ICSE & CBSE Board Question Based on Linear Equations in two Variables of RS Aggarwal

In this question following points are given.

(i)The sum of the areas of two squares

(ii) the difference of their perimeters

You have to find the sides of the squares.

This is the Question Number 51 Of Exercise 3E of RS Aggarwal Solution.

# The sum of the areas of two squares is 640m^2. If the difference of their perimeters is 64m, find the sides of the squares.

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Let the side of one square be x

Perimeter of this square = 4x

Given, Difference of perimeter of 2 squares = 64 m

Thus, Perimeter of the other square = (64 + 4x) m

And, each side of this second square = = (16 +x) m

According to the problem, sum of the areas of two squares is 640

x^2+(16+x)^2=640

x^2+256+32x+x^2=640

2x^2+256+32x-640=0

2x^2+32x-384=0

x^2+16x-192=0

x(x+24)-8(x+24)=0

(x+24)(x-8)=0

x=-24,8

Since side of a square cannot be negative, each side of the square = 8 m.

And, each side of second square = (16+8) m = 24 m