S.chand publication New Learning Composite mathematics solution of class 8 Chapter 5 Algebraic Expressions Exercise 5H

 Exercise 5H

Question 1

(a) a + 4

Sol :

=(a+4)²=a²+2.a.4+4²

=a²+8a+16

(b) 3a + 1

Sol :

=(3a)²+2.3a.1+1²

=9a²+6a+1

(c) 5m + 3

Sol :

=(5m)²+2.5a.3+3²

=25m²+30m+9

(d) 6x + 7y

Sol :

=(6x)²+2.6x.7y+(7y)²

=(36)²m+84xy+49y²

(e) 3ab+2c

Sol :

=(3ab)²+2.3ab.2c+(2c)²

=9a²b²+12abc+4c²

(f) 2xy+7z²

Sol :

=(2xy)²+2.2xy.7z+(7z)²

=4x²y²+28xyz+49z²

(g) m² + n²

Sol :

=(m²)²+2.m²n²+(n²)²

=m⁴+2m²n²+x⁴

(h) x²y + 3

Sol :

=(x²y)²+2.x²y.3+3²

=x⁴y²+6x²y+9

Question 2

(a) k -7

Sol :

=k²-2.k.7+7²

=k²-14k+49

(b) 5y – 1

Sol :

=(5y)²-2.5y.1+1²

=25y²-10y+1

(c) 3x – 8

Sol :

=(3x)²-2.3x.8+8²

=9x²-48x+64

(d) 6b – 7c

Sol :

=(6b)²-2.6b.7c+(7c)²

=36b²-84bc+49c²

(e) 5mn – 2p

Sol :

=(5mn)²-2.5mn.2p+(2p)²

=25m²n²-20mnp+4p²

(f) 3x² – 4z

Sol :

=(3x)²-2.3x.4z+(4z)²

=9x²-24xz+16z²

(g) 5x³ – 3y²

Sol :

=(5x³)²-2.5x³.3y²+(3y²)²

=25x⁶-30x³y²+9y⁴

(h) 9a² – bc

Sol :

=(9a²)²-2.9a².bc+(bc)²

=81a⁴-18a²bc+b²c²

Question 3

Find the product.

(a) (x + 6) (x – 6)

Sol :

=x²-6²=x²-36

(b) (p + 9) (p – 9)

Sol :

=p²-9²=p²-81

(c) (4y + 7) (4y – 7)

Sol :

=(4y)²-7²=16y²-49

(d) (6x + 7y) (6x – 7y)

Sol :

=(6x)²-(7y)²

=36x²-49y²

(e) (3x² – 5y) (3x² + 5y)

Sol :

=(3x²)²-(5y)²

=9x⁴-25y²

(f) (y – x²) (x² + y)

Sol :

=(y-x²)(y+x)²

=y²-x⁴

(g) (x² – y²) (x² + y²)

Sol :

=(x²)²-(y²)²

=x⁴-y⁴

(h) (a² + bc) (a² – bc)

Sol :

=(a²)²-(bc)²

=a²-b²c²

Question 4

Using the identity for square for a binomial, evaluate the following.

(i) (105)²

Sol :

=(100+5)²

=(100)²+2.100.5+5²

=10000+1000+25

=11025

(ii) (401)²

Sol :

=(400+1)²

=(400)²+2.400.1+1²

=160000+800+1

=160801

(iii) (708)²

Sol :

=(700+8)²

=(700)²+2.700.8+64²

=490000+11200+64

=501264

(iv) 98²

Sol :

=(90+8)²

=(90)²+2.90.8+8²

=8100+1440+64

=9604

(v) (199)²

Sol :

=(200-1)²

=200²-2.200.1+1²

=40000+400+1

=40401

(vi) (10.3)²

Sol :

=(10+0.3)²

=(10)²+2.10.0.3+(0.3)²

=100+6.0+0.09

=106.09

(vii) (1.99)²

Sol :

=(1+0.99)²

=1²+2.1.(0.99)+(0.99)²

=1+1.98+0.9801

=3.9601

(viii) (2.1)² + (1.9)²

Sol :

=4.41+3.61

=8.02

Question 5

Using the identity for difference of two squares, evaluate the following.

(a) 18 x 22

Sol :

=(20-2)(20+2)

=(20)²-(2)²

=400-4

=396

(b) 51 x 49

Sol :

=(50+1)(50-1)

=(50)²-1²

=2500-1

=2499

(c) 101 x 99

Sol :

=(100+1)(100-1)

=(100)²-1²

=10000-1

=9999

Question 6

Evaluate:

(a) 298 x 298 – 205 x 205 / 93

Sol :

$=\frac{(285)^2-(205)^2}{93}$

$=\frac{(298+205)(298-205)}{93}$

$=\frac{503\times 93}{93}$

=503

(b) 19.67 x 19.67 – 15.33 x 15.33 / 0.434

Sol :
$=\frac{(19.67)^2-(15.33)^2}{0.434}$

$=\frac{(19.67+15.33)(19.67-15.33)}{0.434}$

$=\frac{35.00\times 4.34}{0.434}$

=350

Question 7

Write a polynomial for the total area of the figure.

Sol :

Total area=Area of large square+Area of small square

=(x+4)²+(y-1)²
=x²+2.x.4+4²+y²-2.y.1+1²

=x²+8x+y²-2y+17sq unit

Question 8

What is the area of the swimming pool?

Sol :

Total area=Area of large rectangle+Area of small rectangle

={(7+x)(7-x)}+x²

=7²-x²+x²

=49 sq unit