1. Use a suitable identity to get each of the following products
(a) (p - 11) (p + 11) (b) (2a-1/2)(2a-1/2) (c) [(p/8)+(3q/4)] [(p/8)+(3q/4)]
2. If a pen cost Rs. (3a -b) and if you buy (3a + b) pens, how much will you
pay to the shopkeeper?
3. Use the identity to find the area of rectangle whose length is (2x + 5y)
and breadth is (2x + 3y)
4. Evaluate (0.99)
2
5. Find the perimeter of the triangle whose sides are 8x
2 + 7xy – 6y
2
,
4x
2 – 3xy + 2y
2 and -4x
2 + xy –y
2
6. Find the perimeter of a rectangle whose length and breadth are
( x + xy + ) unit and (2 - xy +3 ) unit respectively.
2 z
2 x
2 z
2
7. Using suitable identity find:
(i) 231 -
2 131
2
(ii) 97 X 103
(iii) 181 -
2 19
2
8. Find the value of , using suitable identity.
16
38 22
2
−
2
9. If a polygon has (4x+2y) sides of equal length of (16x-4y) unit, find the
perimeter of the polygon.
10. Using identity (a + b) = +2ab+ , find 16 + 9 if 4x + 3y = 10
2
a
2 b
2
x
2 y
2
and xy = 5.
11. Evaluate (x + y)( x - xy + ) if x =3 and y = 4
Answers
Answer:
a) to find the product of (p - 11) (p + 11)
we use the formula (a-b)(a+b)=a^2-b^2
(p - 11) (p + 11)=p^2-11^2=p^2-121
b) (a-b)(a-b)=a^2+b^2-2ab
(2a-1/2)(2a-1/2)=(2a)^2+(1/2)^2 - 2×2a×(1/2)
=4a^2+(1/4)-2a
c) (a+b)(a+b)=a^2+b^2+2ab
[(p/8)+(3q/4)] [(p/8)+(3q/4)]
=(p/8)^2+(3q/4)^2+2×(p/8)(3q/4)
=(p^2/64)+(9q^2/16)+(3pq/16)
2) a pen cost Rs. 3a-b
and we buy 3a+b pens
then total price =(3a-b)(3a+b)
=(3a)^2+b^2 - 2×3a×b
=9a^2+b^2-6ab Rs/-
i have to pay to shopkeeper 9a^2+b^2-6ab Rs/-
3)area of the rectangle =length×breadth
=(2x+5y)(2x+3y)
[(a+b)(a+d)=a^2+ad+ba+bd]
(2x+5y)(2x+3y)
=(2x)^2+(2x × 3y)+( 5y × 2x)+(5y × 3y)
=4x^2+6xy+10xy+15y^2
=4x^2+16xy+15y^2
4)(0.99) ^2
=(1−0.01) ^2
= 1 ^2 −2×1×0.01+(0.01) ^2
=1−0.02+0.0001
=0.9801
9) if a polygon have n sides with length a
then the perimeter is na
so here n=4x+2y a=(16x-4y)
na=(4x+2y)(16x-4y)
=(4x × 16x)-(4x × 4y)+(2y × 16x)-(2y × 4y)
=64x^2-16xy+32xy-8y^2
=64x^2+16xy-8y^2