## Book: RS Aggarwal - Mathematics

#### Subject: Maths - Class 10th

##### Q. No. 63 of Exercise 10E

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63
##### The sum of the areas of two squares is 640 m2. If the difference in their perimeters be 64 m, find the sides of the two squares.

Let the length of first and second square be x and y respectively

According to the question;

x2 + y2 = 640 – – – – (1)

Also 4x – 4y = 64

x – y = 16

x = 16 + y

Putting the value of x in(1) we get

(16 + y)2 + y2 = 640 using (a + b)2 = a2 + 2ab + b2

256 + 32y + y2 + y2 = 640

2y2 + 32y – 384 = 0

y2 + 16y – 192 = 0

Using the splitting middle term – the middle term of the general equation is divided in two such values that:

Product = a.c

For the given equation a = 1 b = 16 c = – 192

= 1. – 192 = – 192

And either of their sum or difference = b

= 16

Thus the two terms are 24 and – 8

Difference = 24 – 8 = 16

Product = 24. – 8 = 192

y2 + 24y – 8y – 192 = 0

y(y + 24) – 8(y + 24) = 0

(y + 24) (y – 8) = 0

(y + 24) = 0 (y – 8) = 0

y = 8 or y = – 24

y = 8 (y cannot be negative)

x = 16 + 8 = 24m

Hence the length of first square is 24m and second square is 8m.

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