# DAV Class 8 Maths Chapter 7 Worksheet 1 Solutions

The DAV Class 8 Maths Solutions and DAV Class 8 Maths Chapter 7 Worksheet 1 Solutions of Algebraic Identities offer comprehensive answers to textbook questions.

## DAV Class 8 Maths Ch 7 WS 1 Solutions

Question 1.
Find the following by using identity – I:

(i) (2x + 5)2
Solution:
(2x + 5)2 = (2x)2 + 2 × 2x × 5 + (5)2
= 4x2 + 20x + 25

(ii) (8x + 3y)2
Solution:
(8x + 3y)2 = (8x)2 + 2 × 8x × 3y + (3y)2
= 64x2 + 48xy + 9y2

(iii) $$\left(\frac{3}{5} a+\frac{2}{3} b\right)^2$$
Solution:
$$\left(\frac{3}{5} a+\frac{2}{3} b\right)^2$$ = $$\left(\frac{3}{5} a\right)^2+2\left(\frac{3}{5} a\right)\left(\frac{2}{3} b\right)+\left(\frac{2}{3} b\right)^2$$
= $$\frac{9}{25} a^2+\frac{4}{5} a b+\frac{4}{9} b^2$$

(iv) (7pq + 4ab)2
Solution:
(7pq + 4ab)2 = (7pq)2 + 2 (7pq) (4ab) + (4ab)2
= 49p2q2 + 56pqab + 16a2b2

(v) (0.2x + 1.5y)2
Solution:
(0.2x + 1.5y)2 = (0.2x)2 + 2 (0.2x) (1.5y) + (1.5y)2
= 0.04x2 + 0.6xy + 2.25y2

(vi) (2m2 + 3n2)2
Solution:
(2m2 + 3n2)2 = (2m2)2 + 2 × 2m2 3n2 + (3n2)2
= 4m4 + 12m2n2 + 9x4

Question 2.
Evaluate the following by using identity—I:

(i) (101)2
Solution:
(101)2 = (100 + 1)2
= (100)2 + 2 × 100 × 1 + (1)2
= 10000 + 200 + 1
= 10201

(ii) (52)2
Solution:
(52)2 = (50 + 2)2
= (50)2 + 2 × 50 × 2 + (2)2
= 2500 + 200 + 4
= 2704

(iii) (8.1)2
Solution:
(8.1)2 = (8 + 0.1)2
= (8)2 + 2 × 8 × 0.1 + (0.1)2
= 64 + 1.6 + 0.01
= 65.61.

(iv) (203)2
Solution:
(203)2 = (200 + 3)2
= (200)2 + 2 × 200 × 3 + (3)2
= 40000 + 1200 + 9
= 41209.

(v) (410)2
Solution:
(410)2 = (400 + 10)2
= (400)2 + 2 × 400 × 10 + (10)2
= 160000 + 8000 + 100
= 168100

(vi) (10.2)2
Solution:
(10.2)2 = (10 + 0.2)2
= (10)2 + 2 × 10 × 0.2 + (0.2)2
= 100 + 4 + 0.04
= 104.04

### DAV Class 8 Maths Chapter 7 Worksheet 1 Notes

Important Identities:

(I) (a + b)2 = a2 + 2ab + b2
(II) (a – b)2 = a2 – 2ab + b2
(III) (a2 – b2) = (a + b)(a – b)
(IV) (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
(V) (x + a) (x + b) = x2 + (a + b) x + ab

Example 1.
Evaluate: (3x + 4y)2.
Solution:
(3x + 4y)2 = (3x)2 + 2 × 3x × 4y + (4y)2
= 9x2 + 24xy + 16y2

Example 2.
Evaluate: (2x + $$\frac{3}{2 x}$$)2.
Solution:
(2x + $$\frac{3}{2 x}$$)2 = (2x)2 + 2 × 2x × $$\frac{3}{2 x}$$ + ($$\frac{3}{2 x}$$)2
= 4x2 + 6 + $$\frac{9}{4 x^2}$$

Example 3.
Evaluate: (102)2 using the identity.
Solution:
(102)2 = (100 + 2)2
= (100)2 + 2 × 100 × 2 + (2)2
[Using (a + b)2 = a2 + 2ab + b2]
= 10000 + 400 + 4
= 10404.

Example 4.
Evaluate: (20.1)2 using the identity.
Solution:
(20.1)2 = (20 + 0.1)2
= (20)2 + 2 × 20 × 0.1 + (0.1)2
[Using (a + b)2 = a2 + 2ab + b2]
= 400 + 4 + 0.01
= 404.01.

Example 5.
Evaluate: (5x – 4y)2.
Solution:
(5x – 4y)2 = (5x)2 – 2 × 5x × 4y + (4y)2
= 25x2 – 40xy + 16y2.

Example 6.
Evaluate: $$\left(\frac{3}{2 x}-\frac{5 x}{4}\right)^2$$.
Solution:
$$\left(\frac{3}{2 x}-\frac{5 x}{4}\right)^2$$
= $$\left(\frac{3}{2 x}\right)^2-2 \times \frac{3}{2 x} \times \frac{5 x}{4}+\left(\frac{5 x}{4}\right)^2$$
= $$\frac{9}{4 x^2}-\frac{15}{4}+\frac{25}{16} x^2$$

Example 7.
Evaluate: (99)2 using identity.
Solution:
(99)2 = (100 – 1)2
= (100)2 – 2 × 100 × 1 + (1)2
[Using (a – b)2 = a2 – 2ab + b2]
= 10000 – 200 + 1
= 10001 – 200
= 9801.

Example 8.
Evaluate: (3pq – 4rs)2.
Solution:
(3pq – 4rs)2 = (3pq)2 – 2 × 3pq × 4rs + (4rs)2
= 9p2q2 – 24pqrs + 16r2s2.

Example 9.
Find the value of (3x – 5y) (3x + 5y).
Solution:
(3x – 5y) (3x + 5y) = (3x)2 – (5y)2
[Using (a + b) (a – b) = a2 – b2]
= 9x2 – 25y2.

Example 10.
Evaluate: 2472 – 1532 using identity.
Solution:
2472 – 1532 = (247 + 153) (247 – 153)
[Using (a2 – b2) = (a + b)(a – b)]
= 400 × 94
= 37600.

Example 11.
Simplify: 408 × 392 using identity.
Solution:
408 × 392 = (400 + 8) (400 8)
= (400)2 – (8)2
= 160000 – 64
= 159936.

Example 12.
Evaluate: (5x + 3y + 2z)2.
Solution:
(5x + 3y + 2z)2 = (5x)2 + (3y)2 + (2z)2 + 2 × 5x × 3y + 2 × 3y × 2z + 2 × 5x × 2z
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 25x2 + 9y2 + 4z2 + 30xy + 12yz + 20zx.

Example 13.
Evaluate: (2x – 5y – 4z)2.
Solution:
(2x – 5y – 4z)2 = (2x)2 + (- 5y)2 + (- 4z)2 + 2 (2x) (- 5y) + 2 (- 5y) (- 4z) + 2 (2x) (- 4z)
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= 4x2 + 25y2 + 16z2 – 20xy + 40yz – 16zx.

Example 14.
Find the product of the following:
(i) (x + 3)(x + 4)
(ii) (x – 5)(x – 7)
(iii) (x – 2)(x + 3)
Solution:
(i) (x + 3) (x + 4) = x2 + (3 + 4) x + 3 × 4
= x2 + 7x + 12

(ii) (x – 5) (x – 7) = x2 + (- b – 7) x + (- 5) (- 7)
= x2 – 12x + 35

(iii) (x – 2) (x + 3) = x2 + (- 2 + 3) x + (- 2) (3)
= x2 + x – 6.

Example 15.
Evaluate: 203 × 205 using identity.
Solution:
203 × 205 = (200 + 3) × (200 + 5)
= (200)2 + (3 + 5) × 200 + 3 × 5
[Using (x + a) (x + b) = x2 + (a + b) x + ab]
= 40000 + 8 × 200 + 15
= 40000 + 1600 + 15
= 41615.

Example 16.
Evaluate: (x2 – $$\frac{1}{6}$$) (x2 + 7) using identity.
Solution:
(x2 – $$\frac{1}{6}$$) (x2 + 7) = (x2)2 + ($$\frac{-1}{6}$$ + 7) x2 + ($$\frac{-1}{6}$$) (7)
= x4 + $$\frac{41}{6}$$ x2 – $$\frac{7}{6}$$.
[Using (x + a) (x + b) = x2 + (a + b) x + ab]

Example 17.
Factorize the following:
(i) 121x2 + 16y2 – 88xy
(ii) x2 + 4y2 + 9z2 + 4xy + 12yz + 6zx.
Solution:
(i) 121x2 + 16y2 – 88xy .
= (11x)2 + (4y)2 – 2 × 11x × 4y
= (11x – 4y)2
[Using a2 + b2 – 2ab = (a – b)2]
= (11x – 4y) (11x – 4y).

(ii) x2 + 4y2 + 9z2 + 4zy + 12yz + 6zx
= (x)2 + (2y)2 + (3z)2 + 2 × x × 2y + 2 × 2y × 3z + 2 × x × 3z
= (x + 2y + 3z)2
[Using (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca]
= (x + 2y + 3z) (x2 + 2y + 3z)

Example 18.
Factorize the following:
(i) x2 – x – 42
(ii) x2 + 22x + 85.
Solution:
(i) x2 – x – 42 = x2 – (7 – 6) x – 42
= x2 – 7x + 6x – 42
= x (x – 7) + 6 (x – 7)
= (x + 6) (x – 7)

(ii) x2 + 22x + 85 = x2 + (17 + 5)x + 85
= x2 + 17x + 5x + 85
= x (x + 17) + 5 (x + 17)
= (x2 + 17) (x + 5).