.. False Justification : 7 75 7 75 9 5 7 5 which is a rational number. [SE Marking Scheme, 0] 5. ( 5 ) ( ) + ( 5 ) 5 [Using (ab) a + b ab] 8 + 5 5 9 5 ( 8 5 ). [SE Marking Scheme, 0] 6. 50 98 + 6 7. 5 5 7 7 + 5 7 + 9 7 [SE Marking Scheme, 0] lternative Method : 50 98 + 6 5 5 7 7 + 5 7 + + 9 7 0 0 5 5 5 5 0 5 5 0 5 5 [SE Marking Scheme, 0] lternative Method : 0 5 5 0 5 5 5 0 0 5 5 5 5 6 5 6 5 5 5. 8. LM of and is 6 6 lternative Method : LM of and is 6 6 6 7 [SE Marking Scheme, 0] \ ( ) 6 and ( ) 6 Now, 6 9 6 8 9. ( ) / 6 lternative Method : 6 6 7 6 79 ( ) / 6 79 8 7 9 [SE Marking Scheme, 0] 79 6 0. Justify, 5 + ( ) 6 79 ( ) 6 6 ( ) ( 5 ) ( ) 5 + 5 ( 5 ) ( ) 0 6 gain, 8 + 98 + 7 0 6 0 6 5 7 + 7 7 + 7 7 7 + 7 + 7 Value : o-operative learning among classmates without any gender and religious bias. P-6 M T H E M T I S - I X T E R M -
TOPI- Laws of Eponents with Integral Powers.. [(6) ] ( ) / / 5. 5( 8 + 7 ) / WORKSHEET-7 / / ( + ) / 5 ( ) ( ) / 5 ( + ) / ( ) / 5 (5) 5 5 6. ( 5 ) + ( + ) ( ) ( ) ( )(5) + (5) + ( ) + ( ) ( ) + ( ) ( ) () +( ) (). ab ( c) ba ( c) S O L U T I O N S b ( ) a ( ) c ab ac bc ba bc [ ] ( abacba+bc ) ( bcac ) 0 ac [SE Marking Scheme, 0] lternative Method : ab ( c) b c ab ac bc ba ( c) a ba bc ac ( abacba+bc ) ( bcac ) ( bcac ) ( bcac ). + + + 0 9 8 0 9 8 ( + + ) 9 ( + ) (9 + + ) (9 + ) [SE Marking Scheme, 0] 7. 8. 8 + 5 0 + 8 + + 6 6 + 5 8 + 6 6 [SE Marking Scheme, 0] + + (6) (56) () 5 + + (6 ) ( ) 5 ( ) 5 + 6 + + 6 6 + 6 + 6 + 8 9 5 6 5 6 9 8 5 5 5 5 P-7
5 5 5 5. ( ). ( 8) ( ) ( ) 8 8 0. { }. 5( + ) { } ( 8) ( 8) ( ). (5 ) 5 [SE Marking Scheme, 0] lternative Method :. 5 8 + 7 8 5 6 9 5 ( ) + ( ) 5( + ) 5(5) (5 ) 5 5. 5 5 5 5 8 5 [SE Marking Scheme, 0] 5 5 8 5 5 5. 6 9 WORKSHEET-8 6 9 8 5 5 5 5 8 5 7 5 8 7 7 8 [SE Marking Scheme, 0, 0] lternative Method : 8 5 5 6 9 5 5 5 5 5 5 8 7 5 7 5 8 8 7 7 8 7 5 5 7 5 7 6. 5 5 7 5 7 7 5 5 7 7 7 7 5 5 5 5 7 5 6 8 7 7 5 5 5 7 5 6 5 7 5 8 5 7 P-8 M T H E M T I S - I X T E R M -
5 7 5 0 5 7 ( ) 7 5 5 7 5 5 7 0 7 5 7 5 75 [SE Marking Scheme, 0] lternative Method : 7 5 5 7 5 7 5 5 7 5 7 7 5 5 7 7 7 7 5 5 5 5 7 5 6 8 7 7 5 5 5 7 5 6 8 7 7 5 5 5 5 7 5 0 5 7 5 7 5 5 7 0 ( ) 7 5 7 5 75 ( a+ b) ( b c) ( c a). +. + 7. a b c ( ) a+ b b+. c. c+ a a b c.. a+ b+ c a+ b+ c [SE Marking Scheme, 0] lternative Method : ( a+b ) (b+c) (c+a).. a b c ( ) ( a+ b) ( b+ c) ( c+ a).. a b c.. a+ b b+ c c+ a.. a b c.. a+ b+ b+ c+ c+ a a+ b+ c a b c 8. 5 a+ b+ c + + 7 5 7 5 6 8 5 7 5 6 8 7 5 5 5 0 [SE Marking Scheme, 0] lternative Method : 7 5 5 7 5 6 8 5 7 5 6 8 7 5 5 0 5 0 5. WORKSHEET-9. 500 5 500 5. 6 6 / / 00 0 5. ( ) 6 + (6) 6. S O L U T I O N S P-9
. ( + + ) / ( + 8 + 7) / (6) / [(6) ] / 6 6 6 [SE Marking Scheme, 0]. Given, a and b. a b + b a + 8 + 9 7 [SE Marking Scheme, 0] 5. ( ab ) a+b. ( bc ) b+c. ( ca ) c+a a b b c c a.. 6. (6) a b + b c + c a 0 (any number to the power 0 is ) [SE Marking Scheme, 0] (56) (6 ) ( ) 6 6 80 [SE Marking Scheme, 0, 0] lternative Method : 7. (6) (56). ab ab + a b+ a ab a (6) 6 () 6 6 80 b ab + b + ab b a b lternative Method : a a + a + a +b a b + a b b a [SE Marking Scheme, 0] a a+ b + ab a a b b a a ab b a+ b + b b a ( b ab ) + bb ( + a ) ( b+ a)( b a) b ab + b + ab b a b ( a b ) b a b 8. LHS + + y + y + + y + y + + y y ( yz z y) y y + + y + y + y + + y y + + y + + y RHS y + y + Hence proved. [SE Marking Scheme, 0] lternative Method : ( + + y ) + ( + y + z ) + ( + z + ) + + + + + y+ + z+ y z + + + + + y + y + + y y ( yz z y) + + y + y + y + y + y + + y y y y y + + y + y + y + y + y + + y ( y + + y ) ( y + y + ). 9. 5 (5 ) 500 5 5 500 5 (5 ) 500 5 500 65 5 6 \ [SE Marking Scheme, 0] P-0 M T H E M T I S - I X T E R M -
lternative Method : Given, 5 (5) 500 5 [(5 )] 500 5 5 500 5 (5 ) 500 5 500 TOPI-5 Rationalisatoin of Real Numbers 65 5 5 comparing the powers of both sides, we get 6.. 50 5 5 5 0 So, rationalising factor is. 6 6 6+ 6 ( 6 ) 6 8 6 + 8 8 8 8 (7 ) 7. +. + 8 + + [SE Marking Scheme, 0] lternative Method : ( + ) + 98 ( ) ( + ) ( + ) + + + \ + WORKSHEET-0 + + + 8 + +. + 5. 6. + Squaring, both sides, we get + ( ) + 5+ + 5 5+ 5 ( ) ( ) + + + + + ( + ) ( + + ) 5 + 5 6 5 5 + 5 6 + + + + 6 + + 6 + 6 6 + + 8 6 ( 6) + + 6 6 S O L U T I O N S P-
5 + 6 + 5 + + 6 7. LHS + + 8 8 7 7 6 6 5 5 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) + 8 8 + 7 7 + 6 + 8 8 7 7 6... ( ) ( + ) ( ) lternative Method : +. 0. [SE Marking Scheme, 0] ( ) ( + ) ( ) ( ) ( ) ( ). 0.. (rationalising). 0.707 + p 0.707 +..88. [SE Marking Scheme, 0] 5+ 6 5 6 5 + 6 5 6 ( 5+ 6) ( ) 5 + 6 5+ 6 5 ( 6) 5 + 6 + 0 6 5 6 + 0 6 9 ( 6 5) ( 5+ ) ( 6 ) ( 5) ( 5 ) ( ) + ( 8 + 7) ( 7 + 6) ( 6 + 5) + 8 + 98 87 76 65 ( 5+ ) + 5 [ a b (a b) (a + b)] + 8 8 7 + 7 + 6 6 5 + 5 + + 5 RHS \ a + b 6 9 WORKSHEET- + 0 9 6 omparing the rational and irrational parts of both sides, we get. and y + 98 + + 98 a 9, b 0 9 + + + + y+ y + + + (+ ) ( ) 6 + 9 8 7 5. (i) We know that between two rational numbers and y, such that < y there is a rational number + y. i.e., < 7 < Now, a rational number between and 7 < is : 7 + 6 + 7 rational number between 7 and is : P- M T H E M T I S - I X T E R M -
7 + 7 + 8 5 + y + + + \ < < 7 < 5 < Further a rational between and is : + + 5 8 rational number between 5 and is : + + 6 + 6 + + 6 + + 6 5 + 6 5 6 + 5 6 5+ 6 ( 5+ 6 ) + ( 5 6 ) ( 5 ) ( 6) S O L U T I O N S 5 + 5 + 6 8 rational number between 8 \ < 5 8 and is : + + 8 8 6 6 < < 7 < 5 < 8 < 6 6 < Hence, si rational numbers between and are (ii) Number system. 5 7 5,,,,, and 6 8 8 6 (iii) Rationality is always welcomed. 6. + and y +. 8 + 5 + 0 ( 5 ) ( 5 + ) 5 ( 0 ) 0 ( ) ( 0 + ) + 8 ( 0 + ) 0 ( 0 + ) 8 + 5 0 8 ( 0 + ) + 5 + + 0 6 + 5 + 0 6 5 98 ( 5 + ) + ( 5 ) 7. LHS ( 5 )( 5 + ) 5+ + 5 + 5+ 5 6 7 \ 6 7. ( 5 ) ( ) 0 + + 5 + 0 + 5 0 6 7 + 5 (0) + 5 (0) a + 5 b RHS comparing both sides, we get a 6, b 0 7 [SE Marking Scheme, 0] WORKSHEET- 0 0 + 0. 7 + ( 7 + ) ( 7 ) ( 7 ) 7 ( 7 ) ( ) 7 79 7 8 7 7 7. P-
[ p+ q + p q]. p+ q p+ q q p q ( p p ) q p q + q (q p) q q p + q 0. [SE Marking Scheme, 0] lternative Method : p+ q + pq p+ q + pq p+ q pq p+ q + pq p+ q + pq + p+ q pq ( ) ( ) ( p+ q) ( pq) q p + p+ p q p+ q p+ q ( p+ p ) q q p q q p p q Squaring both sides, we get q + p pq p q q (q p) q q p + q 0. 0.785. (i) 7 5.000000 9 0 7 0 8 0 60 56 0 5 5 \ and 5 7 0. 785 0.8 9.00 88 0 9 \ 9 0. 8 Thus, three different irrational numbers between 5 7 and 9 are 0.75075007500075000075..., 0.767076700767000767... and 0.80800800080000... (ii) Number system. (iii) Those who are irrational in their approach fail in their efforts. 5. + ( ) ( + ) ( + ) + ( ) ( ) + + 0 \ + 0 \ 7 + 5 ( + ) + ( + ) + 0 + 0 + [SE Marking Scheme, 0] FORMTIVE SSESSMENT WORKSHEET- & Note : Students should do these activities themselves. qqq P- M T H E M T I S - I X T E R M -
SETION HPTER POLYNOMILS TOPI- Polynomials. p (y) y y + p (0) 0 0 +. + is a factor of + + m p ( ) 0 ( ) + ( ) + m 0 6 + m 0 m... Not defined. 5. Linear polynomial. 6. Degree of + 5 S O L U T I O N S Degree of 5 5 Degree of ( 5) ( 5 ) + 5 8. 7. Not a polynomial. 8. ( + ) is a factor of + k p () 0 () + () k () 0 7 + 7 + k 0 k. onstant polynomial is 7.. inomial.. Degree of a polynomial is 0.. Every real number is a zero of the zero polynomial. 5. No. of zeroes of cubic polynomial. 6. ( + ) ( + ) ( + ) [(+) ] One factor of ( + ) ( + ) is ( + ). 7. Given, p() + + k Since is a zero of polynomial p( ) 0 ( ) + ( ) + k 0 WORKSHEET-5 9. ( + ) is a factor of p() k + + 0 p () 0 () k () + () +0 0 6 k 6 + 0 0 k [SE Marking Scheme, 0] 0. p() + k + ( ) is a factor of p(), then p() 0 () + k() + 0 + k + 0 k. f() + 5 then, f(7) 7 + 5 6 then, f(5) 5 + 5 0 f(7) f(5) 6 0 6.. f() 5 + 7 Then, f() 5 + 7 f() ( ) 5 () + 7 and f() f() + f f 5 + 7 9 9 + 9 9 59 9 WORKSHEET-6 6 + k 0 k 8. 8. + or y + y. 9. oefficient of in epression + π 7 is π. 0. p() + 6 Then, p ( ) ( ) ( ) ( ) + 6 6 + 6 0. Linear polynomial +, degree Quadratic polynomial +, degree ubic polynomial, 7, degree P-5
. p() + 6 (i) When then, p ( ) ( ) + 6 + 6 8 (ii) When then, p() + 6 9 9 + 6 6. (i) (I) Possible length and breadth of the rectangle are the factors of its given area. rea 5a 5a + TOPI- Remainder Theorem 5a 5a 0a + 5a(5a ) (5a ) (5a )(5a ) So, possible length and breadth are (5a ) and (5a ) units, respectively. (II) rea 5y + y 5y + 8y 5y 7y(5y + ) (5y + ) (7y )(5y + ) So, possible length and breadth are (7y ) and (5y + ) units respectively. (ii) Factorisation of Polynomials. (iii) Epression of one s desires and news is very necessary.. Let, p(y) 5y y 7y +, then remainder p(0) 0 0 0 +.. Put, + 0 Then remainder is : f( ) ( ) + 0 + 0 00.. p() + 9 + So, + ) + 9 + () (+) 9 () (+) 7 + 6 + (+) () + (Remainder) Quotient + Remainder +. p() + + + Put, 0 Remainder p in p() + + + WORKSHEET-7 8 + + + +++8 8 5 8. [SE Marking Scheme, 0] 5. Factors of (±, ±, ±, ±, ± 6, ± ) p() 6 5 + p() 6() 5() + 8 00 + 6 0 is a zero of p() or ( ) is a factor of p() 6 5 + 6 ( ) ( ) + 6( ) ( ) (6 + 6) ( ) (6 9 + 6) ( ) [( ) ( )] ( ) ( ) ( ). 6. y y + y + y + ) y + y + 6y + 50y + 6 y + 6y + 8y y y + 50y y 6y 8y + + + y + 58y + 6 y + 56y + 8 y So, y must be subtracted. P-6 M T H E M T I S - I X T E R M -
7. Let p() a + and g() 5 + a R p() R g() + p() a() + () R 8a + 8a R () 5() + a 6 0 + a R 6 + a R R 8a 6 + a 7a 7 a [SE Marking Scheme, 0] lternative Method : Let p() a + and g() 5 + a R p() R g() p() a() + () 8a + 8a So, R 8a g() () 5() + a 6 0 + a 6 + a. For zeroes, put p() 0 ( ) ( ) 0, then, 0,,.. Putting, 0 p(0) 0 0 + Putting, p() + 6 + 0 Thus, p(0) + p() + 0.. Here, p () + + +, and the zero of is. So, p () () + () () + + So, by the Remainder Theorem, is the remainder when + + + is divided by.. Here, p() a + 6 a, and the zero of a is a. So, p(a) a a.a + 6a a 5a. So, by the remainder theorem 5a is the remainder when a + 6 a, is divided by a. 5. Let, p() 5 + k, and q() + 7 + k S O L U T I O N S So, R 6 + a ccording to the question, R R then, 8a 6 + a 8a a 6 + 7a 7 a 8. ( + ) quotient + + ) + + + 6 + 8 + 8 6 + 6 + 6 + 6 7 + 6 + + + 7 + 0 Thus, quotient + and remainder 7 + 0 y Remainder theorem, Divide nd (Divisor Quotient) + Remainder + + +6 [( + + ) ( + )] + (7 + 0) [ + 8 + 8 + 6 + 6 ] + 7 + 0 + + + 6 WORKSHEET-8 Put, + 0 or in p () and q () P( ) ( ) 5( ) + k( ) 0 k k 6 q( ) ( ) ( ) + 7 ( ) +k 8 +k 0 +k p() and q() leave the same remainder when divided by +. k 6 k 0 k 6 k 6. p() + 8 + 7 + a p() leave the same remainder when divided by ( + ) and ( +). p() () + 8() + 7() + a 8 + + a 0 + a p() () + 8() + 7() + a() + 8 7 + a 0 a Remainders are equal P-7
So, 0 + a 0 a a 0 a 0 7. R f() () + () a() + b 5 + a + b 5 b a...() R f() () () + () a() + b 9 + + + a + b 9 a + b...() Solving () and (), we get b 8 and a 5 R f() () () + () 5 + 8 0 + 8 0 [SE Marking Scheme, 0] lternative Method : f() + a + b Put, 0 or in f(), we get f() () () + () a + b TOPI- Factor Theorem 5 + a + b 5 a + b a b 5 a b...() gain put, + 0 or in f(), we get f( ) ( ) ( ) + ( ) a ( ) + b 9 + + + a + b 9 6 a + b a + b...() dding equations () and (), a 0 a 5 y equation (), 5 + b b 8 f() + 5 + 8 gain put, 0 or in f() f() () () + () 5 + 8 6 6 + 0 + 8 0 0 0. + is a factor of + + m p() p( ) 0 6 + m 0 m. is a factor of p() 6 + k p 0 6. + k. 0 k. 8 ( y) () ( y) [( y)][() + ( y) + ( y)] Since (a b ) (a b) (a + b ab) y[ + + y y + y] y[ + y 6y]. 9 + 6y + y () + () y + y ( + y) [ a + ab + b (a + b) ] 5. Give Ep ( y) 7( + y) ( y) + ( + y) WORKSHEET-9 ( y) ( + y) ( y) ( + y) ( y) + ( + y) ( y) [ y y] ( + y) [ y y] ( y) [ 5y ] ( + y) [ 5y] ( 5y ) [ y ( + y)] (5y + ) ( y) (5y + ) ( + y) ( + y) (5y + ) 6. pq 7 + ( pq) + 79 9 7 7 7 pq + ( pq) + pq 9 9 9 a + b (a + b) (a + b ab) 7 pq pq + pq 9 7 + 9 8 9 7. a 6 b 6 (a ) (b ) (a b ) (a + b ) (a b) (a + b + ab) (a + b) (a + b ab) (a b) (a + b) (a + b + ab) (a + b ab). [SE Marking Scheme, 0] 8. p() + 6 p() 6 + + 6 0 is a zero of p() P-8 M T H E M T I S - I X T E R M -
p() 5 7 + 6 0 is a zero of p() ( + ) ( ) 6 is a factor of p() p ( ) 6 p() ( + ) ( ) ( ) 9. Factor theorem statement Let p() +. ( ). a b 6ab 6ab(a b). 8a + 8b (a) + (b) (a + b) [(a) + (b) (a) (b)] [ a + b (a + b) (a + b ab)] (a + b) (a + b ab) 8(a + b) (a + b ab). ( + y) + ( y) [ + (y) + y ( + y)] +[() y y ( y)] 5. To factorise, 5a+ a a b (a + b) (a b) S O L U T I O N S ( + ) + + ( + ) a b a b ab a b and ( a b) a b ab ( a b) + 7y + 9 y + 7y + 7 y 7 y + 9y 8 + 6y 8 y + 6y. 5 a+ +a 7 a+ (a + ) 5 a+ a+ [SE Marking Scheme, 0] 6. 9 + y + z 6y + yz 6z ( ) + (y) + (z) + ( )(y) + (y) (z) + ( ) (z) ( + y + z) If, y, z, then ( + y + z) ( + ). is a factor of p(), then p() 0 p() () + k 0 8 + 6 k 0 The factors of the constant term are ±, ± p() () + 0 ( ) is a factor p() () () () + 0 ( + ) is a factor p() () + 0 ( ) is a factor ( ) ( + ) ( ) are the factors of p(). WORKSHEET-0 ( + ). 7. 5 7y + z + 5yz (5) + ( y) + (z) (5) ( y) (z) (5 y + z) [(5) + ( y) + (z) (5)( y) ( y) (z) (5) (z)] a + b + c abc (a + b + c)(a + b + c ab bc ca ) (5 y + z) [5 + 9y + z + 5y + yz 5z] [SE Marking Scheme, 0] 8. We have, y + y z + z 0 ( y ) + (y z ) +(z ) ( y ) (y z ) (z ) a + b + c 0 a + b + c abc ( y ) + (y + z ) + (z ) ( + y) ( y) (y + z) (y + z) (z ) (z + ) Similarly, y + y z + z 0 ( y) + (y z) + (z ) ( y) (y z) (z ) ( y ) + ( y z ) + ( z ) ( y) + ( y ) + ( z ) ( + y)( y)( y z)( z+ )( y+ z)( z ) ( y)( y z)( z ) ( + y) (y + z) (z + ) 9. Let p m(n p ) + n(p m + p(m n ) p(m n) n(n p ) + n(p n ) + p(n n ) n(n p ) n(n p ) + 0 0 m n is a factor of p Similarly p(n p) 0 & p(p m) n p is a factor of p. and p m is a factor of p. WORKSHEET- k.. Since, f 0 P-9
is a zero of polynomial f() So, + or + is a factor of f().. 6a 7b a b + 08ab (a) (b) (a) (b) + (a) (b) (a) (b) a b (a b) (a b). a 9 + b 9 + a 6 b + a b 6 (a ) + (b ) + (a ) (b ) + (a ) (b ) (a ) + (b ) + a b (a + b ) (a + b ) 5. Let, p() 6 a 5 + a + a + ( a) is a factor of the polynomial p(), then p(a) 0 a 6 a a 5 + a a a + a a + 0 a 6 a 6 + a a + a a + 0 a a. 6. a 7 + ab 6 a(a 6 + b 6 ) a[(a ) + (b ) ] a(a + b ) [(a ) + (b ) a b ] a(a + b ) (a + b a b ). [SE Marking Scheme, 0] 7. Let, a Epression a (a ) 0 a a 0 (a 5) (a + ) ( 5) ( + ) ( 5) ( + ) ( ) [SE Marking Scheme, 0]. a (a ) (a + + a ) Then factors of (a ) are (a ) and (a + + a).. + 5 + + + + ( + ) + ( + ) ( )( ) Factors are + and +. + +.. ( + ) + p + p( + ) ( + ) + (p) + ( + ) p ( + + p), [ a + b + ab (a + b) ]. 8 7a 6a + 5a () (a) 8a( a) lternative Method : ( )( ) 0 ( ) ( ) 0 ( ) 5( ) + ( ) 0 ( )[ 5] + [ 5] ( 5) ( + ) ( 5 + 5)[() + () ] [( 5) + ( 5)][ ] ( 5)( + ) ( )( ). 8. p() + 5 + p() + 5 + 0 is a zero of p() p() 8 + 5 + + 0 is a zero of p() ( ) ( + ) + is a factor of p() p ( ), when we divide physically + Hence p() ( ) ( + ) ( ) 9. Factor of 60 (±, ±, ±, ±, ±5, ±6, ±0, ±, ±5, ±0, ±60) p() + 7 60 p() () () + 7() 60 7 08 + 60 68 68 0 is a zero of p() or ( ) is a factor of p() + 7 60 ( ) 9( ) + 0( ) ( ) ( 9 + 0) ( ) ( 5 + 0) ( ) [( 5) ( 5)] ( ) ( ) ( 5) WORKSHEET- () (a) a( a) ( a). 5. p() m n If ( a) is a factor of p(), then p(a) 0 (a) m a n a 0 a[a m n] 0, a 0 a m n 0 a m + n. [SE Marking Scheme, 0] 6. 50 y [5 6y ] [(5) (6y) ] (5 6y) [(5) + (6y) + 5 6y] a b (a b) (a + b + ab) P-0 M T H E M T I S - I X T E R M -
(5 6y) (5 + 6y + 0y). 7. Let, p() 9 5, p ( ) 9 5 0 Since, ( + ) is a factor of 9 5 ( 9 5) ( + ) ( 5) 5 5 + 5 ( 5) + ( + 5) ( + ) ( 5) Factors are : ( + ) ( + ) ( 5) [SE Marking Scheme, 0] 8. p() a + + If ( + ) is a factor of p(), then + 0, is a zero of the polynomial p() So, p 0 p 6 a + 0 a + 8 + 0 TOPI- lgebraic Identities 8 + a + + + 0 8 9 a + 0 8 8 a 9 9. ( ) is a factor of p() + 6 5 + 6, then it completely divide p() + 9 + (quotient) ) + 6 5 + 6 6 + 9 6 9 7 + 5 + + 6 + 6 + 0 Remainder 0. So, ( ) is a factor of p().. +. Given, S O L U T I O N S + + y + y + y y + + + +. + y y + y + y 0 y ( y) ( + y + y) ( y) 0 0.. ( y + z) + 9y + z y 6yz + z [SE Marking Scheme, 0] lternative Method : y using the identity, (a + b + c) a + b + c + ab + bc + ca WORKSHEET- ( + (y) + z) () + (y) + z + ()(y) + (y) (z) + (z)() + y + z y yz + z. y y 7 7 8y 7 8y 7 y y y 9 y y + y 9 5. + y + z 0 + y z ( + y) ( z) + y + y (z) z + y yz z + y + z yz 6. (a + b) (a b) y where a + b and a b y P-
( y) ( + y + y ) [(a + b) (a b)][(a + b) + (a + b) (a b) + (a b) ] 6b[(a + ab + 9b ) + (a 9b ) + (a ab + 9b )] 6b(a + 9b ) 6b (a + b ) 8b(a + b ) 7. (00 + ) (00) + (00) () + (00) () + () 8. 676 + + 6 + + + +. a + b + c abc (a + b + c) (a + b + c ab bc ca) a + b + c abc 0, as a + b + c 0 a + b + c abc.. ( ) () () ( ) 8 6 + oefficient of in the epansion of ( ) 6.. 9 (97) () (97 + ) (97 ) 00 9 900. [SE Marking Scheme, 0]. ( + y) + y + y 7 + 7 + 8 5 ( + y) ± 5 ± 5. 5. ( + ) + ( 5 ) ( ) + ( ) + + ( 5) + ( ) 5 + + 6 + 5 + 0 + 6 0 (6 + 6 0 ) 6. ( + y + z) ( + y + z y yz z) ( + y + z) [ + y + (z) y y z z] (6) + + 6 + + 9. RHS. ( + y + z)[( y) + (y z) + (z ) ] ( + y + z) [ + y y + y + z yz + z + z] ( + y + z) [ + y + z y yz z] ( + y + z). [ + y + z y yz z] + y + z yz [y identity] LHS. WORKSHEET- We know that (a + b + c) (a + b + c ab bc ca) a + b + c abc ( + y + z) ( + y + (z) y y z z) () + (y) + (z) y z + y + 8z 6yz. 7. Given, a + b + c or a b c 0 ( a) + ( b) + ( c) ( a) ( b) ( c) [ a + b + c][( a) + ( b) + ( c) ( a)( b) ( b)( c) ( a)( c)] [ a b c][( a) + ( b) + ( c) ( a) ( b) ( b)( c) ( a) ( c)] 0 [( a) + ( b) + ( c) ( a)( b) ( b)( c) ( c)( a)] 0. 8. + 5 On squaring both sides, we get + + 5 + + 5 [ (a + b) a + b + ab)] + + 5 + 5 P- M T H E M T I S - I X T E R M -
S O L U T I O N S + 9. (i) 0 07 (00 + ) (00 + 7) 00 + ( + 7) 00 + 7 0000 + 000 +. + + () () 6.. rea of rectangle 5a 5a + 5a 0a 5a + 5a(5a ) (5a ) (5a ) (5a ) length breadth Length and breadth are (5a ) and (5a ) respectively.. + y 8 ( + y) 8 8 + 7y + y ( + y) 5 8 + 7y + 8 8 5 8 + 7y 5 88 8 + 7y.. Given, + y 5, then + y + 5y 5 + y 5 + 5y () + (y) + ( 5) y ( 5) ( + y 5) [() + (y) + ( 5) ( 5) y ( 5) y)] (5 5) ( + y + 5 + 5 + 5y y) 0 ( + y + 5 + 5 + 5y y) 0 5. +, (given). 8 + 7 8 8 7 + 7 ( ) ( ) (8+7) 8 8 7 +7 8 8 7 +7 a + b (a+ b) (a + b ab) 8 + 7 00.. a, b, 7 c a + b + c + 7 0 0 (ii) (0) (00 + ) (00) + + 00 (00 + ) 000000 + 8 + 600 0 000000 + 8 + 600 0608 On squaring both sides, we get + WORKSHEET-5 () + + 6 + + 6 +. 6. Given, a + b + c 9 (a + b + c) 9 a + b + c + (ab + bc + ca) 8 [as, (a + b + c) a + b + c + (ab + bc+ca)] 5 + (ab + bc + ca) 8 (ab + bc + ca) 8 5 6 ab + bc + ca 6/ ab + bc + ca. 7. a b 7 and a + b 85 (a b) (7) a ab + b 9 85 ab 9 ab 6 ab 8 Then, a b (a b) (a + b + ab) (7) (85 + 8) (7) (0) 7 WORKSHEET-6 a + b + c abc () + () + ( 7) ( 7) 7.. 9 5 (50 ) (50 + ) (50) () 6500 699. [SE Marking Scheme, 0] P-
. y + + y + + + + 5+ 6 5 6 + 5 6 5+ 6 ( 5+ 6 ) ( 5 6 ) ( 5 6)( 5+ 6) + 5 + + 0 6 5 + 0 6 + 5 0 6 + 0 + 96 + 97. ( a b ) + ( b c ) + ( c a ) 5. ( a b) + ( b c) + ( c a) oth Numerator and Denominator are of the form a + b + c We know that when a + b + c 0 then a + b + c abc For Numerator, a b + b c + c a 0 For Denominator, a b + b c + c a 0 ( ) ( + ) + ( a b b c c a ) ( a b) + ( b c) + ( c a) ( a b ) ( b c ) ( c a ) a b ( b c)( b c) ( ) ( a b)( a+ b)( b c)( b+ c)( c a)( c+ a) ( a b)( b c)( c a) (a + b) (b + c) (c + a). [SE Marking Scheme, 0] 6. (i) (998) (000 ) We know that FORMTIVE SSESSMENT Note : Students should do this activity themselves. (a b) a b ab(a b) Hence, (998) (000) () () (000) () (000 ) 000000000 8 6000 998 000000000 5988008 99099. (ii) lgebraic Identities (iii) Satisfaction solves the identity crisis among people. 7. Epansion of ( + y + z) + y + z + (y + yz + z) 00 0 + (y + yz + z) (y + yz + z) 0 + y + z yz ( + y + z) ( + y + z y yz z) 0(0 0) 00 [SE Marking Scheme, 0] lternative Method : ( + y + z) + y + z + (y + yz + z) (0) 0 + (y + yz + z) 00 0 (y + yz + z) y + yz + z 60 0 and + y + z yz ( + y + z) [ + y + z (y + yz + z)] 0 [0 0] 0 0 00. 8. 8y 6y 6 () + ( y) + ( 6) ()( y)( 6) [ + ( y) + ( 6)][ + ( y) + ( 6) ()( y) ( y)( 6) ()( 6)] ( y 6) ( + y + 6 + y y + 6) 0 [ + y + 6 + y y + 6], ( y + 6 or y 6 0) 8y 6y 6 0. WORKSHEET-7 qqq P- M T H E M T I S - I X T E R M -
SETION HPTER INTRODUTION TO EULID S GEOMETRY TOPI- Euclid s Geometry. system of aioms is called consistent, when it is impossible to deduce from these aioms, a statement that contradicts any aiom or previously proved statement.. + + + 5 + 5 + 5 8. Theorem requires a proof.. Let, First thing Second thing y then, y 5. D D (given) + + D D. [SE Marking Scheme I, 0, 0] 6. Euclid s aioms () Things which are equal to the same thing are equal to one another. () If equals are added to equals, the wholes are equal. + [SE Marking Scheme, 0, 0] 7. D (Given) + + D + D Euclid s aiom used : If equals are added to equals, the wholes are equal. [SE Marking Scheme I, 0, 0]. surface is that which has length and breadth. WORKSHEET-8 8. D (Given) E dding, + D + E DE If equals are added to equals, the wholes are equal. [SE Marking Scheme, 0] 9. Given, So, + + (Equals are added to equals), ( + coincides with ). 0. Given, is the mid-point of. () If possible, let D be the another mid-point of. D D () D Now, subtracting () from (), we get D D D D D 0 D 0 and D coincide. Hence, every line segment has one and only one mid-point.. oncermed, aring. Things equal to same things are equal to one another. Rehman contributed ` 500. ll right angles are, equal to one another (OR) ny postulate of Euclid can be stated. y WORKSHEET-9 Surface breadth ( y) length ( ) S O L U T I O N S P-5
. Dimension of Surface Length and readth (which is ).. Lines are parallel if they do not intersect on being etended. For eample : or Lines and are parallel lines.. D D D Things which are equal to the same thing are equal to one another. [SE Marking Scheme I, 0] 5. Euclid s aiom : If be the mid-point of a line segment, then. and D D D. [SE Marking Scheme, 0] 6. Given : Three lines l, m and n in a plane such that l m and m n.....()...() dding () + (), we get l m n To prove : l n. Proof : If possible, let l be not parallel to n, then l and n should intersect in a unique point, say. Thus, through a point, outside m, there are two lines l and n, both parallel to m. This contradicts the parallel line aiom. So, our assumption is wrong. Hence, l n. 7. (i) If a straight line l falls on two straight lines m and n such that the sum of the interior angles on one side of l is equals to two right angles then by Euclid s fifth postulate the lines will not meet on this side of l. Net, we know that the sum of the interior angles on the other side of line l will also be two right angles. Therefore, they will not meet on the other side also. So, the lines m and n never meet and are therefore, parallel. (ii) Introduction to Euclid s Geometry. (iii) Universal truth. 8. (i) The terms need to be defined are : Polygon : simple closed figure made up of three or more line segments. Line segment : Part of a line with two end points. Line : Undefined term. Point : Undefined term. ngle : figure formed by two rays with a common initial point. Ray : Part of a line with one end point. Right angle : ngle whose measure is 90. Undefined terms used are : Line, Point. Euclid s fourth postulate says that all right angles are equal to one another. In a square, all angles are right angles, therefore, all angles are equal (From Euclid s fourth postulate). Three line segments are equal to fourth line segment. (Given) Therefore, all the four sides of a square are equal (by Euclid s first aiom things which are equal to the same thing are equal to one another. ) (ii) Introduction to Euclid s geometry. (iii) Equality leads to democracy. WORKSHEET-0 + + D Euclid s aiom used : If equals are added to equals, wholes are equal. [SE Marking Scheme, 0] P-6 M T H E M T I S - I X T E R M -
. In a circle having centre at P, we have PR PQ radius In a circle having centre at Q, we have QR QP radius Euclid s first aiom : Things which are equal to the same thing are equal to one another. PR PQ QR.. 5 5 5 + 5 5 + 5 0 If equals are added to equals, the wholes are equal. [SE Marking Scheme, 0]. Here, OX XY, PX XZ XY (OX), XZ (PX) lso, OX PX, (Given) XY XZ, (ecause things which are double of the same things are equal to one another.) 5. Here, and and. Euclid s first aiom says, the things which are equal to same things are equal to one another. So,. TOPI- Euclid s Postulates. If a straight line falling on two straight lines makes the interior angles on the same side of it taken together less than two right angles, then the two straight lines, if produced indefinitely, meet on that side on which the sum of angles is less than two right angles.. Meeting place of two walls.. Only one line passes through two distinct points.. Playfair s iom (statement) : For every line l and for every point P not lying on l, there eists a unique line m passing through P and parallel to l. It is equivalent to Euclid s fifth postulate. 5. (a) Infinite, if they are collinear. (b) Only one, if they are non-collinear. + 6. E S O L U T I O N S (E is the mid-point of ) D DF (F is the mid-point of D) lso, E DF (Given) Therefore, D (Things which are double of the same thing are equal to one another) WORKSHEET- 7. There are two undefined terms, line and point. They are consistent, because they deal with two different situations. (i) Says that given two points be and, there is a point, lying on the line which is in between them. (ii) Says that given and, we can take not lying on the line passing through and. These Postulates do not follow from Euclid s postulates. However, (ii) follow from given postulate (i). 8. (given) X Y (given) If equals are subtracted from equals, then remains are also equal. X Y X Y 9. Since + y 0 and z, therefore, + y z + y 0 z + y Hence, z + y 0. 0. Their sales in July will also be equal as things which are double of the same things are equal to one another. Two other aioms are : (i) The whole is greater than the Part. (ii) Things which are halves of the same thing are equal to one another. P-7
. Two planes intersect each other to form a straight line.. M P N MN M + N. X Y Let has mid-points, say X and Y, then, X and Y + X Y X and Y coincides [SE Marking Scheme 0,, ]. Let be perpendicular to a line l and P is any other line segment. In right P, > P, ( 90 ) P > or < P. FORMTIVE SSESSMENT Note : Students should do this activity themselves. 5. Given, WORKSHEET- + + + + (s, ). 6. Since and, therefore adding both equations + + D D. 7. iom : If equals be subtracted from equals, the remainders are equal. Two more aiom are : (i) Things which are halves of the same thing are equal to one other. (ii) The whole is greater than the part OR any of Eulid s ioms can be stated. WORKSHEET- qqq P-8 M T H E M T I S - I X T E R M -
SETION HPTER LINES ND NGLES TOPI- Different Types of ngles. + 0 + 0 + 60 80 ( Straight line makes an angle of 80 ) 80 60 0 0 Thus, OD 0 80 0 60.. Let the angle be, then ngle omplement of 90 5.. omplementary angle of 65 90 65 5. (s sum of complemetary angles is 90 ). ( 5 ) + ( + 5 ) 90 5 ngles are 60 and 0. 5. Let, a and b Then, a + b 90 + 90 8 a 8 6 b 8 5 c 80 b 80 5 6. [SE Marking Scheme, 0]. ngles (0 a) and (5 + a) are supplementary of each other, then 0 a + 5 + a 80 a 80 55 5.. The complement of (90 a) 90 (90 a) (s sum of complementary angles is 90 ) 90 90 + a a.. Let the angle be, then y given condition, (90 ) 6 90 5 5. S O L U T I O N S WORKSHEET- 6. OP PO 90 Let, POQ a QOR a RO 5a a + a + 5a 90 9a 90 a 0 0 and y 0 0 z 5 0 50. [SE Marking Scheme, 0, 0] 7. b + 75 + b 80 (Linear pair) 5b 80 75 05 b 05 5 b 8 a b 8 (V.O..) c + a 80 (Linear pair) c 80 8 96 c 8. WORKSHEET-5. PO y (Vertically opp. angles) 5y + y + 5y 80 y 80 y 5. [SE Marking Scheme, 0] 5. + y + z + w 60 ( + y) + (z + w) 60 + y + + y 60 ( + y) 60 (Since, z + w + y) P-9
+ y 80 O is a straight line, as straight line makes an angle of 80. [SE Marking Scheme, 0] 6. Let, O and O O + O O + 75 75 5 5 O 0 and O 5. [SE Marking Scheme, 0]. ngle (55 + a) and (5 a) are supplement of each other, then 55 + a + 5 a 80 a 80 70 0.. Here, + 0 + + + 0 80 80 0 50 50 50.. Here, 5 + 80 ( Straight line makes an angle of 80 ) 80 0. 9. Given, a b 6 80 a b 0...() a + b 80 (Linear Pair)...() dding () & () a 0 a 05 b 80 a 80 05 75. [SE Marking Scheme, 0] 5. Let the two supplementary angles are and, then + 80 80 6 5 Hence, the angles are and or 7 and 08. [SE Marking Scheme, 0] 7. Join, D In D Et. +...(i) gain, in DD Et. D + D...(ii) dding (i) and (i), we get + ( + D) + ( + D) + + +. [SE Marking Scheme, 0] 6. P D WORKSHEET-6 O OP and OQ are bisectors of OD and O. Q and O OD (Vertically opp. angles) + O + + OD + ut + O + + + OD + 60 + O + + OD + OP and OQ are in the same line. 7. Given, Let, and POR : ROQ 5 : 7 POR 5 ROQ 7 5 + 7 80 80 80 80 5 POR d b 5 5 5 75 ROQ c a 7 7 5 05. [SE Marking Scheme, 0] P-0 M T H E M T I S - I X T E R M -
TOPI- Transversal Line. Let, l D (by construction) O 0 I F 0 E D O' then, OEF + OE 80 O EF + DOE 80 [orresponding interior angles] OEF + O EF + 0 + 0 60 OEO 60 60 00.. 60 (corr S) + 80 + 60 80 ( 60 ) 0 60. D E 00 0 l Draw l l DE as DE. WORKSHEET-7 + 0 80 (o. int. angles) 0 Similarly, 80. + D + 80 (St. angle) 0 + D + 80 80. D 60. E [SE Marking Scheme, 0, 0] 5 y O 5 y y OD 5y (VO) 5y + 5y + y 80 y 80 y 5 5.. P R O POR + ROQ 80 (linear pair) 5 + 7 80 5 POR 75 QOS, ROQ 05 POS F Q D S WORKSHEET-8 l l. m l n y y l S O L U T I O N S 75 (lternate angles) 80 80 75 05 05 75 + 0 75 + 90. [SE Marking Scheme, 0] (orresponding angles) (orresponding angles) + y 80 + y 80 P-
. y 80 y 90 y O w z D + y + w + z 60 ( + y) 60 ( + y w + z) + y 80 O is a straight line.. PQS + QSF 80 (ngles on the same side of transversal) PQS + RFE 80, as QSF EFR 60 + RFE 80 (orresponding S) RFE 0.. 70 + PRS 80 PRS 0 TRO (Vertically opposite angles) In TRO, + 0 + 0 80 50.. Q L M N M N (Given) Eterior NLQ M + N N NLP N NLP N (lternate interior angles) LP MN. [SE Marking Scheme, 0]. EG is the bisector of EF EG GEF a Similarly, EFH HFD b P. 8 O 75 y E z 8 + 80 (linear pair) 96 8 y + 75 (VO) y 8 75 96 75 D z 8 (VO) WORKSHEET-9 GEF EFH ( a b) ut these are alternate interior angles EG FH gain, EF a and EFD b EF EFD a or b ut these are alternate angles. D. 5. D D D 5 [alternate angles] + y + 80 80 D y 65 [angle sum property] D z 80 [5 + 5 ] 0 [SE Marking Scheme, 0] lternative method : D y + 5 + 80 80 (orresponding interior angles) y 80 5 y 65 In D, D D 5 (lternate angles) 5 + y 0 + z 80 z 80 5 y z 75 65 z 0. P- M T H E M T I S - I X T E R M -
TOPI- ngle Sum Property of a Triangle. We know that + 60 00 (Eterior angle is the sum of the two interior opposite angles) 0.. Since sum of all the eterior angles formed by producing the sides of a polygon is 60. + y + z 60.. D + [Eterior angle is the sum of the two interior opposite angles] S O L U T I O N S 60 + 70 D 0. Given, + 65 Given, + 0 + + + 65 + 0 05 ut, + + 80 (ngle sum prop. of D) 80 + 05 5 0 5 5 [SE Marking Scheme, 0, 0] lternative Method : We know that, + + 80...(i). In, + + 80 lso, in DEF, D + E + F 80 + + + D + E + F 60 90 Hence, k.. PQ 0 + y 8 (lternate angles) (Eterior angle is the sum of the two opposite interior angles) WORKSHEET-0 (ngle Sum prop. of D) 65 + 80 5 gain by (i), + 0 80 0 gain by (i), 0 + + 5 80 (ngle Sum prop. of D) 5. 5. G E 5 5 F D Draw EH. Then GE GEH and FE FEH (lternate int S) GEH + FEH GE + FE...(i) ut GE + 5 80, FE + 5 80 (linear pair) GE 5, FE 55 Hence from (i) 5 + 55 00. H WORKSHEET- 0 + y 8 y 8 0 78.. In, + + 80 (ngle Sum prop. of D) 5 + 5 + 5 + + 50 80 0 0 5 5 + 5 70 + 50 89 [SE Marking Scheme, 0] lternative Method : In, + + 80 (ngle Sum prop. of D) 5 + 5 + 5 + + 50 P-
80 (y given values) 0 + 50 80 0 0 5 6 5 5 + 5 65 + 5 70 + 50 9 + 50 89. In DE, + D + DE 80 (ngle sum property of a triangle) 0 + + 90 80 50 In DF, D + FD FD (Eterior angle is the sum of the two interior opposite angles) 50 + y 0 y 60 5. (i) TOP TO ORS ORD ut, TO ORD (l m and corresponding angles) TOP ORS ut, they are corresponding angles w.r.t. transversal TR and lines OP and RS. Hence, OP RS. T R O (ii) Lines and angles. (iii) isectoring among human beings gives rise to deterioration in the society. S P D l m. QPR 75 (Vertically opposite angles) gain, PQR + QPR 05 (Eterior angle) PQR + 75 05 PQR 0.. 60 0 80 (60 + 0 ) 80 00 80 is the smallest side Reason : Side opposite to smaller angle is shortest. [SE Marking Scheme, 0]. P Q y 8 R 75 T L WORKSHEET- Given : PQ PR PQ RL RQT 8 QTL 75 To find : & y y...() [alt. interior angles] Now, In QRT QTL TQR + QRT [by eterior property of triangles] 75 8 + 7 y 7...() [from eq.()] Now, In QTR QPR + + y 80 [by SPT] 90 + + 7 80 5 [from eq ()]. (i) In, D + [Eterior angle equal to the sum of opposite two interior angles] 6 + + 5 + 6 + 5 + 6 5 P- M T H E M T I S - I X T E R M -
(i) Eterior angle property of a triangle is used in the above problem. (ii) y doing so, students ehibit the importance of water. 5. P + z O + z Q + y O + y O + O + O 80 (ngle sum prop. of D) O + O + O 60 (Multiply by both sides) O + + y + + z 60 O + 80 + 60 O 80 O 90 [SE Marking Scheme, 0]. In, + + 80 (y given conditions) + + 6 80 9 80 0.. Then, + + 80 (Prop. of isosceles D) 50 + + 80 0 65. Eterior angle Sum of opposite two interior angle other angle 80 0 70 0 0 + 80.. 5 55 D, 5 + + 80, 55 (ngle sum prop. of D) + 5 + 55 80 80 0 D + WORKSHEET- (Eterior angle is the sum of the two interior opposite angles) 0 + 55 95 Similarly, D + 5 + 0 85. 5. l y z To prove : Sum of all the angles of is 80. onstruction : Draw a line l parallel to. Proof : Since l, we have y (lternate angles are equal)... (i) Similarly, l z (lternate angles are equal)...(ii) lso, sum of angles at a point on line l is 80. + + 80 i.e., y + + z 80 (from (i) and (ii)) + y + z 80 + + 80 Sum of all angles of a is 80 5 + 6 + 7 80 0 ngles are 50, 60 and 70 respectively. Hence proved. [SE Marking Scheme, 0] FORMTIVE SSESSMENT WORKSHEET-, 5 & 6 Note : Students should do these activities themselves. qqq S O L U T I O N S P-5
SETION HPTER 5 TRINGLES TOPI- riteria for ongruence of Triangles WORKSHEET-7. S congruence : Two triangles are congruent, if two angles and the included side of one triangle are equal to two angles and the included side of other triangle.. Since DE, D and DEF by SS. Therefore DF. D In DD, + + 80...() (ngle sum prop. of ) dding eq. () & (), + + + + + 60 + + + D 60 + + + D 60 Hence proved. 5. Given : D is the mid-point of side To Prove : D E. F onst : Draw, DE Proof : In, D is the midpoint of & ED y mid point theorem, E E...(i) Now, ED (y const) ED (orresponding s) ED 90 ED DE 90...() D Join D. In and D, (Given) D D (Given) D D (ommon) y using SSS ongruency Rule, D D D D (y c.p.c.t.). In D, + + 80...() (ngle sum prop. of ) Now, In DE & DE, E E (from eq....()) ED DE (from eq....()) ED ED (ommon) DE DE (y SS rule) D D...() (y cpct) ut, D D...() ( D is the midpoint of side ) From eq. () & (), D Hence Proved. P-6 M T H E M T I S - I X T E R M -
. O O (Given) OP OP (ommon) OP OP (Given) OP OP (y SS). In Y and X, (Given) Y X (Given) (ommon) y SS, Y X. Proved.. In P and PD, P PD (Given) (P is the mid-point of D) D (Side of a square) P PD 90 y R.H.S., P PD P P (y c.p.c.t.) (ngles opp. to equal sides are equal) P P. Proved.. D (Given) D (Given) (ommon) D (y SS) D (y c.p.c.t.). O O (O is the mid-point of ) O OD (Vertically opposite angles) O OD (O is the mid-point of D) O O (y SS) D. (y c.p.c.t.) Proved. In D and, is common D D (Given) Hence, D (y S rule) D. (y c.p.c.t.) Proved.. D + D 80 (Linear pair) D + 80 D 80 WORKSHEET-8. Proof : In D and, D (Given) D (Given) (ommon side) y SSS congruence aiom, D D (y c.p.c.t.) and D. (y c.p.c.t.) 5. Given, D E dding DE to both sides, we get D + DE E + DE E D In E and D, (Given) E D (Proved) E D (Given) E D (y S cong.) D E. Proved. [SE Marking Scheme, 0] WORKSHEET-9 Similarly, E 80 y Since, y (Given) D E (ommon) E D (S) E D (c.p.c.t.) 5. Given, D [SE Marking Scheme, 0] Let us consider and D (ommon side) So, + + D and, D Then, from the SS ongruence Rule D D Hence Proved S O L U T I O N S P-7
. Let the angles of triangle are 5, and 7, then 5 + + 7 80 5 80 Thus, ngles are 60, 6, 8 Each angle is less than 90 The triangle is an acute angled triangle.. D 90 (Given) D (Given) (ommon) D (y RHS). (i) In OD and O, O O (Given) OD O (Given) OD O (Vertically opposite angles) So, by SS criteria, OD O (ii) D (y c.p.c.t.) D and are two lines intersected by such that D and they form a pair of alternate angles. Hence, D.. 5 P 6 D + 5 80 + 6 (linear pair) ( 5 6) In P and P, (Proved) (D is the bisector of ). PRQ. Interior opposite angles. D +. E F WORKSHEET-50 P P P P (y S) P P. (y c.p.c.t.) Proved. [SE Marking Scheme I, 0] 5. P O Q Proof : QO is bisector of PQR OQR PQR Q RO is bisector PRQ ORQ PRQ R In OQR QOR + OQR + ORD 80 (ngle sum property) QOR + Q + R 80 R QOR 80 ( Q + R) ut in PQR P + Q + R 80 Q + R 80 P QOR 80 (80 P) 80 90 + P 90 + P Hence Proved. These type of rallies spread awareness among people for not to kill girl child and helping in equalising se ratio. WORKSHEET-5 In ED and FD, DE DF 90 D D (D is the midpoint) ED FD (Given) ED FD (y RHS). (y c.p.c.t.) Proved. D P-8 M T H E M T I S - I X T E R M -
. PQRS is a square. (given) T PQ (ssumed) PQR (y SS rule) P S R P Q (i) SRT is an equilateral triangle. (given) PSR 90, TSR 60 PSR + TSR 50. Similarly, QRT 50 In PST and QRT, we have PS QR (S) PST QRT 50 () and ST RT (S) y SS, PST QRT PT QT (y c.p.c.t.) Proved. (ii) In TQR, QR RT (Square and equilateral D on same base) TQR QTR + + QRT 80 + 50 80 0 5. Proved. 5. Proof : We are given two triangles and PQR in which Q, R and QR We need to prove that PQR There are three cases. ase I : Let PQ In and PQR, Q (Given) QR (Given) TOPI- Some Properties of Triangles ase II : Suppose PQ and < PQ Take a point S on PQ such that QS P Q Q Join RS. In and SQR, SQ (y construction) QR (Given) Q (Given) SQR (y SS rule) QRS (y c.p.c.t.) ut, QRP QRP QRS Which is impossible unless ray RS coincides with RP. must be equal to PQ. So, PQR ase III : If > PQ. We can choose a point T on such that T PQ and repeating the arguments as given in ase II, we can conclude that PQ and so, S PQR [SE Marking Scheme, 0, 0] R R. In PEQ and PER, PEQ PER 90 QPE RPE (Given) PE PE (ommon) PEQ PER (by S) S O L U T I O N S WORKSHEET-5 PQ PR. (y c.p.c.t.) P Q 90 90 E R P-9
. D D D...(i) D D D...(ii) dding eqns. (i) and (ii), we get D + D D + D D. Proved. [SE Marking Scheme, 0, 0]. P In P and P, (Given) P P (ommon) P P 90, (P ) y RHS rule, P P. (y c.p.c.t.). X Y. y angle sum property, 5. ut X X (ii) ut Y (i) From (i) and (ii) X Y E 5 O 6 D + 80 90 + 5 90 + (Eternal angle) 90 + (Eternal angle) dding, + 80 + + 80 + 5 (Eternal angle) 5 WORKSHEET-5 PQS PRS QPS RPS (y c.p.c.t.) Hence, S bisects P. P 90 + + 80 90 + + 80 90 5.. In PQS and PRS, PQ PR (Given) PS PS (ommon) PSQ PSR 90 (PS is altitude) y R.H.S. rule, Q S R. From DF, we have + y FD 0...() lso FE 80 0 70 FE 90 70 0 From y FE + 0 + 0 60 Then () gives 50. P-0 M T H E M T I S - I X T E R M -
. D In,...() ngles opp. to equal sides are equal.. Proof : In F and E, (Given) (ommon) F E F E (by SS cong.) y c.p.c.t., F E. [SE Marking Scheme, 0] lternative method : E F, Since E and F are the mid-points of and. In F and E, (Given) (ommon) F E (Proved) F E (y SS cong.) F E. (y c.p.c.t). In D., D D In D,...() + ( + ) + 80 + + + 80 ( + ) 80 + 90 D is a right angle. [SE Marking Scheme, 0, ] lternative Method : WORKSHEET-5 Let be an equilateral triangle, so that. Now,...() ( ngles opp. to equal sides are equal)...() ( ngles opp. to equal sides are equal) From () and (), we have lso, + + 80, (ngle sum property) + + 80 80 60 60. Thus, each angle of an equilateral triangle is 60.. lso, ut, + + 80 80 60 60. [SE Marking Scheme, 0] D E We have here E E (given)...() lso D + DE E E [Use ()]...() Now D + 90 DE + E + D + DE + D + DE + [Use ()] DE + D + DE S O L U T I O N S P-
DE ( ). Proof and D D D D D D D D D D by SS P P P P by SS TOPI- Inequalities of a Triangle P P, P P P D 90 P is perpendicular to P is perpendicular bisector of D P. If D is a point on the side of a such that D bisect, then > D.. In, > (Given) > (ngles opposite to larger side is greater) + > + (dding on both sides) + > + (D bisects, ) D > D. (Eterior angle property of triangle). Proof : In O, > < O < O...() In OD, > D D > O < OD...() dding () + (), O + O > O + OD > D or, D < [SE Marking Scheme, 0]. No, ecause,. +. 5. cm (third side) Not possible to construct a triangle.. Since > >, > >. WORKSHEET-55. In PQS, PQ + QS > PS...() (Sum of any two sides is greater than the third side) In PSR, PR + SR > PS...() dding () & (), PQ + QS + PR + SR > PS ( QR QS + SR) PQ + QR + RP > PS. [SE Marking Scheme, 0] 5. D D D D 59 (ngles opp. to equal sides are equal) In D, 59 + 59 + D 80 D 80 8 6 and D 6 0 (Eterior angle is equal to the sum of interior opposite angles) In D, > D (Side opp. to greatest angle is the longest) lso in, < D <. 70 0 80 WORKSHEET-56 P- M T H E M T I S - I X T E R M -
. P S T Q R onstruction : Produce QS to meet PR in T In PQT, PQ + PT > QT PQ + PT > QS + ST...() In SRT, TR + ST > SR...() dding () and (), we get PQ + PT + TR + ST > QS + ST + SR PQ + PR > QS + SR QS + SR < PQ + PR. [SE Marking Scheme, 0] DO O 90 (Given) and D (Given) OD O (y S congruence criterion) O O (y c.p.c.t.) i.e., O is the mid-point of Hence, D bisects. (ii) ongruency of triangles. (iii) Equality is the sign of democracy. 6. D. In, as is the greatest side > >...() > >...() On adding () and (), we get > + + > + + > 80 > 60. [SE Marking Scheme, 0] 5. (i) and D intersect at O OD O In OD and O, we have OD O (Vertically opp. angles)...(i) From...(ii) onstruction : E Produce D to E such that In triangles D and ED, D DE. Join E. D DE D D D ED D ED (onst) (Given) (V.O.) (SS congruence aiom) E (y c.p.c.t.) In E, + E > E [Triangle Inequality Property] + > E ( E ) + > D + DE + > D + D ( DE D) + > D. Proved. [SE Marking Scheme, 0, ] FORMTIVE SSESSMENT WORKSHEET-57 & 58 Note : Students should do these activities themselves. qqq S O L U T I O N S P-
SETION HPTER 6 OORDINTE GEOMETRY TOPI- artesian System. P (, ) and Q (, ) lie in IV and II quadrants.. (, ). In II quadrant, < 0 Points (, 0), (, 0). The distance of a point from the y-ais is called its -co-ordinate, or abscissa. 5. The P is on -ais y 0 P is at a distance of units from y-ais to its left. In second quadrant, the co-ordinates of the point P (, 0). 6. II Quadrant. 7. () P(0, 5) () Q(0, ) () R(5, 0) (D) S(, 0) ' y 6 5 P(0, 5) (, 0) R(5, 0) 5 O 5 6 Q(0, ) WORKSHEET-59 8. () II quadrant () III quadrant () IV quadrant (D) I quadrant. 9. y (, ) D (, ) ' 0 (, ) (, ) y' o-ordinates of D are (, ). 0. In a point ( 5, ), < 0 and y > 0 Point ( 5, ) lies in II quadrant. In a point (, ), > 0 and y < 0 Point (, ) lies in IV quadrant In a point (5, 0), > 0 and y 0 Point (5, 0) lies on -ais In a point (6, 6), > 0 and y > 0 Point (6, 6) lies in I quadrant In a point ( 5, ), < 0 and y < 0 Point ( 5, ) lies in III quadrant y'. (, y) (y, ). 7 WORKSHEET-60. The co-ordinate of -ais (, 0). Negative ordinate i.e., (, y) P- M T H E M T I S - I X T E R M -
5. The point on y-ais has -co-ordinate 0. Since it lies at a distance of units in the negative direction of y-ais. The point is (0, ). 6. (a) point which lies on and y-aes is (0, 0) i.e., origin (b) point whose abscissa is 5 and ordinate is 6 i.e., 5 and y 6 is (5, 6) (c) point whose ordinate is 6 i.e., y 6 and lies on y-ais is (0, 6) (d) point whose ordinate is and abscissa is 7 i.e., y and 7 is (7, ) (e) point whose abscissa is i.e., and lies on -ais is (, 0) (f) point whose abscissa is and ordinate is i.e., and y is (, ) 7. Firstly, we plot all the points i.e., (, 0), (5, 0), (5, ) and D(, ) on a graph paper and join all these points.. (0, 0). Origin. Point (, a ) lie on -ais a 0 a.. () II quadrant () III quadrant () I quadrant (D) II quadrant. 5., 0,,. 6. The vertices of the rectangle O are O(0, 0), ( 6, 0), ( 6, ), (0, ) 7. (i) Draw X OX and Y OY as the co-ordinate aes and mark their point of intersection O as the origin (0, 0). In order to plot the points (, 8), we take units on OX and then 8 units parallel to OY to obtain the point (, 8). Similarly, we plot the point (, 7). In order to plot (0,.5), we take.5 units below the -ais on the y-ais to obtain (0,.5). S O L U T I O N S Y 5 D(, ) (5, ) X' (, 0) (5, 0) 5 0 X 5 6 5 Y' The obtained figure D is a rectangle, since D and D and all lines are perpendicular to each other. 8. (i) E(, ) (ii) D(, ) (iii) o-ordinates of (, ) o-ordinates of (, ) The abscissa of abscissa of (iv) o-ordinates of (, ) o-ordinates of F (, ) The ordinate of + ordinate of F + ( ) + + + + + WORKSHEET-6 In order to plot (, ), we take unit on OX and then units parallel to OY to obtain the point D(, ) Y (, 8) 8 (, 7) 7 6 5 D(, ) X' X 5 0 5 6 7 8 E(, ) (0,.5) 5 Y' In order to plot (, ), we take units on OX and then unit below -ais parallel to OY to obtain the point E(, ) (ii) o-ordinate geometry. (iii) o-ordination among people is good for progress. P-5