TOPIC-1 REAL NUMBERS. Rational Numbers WORKSHEET-1 WORKSHEET-2 P-1 SECTION. \ x = [CBSE Marking Scheme, 2012] S O L U T I O N S CHAPTER

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1 SECTION CHAPTER REAL NUMERS TOPIC- Rational Numbers WORKSHEET So, it is a rational number Yes, zero is a rational number. Zero can be epressed as 0 0, 0 6, 00 etc, which are in the form of, p q where p and q are integers and q ) \ 0.8 [CSE Marking Scheme, 0] [CSE Marking Scheme, 06] 6. Since LCM of and is, \ and Hence, three rational numbers between and 9 are : 6, 8,. Any eample & verification of eample : Let m /, n 9/ Difference 9 0 (Rational Number) Sum (Rational Number) Product (Rational Number) Division 9 8 (Rational Number) WORKSHEET (Decimal point is shifted three 000 places to the left) [CSE Marking Scheme, 0] S O L U T I O N S P-

2 .. 6 Terminating [CSE Marking Scheme, 0] Alternative Method :. 6 6) (Terminating). Two rational numbers between and are 0. and 0. i.e. and and 0. [CSE Marking Scheme, 06] 6. LCM of and is \ and so, < The required three rational numbers are, and. [CSE Marking Scheme, 0]. Let a & b Here, we find si rational numbers, i.e., n 6 So d b a n + 6+ st rational number a + d + nd rational number a + d + rd rational number a + d + th rational number a + d + th rational number a + d + 6 th rational number a + 6d So, si rational numbers are,,,, &. 8. Since LCM of and 6 is 0 \ and Hence, four rational numbers between 6 and are : 6 0, 8 9 0, 0, [CSE Marking Scheme, 0] WORKSHEET-. and 8 So three rational numbers and are 6,, [CSE Marking Scheme, 06]. and i.e., 0 and The numbers are and [CSE Marking Scheme, 06] P- M A T H E M A T I C S IX

3 . Let, ( ) ( ) 9 9. Let, ( ) ( ) 9 9. Let, (...) (0...) Alternative Method : Let [CSE Marking Scheme, 0] (...) (0...) 999 \ Let Let, (i) multiplying 0 on both the sides, we get, (ii) From (ii) (i), we get [CSE Marking Scheme, 0] TOPIC- Irrational Numbers WORKSHEET-. 0. is a terminating number. So, it is not an irrational number , is repeating continuously, so it is not an irrational number , is repeating continuously, so it is not an irrational number , non-terminating and non recurring decimal. Hence, it is an irrational number. So, is an irrational number.. No, it may be rational or irrational.. Required two irrational number are : (i) (ii) and [CSE Marking Scheme, 0] S O L U T I O N S P-

4 Hence three inrrational numbers between and 9 can be : [CSE Marking Scheme, 06] Let A C unit length Using Pythagoras theorem, we see that OC + Construct CD unit length perpendicular to OC, 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 point Q, then Q corresponds to. Let two irrational numbers are : 6 and, (i) 6 Difference is an irrational number. (ii) 6 + sum is an irrational number. (iii) 6 8 product is an irrational number. (iv) 6 / division is an irrational number. WORKSHEET-. Sum of and +.. Since,.6 Hence, the irrational number between and. is. ( ) ( ) + ( ) Given, (Irrational Number) Hence, required number can be [CSE Marking Scheme, 0] 6. Marks the distance 9. units from a fied point A on a given line to obtain a point such that A 9. units from, marks a distance of unit and mark the new point as C. Find the mid point of AC and mark that point as O. Draw a semi-circle with centre O and radius OC. Draw a line perpendicular to AC passing through and intersecting the semi-circle at D. then D 9. To represent 9. on the number line, let us treat the line C as the number line, with as zero (c) as and so on. Draw an arc with centre and radius D which intersect the number line at E \ E represents 9. [CSE Marking Scheme, 06]. Mark the distance. units from a fied point A on a given line to obtain a point such that A. units. From, mark a distance of unit and mark the new point as C. Find the mid-point of AC P- M A T H E M A T I C S IX

5 . and mark that point as O. Draw a semi-circle with centre O and radius OC. Draw a line perpendicular to AC passing through and intersecting the semicircle at D. Then, D. To represent. on the number line, let us treat the line C as the number line, with as zero, C as, and so on. Draw an arc with centre and radius D, which intersects the number line at E. \E represents. TOPIC- n th Root of Real Number ` Mark the distance 9. units from a fied point A on a given line to obtain a point such that A 9. units from, mark a distance of unit and mark the new point as C. Find the mid-point of AC and mark that point as O. Draw a semicircle with centre O and radius OC. Draw a line perpendicular to AC passing through and intersecting the semi-circle at D. Then, D 9. To represent 9. on the number line,let us treat the line C as the number line, with as zero, C as, and so on. Draw an arc with centre and radius D, which intersects the number line at E. \ E represents 9. [CSE Marking Scheme, 0] WORKSHEET-6 ( )( ). + S O L U T I O N S { ( ) } { } 0 [CSE Marking Scheme, 0] ( 8 + ) [CSE Marking Scheme, 0] [CSE Marking Scheme, 0] Alternative Method : 0 0 [CSE Marking Scheme, 0] P-

6 8. LCM of and is [CSE Marking Scheme, 0] Alternative Method LCM of and is 6 \ and ( ) ( ) Now, ( ) ( ) 6 [CSE Marking Scheme, 0] WORKSHEET-. b a a b. ( )( ) ( ). a+ b a b ( a) b a b [CSE Marking Scheme, 06] \ ( + ) ( ) ( ) ( ) [CSE Marking Scheme, 0, 0] 6. ( ) ( ) + ( ) [Using (a b) a + b ab] ( 8 ). [CSE Marking Scheme, 0] ( )( + ) [CSE Marking Scheme, 0] Alternative Method : ( )( + ) ( ) ( ) ( ) ( ) ( + ) ( ) P-6 M A T H E M A T I C S IX

7 Alternative Method : [CSE Marking Scheme, 0] ( ) + ( ) + + TOPIC- Laws of Eponents with Integral Powers + ( ) + ( ) ( ) + ( ) WORKSHEET (6) 0.6 (6) 0.09 (6) 0. (6) / ( ) /. ( + + ) / ( ) / (6) / [(6) ] / [CSE Marking Scheme, 0]. Given, a and b. a b + b a [CSE Marking Scheme, 0] ( ) + ( + ) ( ). ( ) ( )() + () + ( ) + ( ) ( ) + ( ) ( ) () +( ) () Alternative Method : ( y ) y ( ) y ( y) y y y 9 y 9 [CSE Marking Scheme, 0] y. y ( ) ( ) y ( y) y ( y) + + y 9 y 9 y [CSE Marking Scheme, 0] S O L U T I O N S P-

8 . ( a b ) a+b. ( b c ) b+c. ( c a ) c+a a b b c c a.. a b b c c a (Any number to the power 0 is ) [CSE Marking Scheme, 0] ( ) ( ) + ( ) + ( ) [CSE Marking Scheme, 06] WORKSHEET-9. [(6) ] ( ) [ ] / /. 6 ( 6 ) ( 6 ) / ( 6 ) / 6 [CSE Marking Scheme, 06]. Given, a and b. (i) (a b + b a ) ( + ) (8 + 9). (ii) (a a + b b ) ( + ) ( + ) [CSE Marking Scheme, 0] 8 [CSE Marking Scheme, 0]. ( 6) ( 6) 6 ( ) ( ) [CSE Marking Scheme, 0, 0] Alternative Method : 6 ( ) ( ) 6 ( ) ( ) (i) ( + ) (ii) ( + ) 9 [CSE Marking Scheme, 0] Alternative Method : (i) ( y +y ) ( + ) ( + ) (). (ii) ( +y y ) ( + ) ( + ) (9) 9 P-8 M A T H E M A T I C S IX

9 0 ( ) [CSE Marking Scheme, 0] WORKSHEET ( ) ( ) ( ) ( 8 ) 6 / + ( ) ( ) ( ) / + ( ) [CSE Marking Scheme, 0] / 6 +. ( 6) ( 6) 6. / 6. [(8 / + 6 / ) / ] [( + ) / ] [ / ] [ / ] [ / ]. 8 + ( + ) ( ) [CSE Marking Scheme, 0] Alternative Method : 8 + ( ) + ( ). ( + ) ( ) ( ). ab c ( b ) c ( ) ba ( c) a ab ac ( ) ( ab ac ba+bc ) ( bc ac ) ac ba bc bc Alternative Method : ab ( c) b a ba ( c) c ab ac ac ba bc bc ( ab ac ba+bc ) ( bc ac ) ( bc ac ) ( bc ac ). ( ) ( ) ( ) \ [CSE Marking Scheme, 0] Alternative Method : Given, ( 8) ( ) ( ) ( ) ( ). + + comparing the power of both sides, we get 0 [CSE Marking Scheme, 0] S O L U T I O N S P-9

10 \ [CSE Marking Scheme, 0] Alternative Method : 8 6 Comparing the eponents, we get.. abc ( a ) bc (y) bc (y b ) c (z) c abc 8. ab ab a b+ a + a b a Alternative Method : WORKSHEET- abc [CSE Marking Scheme, 0] b ab+ b + ab b b a b a [CSE Marking Scheme, 0] 9. a a + b + a a b a + a + a b a b a a a+ b + b a ab ab b b a+ b + b a ( b a ) + bb ( + a ) ( b+ a)( b a) b ab+ b + ab b a b b ( a b ) a b [CSE Marking Scheme, 06]. ( 8) ( 8) ( ) ( ). 8. P-0 M A T H E M A T I C S IX

11 . a b [CSE Marking Scheme, 06] a b 8 b a [ 8] a b [ 8] 9 [CSE Marking Scheme, 0].,, 6,, ( ),( ),( 6 ) 9, 6, 6 In ascending order ( 6),( 6),( 9) i.e., 6,,.. [CSE Marking Scheme, 0] Alternative Method : Since LCM of,, is.,, 6,, 6 6, 6, 6, 6,, 6 6, 9, 6. In ascending order ( 6),( 6),( 9) a b ( b c) ( c a) ( + ) a b c ( ) 6,, + a+ b b+ c c+ a.. a b c.. a+ b+ c a+ b+ c [CSE Marking Scheme, 0] 6. a b 0, + 0 ( ) ( + ) [CSE Marking Scheme, 06]. ( ) ( ) \ [CSE Marking Scheme, 0] TOPIC- Rationalization of Real Numbers WORKSHEET So, rationalizing factor is ( 6 ) 6 8 S O L U T I O N S P-

12 ( ) ( ) ( + ) ( ) ( ) ( ) 9 8. ( ) ( ) ( ) ( ) + [CSE Marking Scheme, 0] Alternative Method : 0 ( ) ( ) ( ) ( ) ( + ) ( + ) (rationalizing) ( ) 0 + ( ) ( ) ( ) + + ( ) ( ) + ( ( + ) ( ) ( + ) ( ) ( ) ( ) ( + + ) + + ( + )+ ( + ) ( + ) ( + + 6) ( ) ( 6) WORKSHEET-. ( ) ( + ) ( ) [CSE Marking Scheme, 0] Alternative Method : + ( ) ( + ) ( ) + 6 ( ) ( ) () + 6 P- M A T H E M A T I C S IX

13 \ Given epression [CSE Marking Scheme, 0] Alternative Method : ( ( ) + ( ) ( ) ( 6 ) ( ) ( ( + ) ( 6 ) ( 6) S O L U T I O N S ( ) and ( )( ) ( ) ( + ) + + ( + ) + ( ) + 0 8( + ) [CSE Marking Scheme, 0] Alternative Method : ( + ) ( + ) ( + ) ( ) ( ) ( + ) LHS 8 88 ( + ) ( ) ( 8) ( 8) ( ) + 6 ( + 6) + ( ) ( 6) ( 6 + ) ( 6) ( ) ( 6 + ) + ( ) ( ) + 8 ( 8 + ) ( + 6) ( 6 + ) ( + ) [ a b (a b) (a + b)] RHS 6. Denominator \ ( 0 ) ( 0 + ) ( 0 ) [CSE Marking Scheme, 0] Alternative Method : ( 0 + ) 0 ( 0 + ) ( 0 + ) 0 (. +. ). P-

14 WORKSHEET-.... (rationalizing) p [CSE Marking Scheme, 0] ( ) ( + ) ( 0 ) ( ) ( 0 + ) 8 + ( 0 + ) 0 ( 0 + ) ( 0 + ) ( ) + ( ) [CSE Marking Scheme, 06] 9. ( + ) ( )+ ( + ) 9 ( ) ( 6+ )+ ( + ) ( ) (Rationalizing) 9( ) ( 6+ )+. +. [CSE Marking Scheme, 0] + ( + ) ( ) ( ) + ( ) + ( ) ( ) ( + ( + ). (appro). 6. a a a b [CSE Marking Scheme, 0] ( ) ( ( + ) ) ( ) ( ) ( ) b b 9 9 On comparing both sides, we get a 9, b 9. [CSE Marking Scheme, 0] P- M A T H E M A T I C S IX

15 WORKSHEET Alternative Method : (. ) + (. ) ( + ) + ( ) ( )( + ) +.06 [CSE Marking Scheme, 0] Squaring, both sides, we get b and y y+ y (+ ) ( ) a ( + ) 9+ ( ) b + 9 \ a + b [CSE Marking Scheme, 06] \ + ( + + ) (8) 6 S O L U T I O N S P-

16 SECTION CHAPTER POLYNOMIALS TOPIC- Polynomials WORKSHEET-6... Not defined. Linear polynomial.. Not a polynomial.. Constant polynomial is. 6. Linear polynomial +, degree Quadratic polynomial +, degree Cubic polynomial,, degree. f() + then, f() + 6 then, f() + 0 f() f() p() + 6 (i) When (ii) When then, p() f() + Then, f() + f( ) ( ) ( ) + and f f() f( ) + f then, p( ) ( ) + 6 WORKSHEET-. inomial. Degree of a polynomial is 0. Degree of + Degree of Degree of ( ) ( ) Every real number is a zero of the zero polynomial.. No. of zeroes of cubic polynomial. 6. p (y) y y + p (0) or y + y. 8. Co-efficient of in epression + π is π. 9. p() + 6 Then, p( ) ( ) ( ) ( ) f(y) y y + ay + b f() () () + a() + b 0 Þ a + b 0 Þ a + b...(i) f(0) b 0 from (i) a + 0 Þ a \ a, b 0 [CSE Marking Scheme, 06]. f() + + f(0) f( ) ( ) ( ) + ( ) ( ) f() () () + () () \ f(0) f( ) ¹ f() [CSE Marking Scheme, 06] P-6 M A T H E M A T I C S IX

17 . f() + f() () () + () f() f( ) ( ) ( ) + ( ) 8 6 f( ) 0 f(0) \ f() + f( ) + f(0) 0 6 [CSE Marking Scheme, 06]. p() + + p() () + () () p( ) ( ) + ( ) ( ) p(0) p() + p( ) p(0) [CSE Marking Scheme, 0] TOPIC- Remainder Theorem WORKSHEET-8. Let, p(y) y y y +, then remainder p(0) p() Put, 0 Remainder p S O L U T I O N S in p() [CSE Marking Scheme, 0]. Factors of (±, ±, ±, ±, ± 6, ± ) p() 6 + p() 6() () is a zero of p() or ( ) 6 + is a factor of p() 6 ( ) ( ) + 6( ) ( ) (6 + 6) ( ) ( ) ( ) [( ) ( )] ( ) ( ) ( ).. Let p() a + and g() + a R p() R g() + p() a() + () R 8a + 8a R () () + a a R 6 + a R R 8a 6 + a a a [CSE Marking Scheme, 0]. p() a p() leave the same remainder when divided by ( + ) and ( +). p( ) ( ) + 8( ) + ( ) + a ( ) 8 + a 0 a p( ) ( ) + 8( ) + ( ) + a( ) + 8 a 0 a Remainders are equal So, 0 a 0 a a 0 a 0 6. f() + a + b Put, 0 or in f(), we get f() () () + () a + b + a + b a + b a b a b...() P-

18 Again put, + 0 or in f(), we get f( ) ( ) ( ) + ( ) a ( ) + b a + b 9 6 a + b a + b...() Adding equations () and (), a 0 a y equation (), + b b 8 f() Again put, 0 or in f() f() () () + () WORKSHEET-9. Let, f() + 0 Put, + 0 Then remainder is : f( ) ( ) Here, p() a + 6 a, and the zero of a is a. So, p(a) a a.a + 6a a a. So, by the remainder theorem a is the remainder when a + 6 a, is divided by a.. Let, p() + k, and q() + + k Put, + 0 or in p () and q () P( ) ( ) ( ) + k( ) 0 k k 6 q( ) ( ) ( ) + ( ) +k 8 +k 0 +k p() and q() leave the same remainder when divided by +. k 6 k 0 k 6 k. Let p() a + + and q() + a Put, 0 or in p() and q() p() a() + () + a a + q() () () + a + a + a According to the question, p() q() a + + a a a 6a 6 a. [CSE Marking Scheme, 0]. p() + Put, 0 or in p() p() () + + Hence, must be subtracted from + so that it is eactly divisible by ( ). Resultant polynomial to be divisible by ( ) quotient + ) Thus, quotient + + and remainder. Let, p() Zero of + is / y remainder theorem, Remainder p P-8 M A T H E M A T I C S IX

19 WORKSHEET-0. 0 Þ y remainder theorem, if f() is divided by, the remainder is f \ f Hence required remainder is.. p() So, [CSE Marking Scheme, 06] + ) ( ) (+) 9 ( ) (+) + + (+) ( ) (Remainder) Quotient + Remainder. p() + p + p Put, + 0 or in p(), we get p( ) ( ) ( ) + ( ) p( ) + p p + p 9 p 9 p 0 p The polynomial p() Put, + 0 or in p() p( ) ( ) ( ) + ( ) ( ) ) ( Hence, Quotient + + and Remainder (Remainder) Verification : Let, f() + 0 y remainder theorem, remainder f( ) ( ) + ( ) ( ) [CSE Marking Scheme 0] S O L U T I O N S P-9

20 TOPIC- Factor Theorem WORKSHEET-. ( ). a b 6ab 6ab(a b). Since, f 0 is a zero of polynomial f() So, + or + is a factor of f().. f() + k ( + ) is a factor of f() + k Þ f( ) 0 Þ ( ) + ( ) k( ) 0 Þ + + k 0 Þ k 0 Þ k. ( ) is a factor of f() + k + k f() 0 Þ () + k() + k 0 Þ + k + 0 Þ + k 0 Þ k [CSE Marking Scheme, 06] 6. Let, p() 9, p ( ) Since, ( + ) is a factor of 9 ( 9 ) ( + ) ( ) + ( ) + ( + ) ( + ) ( ) Factors are : ( + ) ( ) [CSE Marking Scheme, 0]. 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 ) m n is a factor of p Similarly p(n p) 0 & p(p m) 0 n p is a factor of p. and p m is a factor of p. [CSE Marking Scheme, 0] 8. Let p() + a + 6 p() is divisble by Þ p 0 p p a Þ a 0 Þ a Þ a 9 0 Þ a 9 \ p() ) \ ( + )( + ) Hence p() ( )( + )( + ) WORKSHEET-. + is a factor of + + m p() p( ) m 0 m. is a factor of p(), then p() 0 p() () + k k 0 k. P-0 M A T H E M A T I C S IX

21 . is a factor of p() 6 + k p k. 0 k [CSE Marking Scheme, 0]. ( ) is a factor of f() a a + a f() 0 Þ a () a() + a 0 Þ a a + a 0 Þ a 0 Þ a ± [CSE Marking Scheme, 06]. Given, p() k ( + ) is a factor of p(), then p( ) 0 k ( ) ( ) 0 k + 0 k [CSE Marking Scheme, 0] 6. 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. [CSE Marking Scheme, 0]. p() 9 Factor of ± p() 9 0 Þ is a factor of p() \ 9 9 ( ) + 6( ) + ( ) ( )( ) ( )( + ) \ p() ( )( + )( + ) [CSE Marking Scheme, 06] 8. Let f() The factors of the constant term are ±, ±, ± and ± 6, The factor of coefficient of is Hence possible rational factors are ±, ±, ± ±, ± We have f( ) ( ) + ( ) ( ) 9( ) andf( ) ( ) + ( ) ( ) 9( ) So, + and + are factors of f() Þ ( + )( + ) is a factor of f() Þ + + is a factor of f() Let us now divide f() by + + to get the other factos of f(). + + ) \ ( + +)( ) ( + )( + )( 6 + ) ( + )( + )[( ) + ( )] ( + )( + )( )( + ) WORKSHEET-. Since, a + ab 6 a (a 6 + b 6 ). Factors are a and (a 6 + b 6 ) ( ) ( ) ( ) ( ) ( ) + ( ) ( )( + ) [CSE Marking Scheme, 06] S O L U T I O N S. f() + 6 Put, 0 or in f() Thus, f() Hence, ( ) is a factor of f(). [CSE Marking Scheme, 0] P-

22 . ( + ) is a factor of p() k p ( ) 0 ( ) k ( ) + ( ) k k [CSE Marking Scheme, 0] 6. Let, p() 6 a + a + a + ( a) is a factor of the polynomial p(), then p(a) 0 a 6 a a + a a a + a a + 0 a 6 a 6 + a a + a a + 0 a a.. Factors of (±, ±) p() () () 9() 0 is a zero of p() or ( ) is a factor of p() Then, 9 ( ) + ( ) + ( ) ( ) ( + + ) ( ) ( ) ( ) (( + ) + ( + )) ( ) ( + ) ( + ) 8. Factor of 6 (±, ±, ±, ± 6) p() + 6 p( ) ( ) + ( ) ( ) is zero of p() of ( + ) is a factor of p() \ + 6 ( + ) + ( + ) 6( + ) ( + )[ + 6] ( + )[ + 6] ( + )[( + ) ( + )] ( + )( + )( ) TOPIC- Algebraic Identities WORKSHEET-. y ( ) 9 y y y + [CSE Marking Scheme, 0] (9) () (9 + ) (9 ) [CSE Marking Scheme, 0]. 9 (0 ) (0 + ) (0) () [CSE Marking Scheme, 0]. ( y + z) + y + z y + yz + z [CSE Marking Scheme, 0] Alternative Method : y using the identity, (a + b + c) a + b + c + ab + bc + ca ( + ( y) + z) () + ( y) + z + ()( y) + ( y) (z) + (z)() + y + z y yz + z 6. (a b) [ (a b) ] [() {(a b)} ] [ + (a b)] [ (a b)]. [CSE Marking Scheme, 0]. 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). [CSE Marking Scheme, 0] 8. a + ab 6 a(a 6 + b 6 ) a[(a ) + (b ) ] a(a + b ) [(a ) + (b ) a b ] a(a + b ) (a + b a b ). [CSE Marking Scheme, 0] P- M A T H E M A T I C S IX

23 9. (i) 0 0 (00 + ) (00 + ) 00 + ( + ) (ii) (0) (00 + ) (00) (00 + ) WORKSHEET-. ( ) () () ( ) Co-efficient of in the epansion of ( ) 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.. ( y + z) [ + ( y) + z] () + ( y) + z + ( y) + ( y) z + z + 9y + z y 6yz + z.. a b + a b + + a b + + () + a b + b () + a () a b ab a + + b Area of rectangle a a + a 0a a + a(a ) (a ) (a ) (a ) length breadth Length and breadth are (a ) and (a ) respectively y [ 6y ] [() (6y) ] ( 6y) [() + (6y) + 6y] a b (a b) (a + b + ab) ( 6y) ( + 6y + 0y).. pq + ( ) 9 8. (i) + pq pq ( pq) a + b (a + b) (a + b ab) 9 pq + pq , + pq 9 9 pq 9 () (ii) y ( ) (y ) ( y ) ( + y ) ( y) ( + y) ( + y ). ++ WORKSHEET-6.. Given, + + y + y + y y ( ) () 6. + y y + y + y 0 y ( y) ( + y + y) ( y) ( ) ( 8+ ) ( ) a + b (a+ b) (a + b ab) y y y y S O L U T I O N S P-

24 8 y y y 8y y y (00 + ) (00 + ) [CSE Marking Scheme, 0] 6. ( + y) + y + y ( + y) ± ±.. Given Ep ( y) ( + y) ( y) + ( + y) ( y) ( + y) ( y) ( + y) ( y) + ( + y) ( y) [ y y] ( + y) [ y y] ( y) [ y ] ( + y) [ y] ( y ) [ y ( + y)] (y + ) ( y) (y + ) ( + y) ( + y) (y + ) 8. + Cubing both sides ( ) ( ) + 0 [CSE Marking Scheme, 06] WORKSHEET as(a + b) (a b) a b [CSE Marking Scheme, 0, 0]. Given,. p 6 abc abc a+ b+ c 0 a + b + c abc. 9 p + p (p) 6.(p) + (p) 6 6 p 6 p p p [CSE Marking Scheme, 00, 0, 0]. On squaring both sides, we get + + a bc ( ) +. [CSE Marking Scheme, 0] b c + + a + b + c ac ab abc. y + z + yz () + ( y) + (z) () ( y) (z) ( y + z) [() + ( y) + (z) ()( y) ( y) (z) () (z)] a + b + c abc (a + b + c)(a + b + c ab bc ca ) ( y + z) [ + 9y + z + y + yz z] [CSE Marking Scheme, 0] P- M A T H E M A T I C S IX

25 + (6) WORKSHEET-8. Let, + p and q, then given epression p 8pq q p 8pq + 0pq q 6p(p q) + q(p q) (p q) (6p + q) [( + ) ( )] [6( + ) + ( )] ( ) ( ) ( 6 + ) ( ).. a, b, c a + b + c + 0 a + b + c abc () + () + ( ) ( ). + y 8. ( + y) y + y ( + y) 8 + y y y.. a + b a b (a b) () a b ab(a b) a b ab a b 9ab.. (00 + ) (00) + (00) () + (00) () + () 66 [CSE Marking Scheme, 0] 6. ( + ) + ( ) ( ) + ( ) + + ( ) + ( ) ( ). Add, and [CSE Marking Scheme, 0] S O L U T I O N S P-

26 SECTION CHAPTER LINEAR EQUATIONS IN TWO VARIALES TOPIC- Introduction of Linear Equation WORKSHEET-9. a + by + c 0, where a, b, c are real numbers and both a, b 0.. No. ( a, b 0 for a linear equation). True.. Let the runs scored by Raina & Dhoni are & y respectively then, + y 98. So, when y 0 0 y 0 6. The equation of two lines on the same plane which are intersecting at point (, ) are : + y y. Let father s present age years Son s present age y years After years father s age will be ( + ) years After years son s age will be (y + ) years According to the question, + (y + ) + y + 0 y 0 WORKSHEET-0. Let the number of goats & hens in herd are & y respectively. then, + y 0. 6y. Given equation : y The point (0, 0) satisfies the given equation, hence answer is yes.. No. ( 0 0). According to question, 00 + y y Put and y, then y () () So, (, ) is the solution of the equation. Again put and y, then, y. 8. The line passing through (, ) is y or, y Infinitely many lines are there. The equation in the form a + by + c 0 is y y y + 0 Putting 0, y + 0. \ (0, ) is not the solution of given equation. Putting, y , which is correct \ (, 9) is a solution of the given equation. 0. Let larger number be, then times of larger number and smaller number be y Quotient and remainder 9 So, according to the question, y + 9 y 9 0. y ( ) ( ) So, (, ) is not a solution of the given equation.. (, ) lies on the graph y + k 0 \ ( ) + k 0 k + k 8. Given, A(, ) and C(, ) Then, (, ) and D(, ) Also, equations of sides of square are, A : Y C : X CD : Y DA : X P-6 M A T H E M A T I C S IX

27 9. (, ) lies on the graph of ay + \ ( ) a ( ) + a + a 0 a. 0. (a) when y, then + (b) when, then () + y y 8 (c) when, then + y y \ One more solution is (, ). WORKSHEET-. Sol y. + y 0. As the line intersects y-ais, put 0 in the given equation, we get (0) + y \ y 6 The required point is (0, 6).. Given, + \ ( ) \. + y + k 0 If, and y is the solution of the linear equation + y + k 0 then () + () + k 0 k. 6. y 8 y 8 For y 8 8 \ (, ) lies on the line.. Equation y or y \ y On -ais y 0 \ 0 6 At point (6, 0) the given line cuts the -ais. On y-ais 0 \ y 0 y At point (0, ) the given line cuts the y-ais. 8. The linear equation is + ky 8 At, y, () + k() 8 + k 8 \ k If, then () + y y 8 y 0 \ y 0 9. The equation is y 9 A (, ); 9() ; True (, 6); 6 9 ( ) 9 6; True C (0, ); 9 (0) 0 ; True. WORKSHEET-. Given point lies on the line i.e., () a() + a a... (0, ) is the solution of given equation \ it satisfies the equation S O L U T I O N S \ (0) + () k \ k 6. y 6. (, ). 6. If point (, ) lies on y a + \ a + a a. P-

28 . Equation : + y For intersection with -ais y 0 \ or \ Co-ordinates are (, 0) For intersection with y-ais or y 0 \ y \ Co-ordinates are (0, ). 8. Substituting and y in 9k + ky 6, we get 9k ( ) + k ( ) 6 9k k 6 k 6 k 9. Writing in standard form On -ais, y 0 + y 0 0 \ Point on the -ais (, 0) On y-ais, 0 y 0 y 6 \ Point on the y-ais (0, 6). 0. Given equation is (p + ) (p + )y 0...(i) If, y is the solution of the equation (i), then (p + ) (p + ) 0 p + 6p 9 0 p 8 0 p Put the value of p in the equation (i), then + y 0 or y + 0, is the required equation. TOPIC- Graphical Representation of Linear Equation in Two Variable WORKSHEET-. y k.. k.. y-ais. Given y + y At y-ais 0; \ (0) y 0 y \ y Given line intersects the y-ais at y.. Given equation + y Put 0, then 0 + y y Hence point on y-ais is (0, ) Now put y 0, then + 0 Hence point on -ais is (, 0) y O (0, ) (, 0) 6. Let Sita contribute ` and Gita contribute ` y. According to the question, + y y y 00 P-8 M A T H E M A T I C S IX

29 (0, 00) y y The points are (0, ) & (, ) y (0, ) (, ) y (00, 00) y'. (i) One Variable (00, 0) y 0 (ii) Two Variables 0 + y y 0 y ' y' WORKSHEET-. a > 0.. (0, ). Parallel, 6. Let the cost of a toy elephant, ball y \ y y 6 9 y P O 0 (i) Equation in one variable (Number line) : A point P at a distance of units to left of O on the number line. (ii) In two variables (Cartesian plane) : A line A parallel to y-ais at a distance of units to the left of y-ais.. y 0 Þ y 6 9 y 0 S O L U T I O N S P-9

30 WORKSHEET-., y 0 i.e., (, 0). (, ) O X. + 6 y ais From graph it is clear that line meets -ais at (, 0) and y-ais at (0, ).. + y 8 y y From the graph it is clear that Line intersects -ais at (, 0) and y-ais at (0, 8). -ais Plot on the cartesian plane. On cartesian plane, is a line parallel to y-ais.. y + ; three solutions are, y 0;, y and 0, y. WORKSHEET-6. y + 0. The given equation can be written as, + 6y 0 y ( + 0)/6 0 y 0 0 P-0 M A T H E M A T I C S IX

31 . 8 6 (i) Point P (, 0) represents the solution on the number line p 0 (ii) Line A represents the solution in the cartesian plane. y. Equation of the sides are, A : Y 0 C : X CD : Y DA : X Area sq. units ' (, 0) 0 y' S O L U T I O N S P-

32 SECTION CHAPTER COORDINATE GEOMETRY TOPIC- Cartesian System WORKSHEET-. Origin. P (, ) and Q (, ) lie in IV and II quadrants.. (a, b) (0, ) a 0, b., 0,,. [CSE Marking Scheme, 0]. 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) II quadrant () III quadrant (C) IV quadrant (D) I quadrant.. (A) IV quadrant () II quadrant (C) III quadrant (D) I quadrant 8. C(, ), distance from ais distance from y ais D(, ), distance from ais distance from y ais WORKSHEET-8. Since P(a, b) lies in IV quadrant \ a > 0, and b < 0 \ a > b [CSE Marking Scheme, 06]. In II quadrant, < 0 Points (, 0), (, 0). II Quadrant.. (A) II quadrant () III quadrant (C) I quadrant (D) II quadrant.. (A) P(0, ) () Q(0, ) (C) R(, 0) (D) S(, 0) 9. In a point (, ), < 0 and y > 0 Point (, ) lies in II quadrant. In a point (, ), > 0 and y < 0 Point (, ) lies in IV quadrant In a point (, 0), > 0 and y 0 Point (, 0) lies on -ais In a point (6, 6), > 0 and y > 0 Point (6, 6) lies in I quadrant In a point (, ), < 0 and y < 0 Point (, ) lies in III quadrant 0. (i) E(, ) (ii) D(, ) (iii) Co-ordinates of A (, ) Co-ordinates of (, ) The abscissa of A abscissa of (iv) Co-ordinates of C (, ) Co-ordinates of F (, ) The ordinate of C + ordinate of F + ( ) ' y 6 P(0, ) (, 0) R(, 0) O 6 Q(0, ) y' 6. (i) A point which lies on and y-aes is (0, 0) i.e., origin (ii) A point whose abscissa is and ordinate is 6 i.e., and y 6 is (, 6) (iii) A point whose ordinate is 6 i.e., y 6 and lies on y-ais is (0, 6) (iv) A point whose ordinate is and abscissa is i.e., y and is (, ) P- M A T H E M A T I C S IX

33 (v) A point whose abscissa is i.e., and lies on -ais is (, 0) (vi) A point whose abscissa is and ordinate is i.e., and y is (, ). (i) Required points are A(0, ) and L(0, ). (ii) Required points are G(, 0) and I(, 0). (iii) Required points are D(, ) and H(, ). WORKSHEET-9. The distance of a point from the y-ais is called its -co-ordinate, or abscissa.. 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).. (, y) (y, ).. The co-ordinate on -ais (, 0) 6. Co-ordinates of A (, 0) Co-ordinates of (, ) Co-ordinates of C (, ) Co-ordinates of D (0, ).. (i) Since the point lies on -ais at a distance of 9 units from y-ais. Hence its co-ordinates are (9, 0) (ii) According to the question, the required coodinates are (0, 9). 8. The vertices of the rectangle OAC are O(0, 0), A( 6, 0), ( 6, ), C(0, ) 9. (i) (, ) lies in III quadrant, as < 0 and y < 0. (ii) (, ) lies in IV quadrant, as > 0 and y < 0. (iii) (, ) lies in II quadrant, as < 0 and y > Verification : The points (, ), (, ) and (, ) are plotted as shown in figure : Result is verified : (i) (, ) lies in III quadrant. (ii) (, ) lies in IV quadrant. (iii) (, ) lies in II quadrant. Y X' (, ) (II) O Y' (I) (III) (IV) (, ) (, ) X TOPIC- Plotting a Point in a Plane WORKSHEET-0. The co-ordinate of point Q (, ). on ais C, F, G on y ais A, D, E [CSE Marking Scheme, 06]. For plotting a point A(, ), we will take a distance of units in the negative direction of y-ais and a distance of units in the positive direction of -ais, which is shown in the figure given below. Similarly, we plot all the points i.e., ( 6, 0), C(, ) and D(, ). D(, ) ( 6, 0) X' X 6 O 6 A(, ) C(, ) Y' S O L U T I O N S Y. From the graph it is clear that the verte of C(, ) Y (, ) 6 A(, ) X 6 6 X C(, ) 6 D(, ) Y [CSE Marking Scheme, 06] P-

34 . Y D(0, ) 6 C(, ) X A(0, 0) (, 0) X Y [CSE Marking Scheme, 06] 6. (i) Plotting of points M(, ) and N(, ) on the graph paper is shown in the diagram. (ii) Length of MN + 8 units (iii) From figure, A(, ), (, ), C(, ) + + y ' 6 N(, )C 6 8 A M(, ) y' WORKSHEET-. Plotting of P(, 6) y ' 6 M (, 6) N 6 P Co-ordinates of M(, 0) Co-ordinates of N(0, 6). y'. Middle point of line segment A is (0, 0) Y A(, ) X X (, ) O(0, 0). Plotted graph shows that P, Q and R are collinear. Y K(, ) X Q(0, ) P(, 0) X Y [CSE Marking Scheme, 06]. Point of intersection of diagonals is (, ) Y 6 P(, ) S(, ) A(, ) X Q(, ) R(, ) X Y [CSE Marking Scheme, 06] Y [CSE Marking Scheme, 06] P- M A T H E M A T I C S IX

35 . Y (, )C D (, ) A(, ) X' O X 6. Plot the three vertices of the rectangle as A(, ), (, ), C(, ). Y' We have to find the co-ordinates of the fourth verte D so that ACD is a rectangle. Since the opposite sides of a rectangle are equal, so the abscissa of D should be equal to abscissa of A, i.e., and the ordinate of D should be equal to the ordinate of C, i.e.,. Co-ordinates of D are given by (, ). WORKSHEET-. Drawing the points, A(, ), (, ), C(, ) and D(, ). Joining A, C, CA, AD, we get a quadrilateral ACD. y. (,) 6 C ' 6 y' 6 (,) A (,) 6 8 (, ) D After plotting these two points A line segment is formed. + + Mid-point, (, ). The points A(, ), (, 0), C(, ) and D(, 0) are plotted as shown below : (II) C (, ) Y (, 0) D (, 0) X' X 0 (I) X' Y (,) 0 0 A(,) 0 0 O 0 0 X 0 0 Y' (III) (IV) A (, ) Y' (i) A(, ) lies in III quadrant. (ii) (, 0) lies on -ais. (iii) C(, ) lies in II quadrant. (iv) D(, 0) lies on -ais S O L U T I O N S P-

36 . y 6. Y D(, ) C(, ) A( p, 0) O (0, 0) X' ( p) ( q) X ' A(, ) (, ) y' From figure length of line segment A is 6 units.. 6 (,) (,) (, ) (0,.) 6 (6, ) 6 (, ) 6 ( p, q) Y' C (0, q) From figure the co-ordinate of vertices of a rectangle are : TOPIC- Graph of a Linear Equation O(0, 0) A( p, 0) ( p, q) C(0, q) WORKSHEET-. -ais.. Given equation + y 0 Y y +. The given equation is : Putting 0, we get y 0 Putting, we get y Putting, we get y Thus, we have the following table : 0 y 0 Now, plot the points R(0, 0), P(, ) and a(, ) on the graph paper. X' a (, ) (0, 0) R P(, ) Y' X P-6 M A T H E M A T I C S IX

37 . OA O 6 units and C AD units. Co-ordinates of A (, 0) Co-ordinates of (, 0) Co-ordinates of C (, ) Co-ordinates of D (, ).. (i) y ' y' No. of Pairs of Shoes Price in Hundred of Rupees (ii) Graph is a straight line. 6. (i) Plot the points A(, ) and (, ). WORKSHEET-. Given equation, + 6y 0... Line intersect the -ais... y (0) 0 S O L U T I O N S So, the point where line intersect the -ais is (, 0).. Given, y + 6 Put 0, we get y Similarly at, y 9 Points (0, 6), (, 9) lies on the line y The given equation is : y + Putting 0, we get y Putting, we get y Putting, we get y Putting, we get y Thus, we have the following table : 0 y Now Let plot the points A(0, ), (, ), C(, ) and D(, ) on the graph paper. X' Joining these points, we get the line segment A. We can take the point C(, ) between the points A and. (ii) We can take the point D(, ) lying on the A produced. Y X' 8 D(, ) 6 (, ) C(, ) O 6 8 A(, ) 6 Y' Y D(, ) 6 C(, ) (, ) (0, )A 6 X Y'. The given equation is: y Putting 0, we get : y ( ) Putting, we get y Putting, we get y Putting, we get y Thus, we have the following table 0 y Now plot the points A(0, ), (, ), C(, ) and D(, ) on the graph paper. Join A,, C, D and etend it in both the directions. Then, line AD is the graph of the equation y. X P-

38 ' y D(, ) 6 C(, ) (, ) 0 6 A(0, ) y'. (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 A(, 8). Similarly, we plot the point (, ). In order to plot (0,.), we take. units below the -ais on the y-ais to obtain C(0,.). In order to plot (, ), we take unit on OX and then units parallel to OY to obtain the point D(, ) Y A(, 8) 8 (, ) 6 D(, ) X' X C E(, ) (0,.) Y' In order to plot (, ), we take units on OX and then unit below -ais parallel to OY to obtain the point E(, ) (ii) Co-ordinate geometry. (iii) Co-ordination among people is good for progress. P-8 M A T H E M A T I C S IX

39 SECTION CHAPTER INTRODUCTION TO EUCLID S GEOMETRY TOPIC- Euclid s Geometry WORKSHEET-. Theorem requires a proof.. 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. + S O L U T I O N S [CSE Marking Scheme, 0, 0] If equals are added to equals, the wholes are equal. [CSE Marking Scheme, 0]. Given, and Again, Aiom : Things which are equal to the same or equal things are equal to one another... AC DC (Given) C CE Adding, AC + C DC + CE A DE If equals are added to equals, the wholes are equal. WORKSHEET-6. Let, First thing [CSE Marking Scheme, 0] Second thing y then, y. D A C 6. Here, and and. Euclid s first aiom says, the things which are equal to same things are equal to one another. So,.. Since and, therefore adding both equation + + Þ AD CD Þ A C If equals are added to equal, the wholes are equal Two more aioms : Things which are equal to the same thing are equal to one another e.g., if A PQ and PQ XY, then A XY If equals are subtracted from equals, the remainders are equal. e.g., if m m then m m m m [CSE Marking Scheme, 06] A AD AC AD A AC Things which are equal to the same thing are equal to one another. [CSE Marking Scheme 0]. Let kg be the weight of Ram and Ravi each. On gaining kg, weight of Ram and Ravi will be ( + ) kg each. According to Euclid s second aiom, when equals are added to equals, the wholes are equal. So, weights of Ram and Ravi are again equal. P-9

40 . + 0 Þ If equals are subtracted from equals, the remainder are equal.. Here PQ QR P X [CSE Marking Scheme, 06] Q QX QY Y If equals are subtracted from equals, the remainders are also equal We have R PQ QX QR QY Q PX RY [CSE Marking Scheme, 06] 6. A C (given) X Y (given) If equals are subtracted from equals, then remains are also equal. A X C Y AX CY. Here OX XY Also Þ PX XZ OX PX XY XZ XY XZ Things equal to half of equals, are equal to one another. Two other aioms : Things coincide with one another are equal to one another. e.g., if A coincide with XY, such that A falls on X and falls on Y, then A XY The whole is greater than the part e.g., if mð mð + mð, then mð > mð and mð > mð. [CSE Marking Scheme, 06] WORKSHEET-. A 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. [CSE Marking Scheme, 0]. Since, A C AX + X Y + CY Since, X Y AX + X X Y + CY Y \ AX CY Aiom : If equals are subtracted from the equals, the remainders are equal. [CSE Marking Scheme, 0]. 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.....() Adding () + (), we get...() + + AC DC Euclid s aiom used : If equals are added to equals, wholes are equal.. Given, AC AC [CSE Marking Scheme, 0] (As, ). P-0 M A T H E M A T I C S IX

41 TOPIC- Euclid s Postulates WORKSHEET-8. Only one line passes through two distinct points. A. Two planes intersect each other to form a straight line.. (i) Infinite, if they are collinear. (ii) Only one, if they are non-collinear. +. Let A be perpendicular to a line l and AP is any other line segment. In right AP, > P, ( 90 ) AP > A or A < AP.. Playfair s Aiom (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. 6. A C D AC D (given) A + C C + CD A CD. [CSE Marking Scheme 0, 0]. Concermed, Caring. Things equal to same things are equal to one another. Rehman contributed ` 00. All right angles are, equal to one another (OR) Any postulate of Euclid can be stated. 8. Polygon : A simple closed figure made up of three or more line segments. Line Segment : Part of a line with two end points. Angle : A figure formed by two rays with a common initial point Right angle : Angle whose mcahre in 90 [CSE Marking Scheme, 06] WORKSHEET-9. Meeting place of two walls.. A surface is that which has length and breadth. y. A AE S O L U T I O N S Surface breadth ( y) length ( ) (E is the mid-point of A) CD DF (F is the mid-point of CD) Also, AE DF (Given) Therefore, A CD (Things which are double of the same thing are equal to one another). (i) Since it is true for things in any part of universe so this is a universal truth. (ii) If the sum of the cointerior angles made by a transversal intersect two straight lines at distinct points is less than 80, then the lines cannot be parallel.. (i) False : ecause infinitely many lines can pass through a single point. (ii) False : ecause only one line can pass through two distinct points. (iii) True : Two circles are equal if : (a) their circumferences are equal, or (b) their radii are equal. (iv) True : Things which are equal to the same thing are equal to one another (Euclid s aioms). 6. 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-

42 WORKSHEET-0. Dimension of Surface Length and readth (which is ).. Lines are parallel if they do not intersect on being etended. For eample : A or A Lines A and are parallel lines.. 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 A and, there is a point C, lying on the line which is in between them. (ii) Says that given A and, we can take C not lying on the line passing through A and. These Postulates do not follow from Euclid s postulates. However, (ii) follow from given postulate (i).. Euclid s aiom : If C be the mid-point of a line segment A, then AC A. AC A and AD AC AD A AD A. [CSE Marking Scheme, 0]. Given, AC C A C So, AC + AC AC + C (Equals are added to equals) AC A, ( AC + C coincides with A) AC A. 6. Here, OX XY, PX XZ XY (OX), XZ (PX) Also, OX PX, (Given) XY XZ, (ecause things which are double of the same things are equal to one another.). (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. P- M A T H E M A T I C S IX

43 SECTION CHAPTER 6 LINES AND ANGLES TOPIC- Different Types of Angles WORKSHEET ( Straight line makes an angle of 80 ) Thus, COD Let the angle be, then Angle Complement of 90.. ( ) + ( + ) 90 Angles are 60 and 0.. Let the two supplementary angles are and, then Hence, the angles are and or and 08.. ÐPOR + ÐROQ [CSE Marking Scheme, 0] 6 \ ÐPOR, ÐROQ 08 [CSE Marking Scheme, 0, 0] 6. Let normals at A and meet at P. m D C As mirrors are perpendicular to each other there fore P OA and AP O. So P ^ PA, i.e, ÐPA 90 Therefore P Ð + Ð 90 O A...(i) (Angle sum property) Also Ð Ð and Ð Ð (Angle of incidence Angle of reflection) Therefore Ð + Ð 90...(ii) [form (i)] Adding () and (), we have Ð + Ð + Ð + Ð 80 i.e. ÐCA + ÐDA 80 Hence CA D [CSE Marking Scheme, 06] WORKSHEET-. Angles (0 a) and ( + a) are supplementary of each other, then 0 a + + a 80 a 80.. The complement of (90 a) 90 (90 a) (As sum of complementary angles is 90 ) a a.. Let the angle be, then y given condition, (90 ) Let, a and b Then, a + b a 8 6 S O L U T I O N S P-

44 b 8 c 80 b [CSE Marking Scheme, 0]. POC y (Vertically opp. angles) y + y + y 80 y 80 y. [CSE Marking Scheme, 0] 6. ÐCOF (vertically opposite angles) \ Þ 0 80 Þ 8 ÐAOC ÐOF 90 ÐDOE 6 [CSE Marking Scheme, 06] WORKSHEET-. Angle ( + a) and ( a) are supplement of each other, then + a + a 80 a Here, + 80 ( Straight line makes an angle of 80 ) Here, 0 + y + y 80 ( Straight line makes an angle of 80 ) y 0 y 8.. a + b 80 (Linear pair) a b 80 (Given) Adding, a 60 a 0 and b 80 a [CSE Marking Scheme, 0, 00, 0]. Let, AO and OC AO + OC AOC + AO 0 and OC. [CSE Marking Scheme, 0] 6. b + + b 80 [Linear pair] Þ b 0 Þ b \ b 8 [Vertically opposite anlge] a a 8 \ c 80 a c 80 8 c 96 c 8 [CSE Marking Scheme, 06] WORKSHEET Þ 0 So the angles are So two angles are 0 and 0. [CSE Marking Scheme, 06] q 0. (Vertically opposite angles). E C O Etend AO to E, AO + OE 80 (Linear pair)...() EOC + COD + DOA 80 (Adjacent angles)...() D A P- M A T H E M A T I C S IX

45 Adding () and (), we get AO + OE + EOC + COD + DOA AO + OC + COD + DOA 60. [CSE Marking Scheme, 0]. A C 8 O y z D. Given, COA 8 CO + OA 8 CO + 8 ( OA ) CO 8 Similarily, DO DOC + CO 8 DOC + DOC 8 0. [CSE Marking Scheme, 0] TOPIC- Transversal Line E (linear pair) 96 8 y + (VOA) y 8 96 z 8 (VOA) WORKSHEET-. m and n lines are parallel [CSE Marking Scheme, 06] z 0 (V.O.A.) Pair of alternate interior angles. [CSE Marking Scheme, 06]. of a right angle 90 0 (Sum of supplementary angles is 80 ) Supplement of [CSE Marking Scheme, 0] 6. Q P 6 S R A C l m. 0 (V.O.A.) y (Linear pair) y 0 y 0 ut and y are alternate angles. l m. [CSE Marking Scheme, 0, 00, 0] S O L U T I O N S (Corresponding interior angles) y 80 ( 0 ) 80 (0 0 ) y T Given : l m line t is a transversal intersecting them at P and Q rerpectively. To prove : PR QS Proof : Þ Ð Ð6 (Corresponding angles and l m) Ð Ð6 Þ Ð Ð \ PR QS [CSE Marking Scheme, 06] P-

46 WORKSHEET-6. y + y (Corresponding interior angles) (Vertically opposite angles) 0. ÐDS ÐCQ [Conresponding angles] ÐRD 80 ÐDS 80 [CSE Marking Scheme, 06]. 60 (corr S) + 80 WORKSHEET ( 60 ) Let, a, b Then, a h + a 80 (Corresponding eterior angles) h Let, l A CD (by construction) A O 0 I F 0 E C D O' then, OEF + OE 80 O EF + DOE 80 [Corresponding interior angles] OEF + O EF OEO z z + (Alternate interior angles) z + z DNM z A DC CD AD [alternate angles] + y AD y [angle sum property] DC z 80 [ + ] 0 [CSE Marking Scheme, 0] Alternative Method : A DC y (Corresponding interior angles) y 80 y 6 AD CD (Alternate angles) In CD, + y 0 + z 80 z 80 y z 6 z 0. CNM 80 DNM. ÐEFD 60 Þ + y 60 Þ y (Linear pair) 80. [CSE Marking Scheme, 0] Þ ÐEF + ÐEFD 80 [Sum of co-interior angles on same side of transversal is supphymentry] Þ ÐEF Þ ÐEF 0 Þ ÐPEF Þ ÐPEF 80 Now in DPEF Þ [Angle sum property] Þ [CSE Marking Scheme, 06] P-6 M A T H E M A T I C S IX

47 WORKSHEET-8. PQS + QSF 80 (Angles on the same side of transversal) PQS + RFE 80, as QSF EFR 60 + RFE 80 (Corresponding S) RFE 0.. Let, k, y k Then, + y k + k 80 (Angles on the same side of transversal) k 80 k 6 k 08 y k Thus, z 08. (Alternate interior angles). Ða (Linear pair) Ðb 0 (Since Vertically opposite angles are equal) This showes that Ða Ðb ut the are alternate interior angles, \ LM XY [CSE Marking Scheme, 06]. Let DAF CFA, FE, EF Since, AE DC D + 80 (Angles on the same side of transversal) (Alternate angles) Again, (Angle on the same side of transversal) (Vertically opp. angles) D C A 0 9 F In EF, + + EF 80 (Angle sum property) EF 80 EF 80 EF. [CSE Marking Scheme, 0] E TOPIC- Angle Sum Property of a Triangle WORKSHEET-9. Let, equal angles are and, then (Eterior angle is the sum of the two opposite interior angles). Eterior angle Sum of opposite two interior angle other angle ACD A + [Eterior angle is the sum of the two interior opposite angles] S O L U T I O N S ACD 0. ÐCD (Eterior Ð of DAC) ÐED + ÐED [CSE Marking Scheme, 06]. In AC, A + + C 80 (Angle sum prop. of D) A + C ( A + C) P-

48 6.. In AOC, A + C + AOC 80 A WORKSHEET-60 AOC 80. [CSE Marking Scheme, 0] C. We know that (Eterior angle is the sum of the two interior opposite angles) 0.. Given, A + 6 Given, + C 0 A C ut, A + + C 80 (Angle sum prop. of D) C 0 [CSE Marking Scheme, 0, 0] Alternative Method : We know that, A + + C 80...(i) (Angle Sum prop. of D) 6 + C 80 C Again by (i), A A 0 Again by (i), (Angle Sum prop. of D).. In AC, A + + C (Angle Sum prop. of D) A First prove the theorem Let A, & C are the s of Then, A + C + A + + C 80 ( sum prop. of ) C [CSE Marking Scheme, 0] (Linear pair) A a + b + (Eterior angle) +. [CSE Marking Scheme, 0] y z C To prove : Sum of all the angles of AC is 80. Construction : Draw a line l parallel to C. Proof : Since l C, we have y (Alternate angles are equal)... (i) Similarly, l C z (Alternate angles are equal)...(ii) l P-8 M A T H E M A T I C S IX

49 Also, sum of angles at a point A on line l is (linear pair) i.e., y + + z 80 (from (i) and (ii)) + y + z 80 A + + C 80 WORKSHEET-6. Since sum of all the eterior angles formed by producing the sides of a polygon is y + z 60.. Given, z + y z + y + y + z 80 (Angle sum property) z + z 80 z 60.. A 60 0 C C 80 ( ) AC is the smallest side Reason : Side opposite to smaller angle is shortest. [CSE Marking Scheme, 0] WORKSHEET-6. In AC, A + + C 80 (y given conditions) A + A + 6 A 80 9 A 80 A 0.. In AC, A + + C 80 Also, in DEF, D + E + F 80 A + + C + D + E + F Hence, k.. APQ 0 (Alternate angles) + y 8 (Eterior angle is the sum of the two opposite interior angles) 0 + y 8 y S O L U T I O N S Sum of all angles of a is Angles are 0, 60 and 0 respectively. Hence proved. [CSE Marking Scheme, 0]. Since A CD AC DCE (Corresponding angles) DCE CEF + CFE 0 + CFE (Eterior angle) CFE. [CSE Marking Scheme, 0]. y (Corr. angles) y PRQ 80 (8 + ) (Linear pair) ( ) (Angle sum property) [CSE Marking Scheme, 0]. A + C A + + C 80 A 80 A 90. [CSE Marking Scheme, 0] Alternative Method : Given, A + C A + + C 80 A + A 80 A 90. Hence, AC is a right angled triangle.. P Q A C P-9

50 Through verte A, draw PAQ C PA AC QAC AC Adding above equalities, (Alternate angles) (Alternate angles) PA + QAC AC + AC Adding AC to both sides, we get PA + QAC + AC AC + AC + AC 80 AC + AC + AC (Linear pair) [CSE Marking Scheme, 0] 6. P D y y C ACD AC CP DCP y Et APC + y ACP + y ACP APC. [CSE Marking Scheme, 0] A P-0 M A T H E M A T I C S IX

51 SECTION CHAPTER TRIANGLES TOPIC- Criteria for Congruence of Triangles WORKSHEET-6. ASA 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.. A A A, AE AFC 90 AE ACF, E CF AE AFC (ASA) A AC (c.p.c.t) [CSE Marking Scheme, 0] Alternative method : A C F E D Join AD. In AC and ACD, A AC (Given) D CD (Given) AD AD (Common) y using SSS Congruency Rule, AD ACD AD ACD (y c.p.c.t.). In AD and CD, A CD (Given) AD CD (Given) D D (Common) AD CD (y SAS) AD C. (y c.p.c.t.) [CSE Marking Scheme, 0]. In AE & ACF, A F E C S O L U T I O N S D C In CE and CF, EC FC 90 (Given) E CF (Given) C C (Common) CE CF (y RHS) C AC A (y c.p.c.t.) Similarly, AD AE AC C Therefore, A C AC Thus, AC is an equilateral triangle.. Given : D is the mid-point of side AC To Prove : D AC Const : Draw, DE C Proof : In AC, D is the mid point of AC & ED C y mid point theorem, AE E...(i) Now, ED C (y const) AED AC (Corresponding s) AED 90 AED DE 90...() Now, In ADE & DE, AE E (from eq....()) AED DE (from eq....()) ED ED (Common) ADE DE (y SAS rule) P-