#### Question

Sum of the areas of two squares is 640 m^{2}. If the difference of their perimeters is 64 m. Find the sides of the two squares.

#### Solution

Let the sides of the squares are *x *m and = y m.Then

According to question,

Sum of the difference of their perimeter=64 m

4x - 4y = 64

x - y = 16

y = x - 16 .................... (1)

And sum of the areas of square = 640 m^{2}

x^{2} + y^{2} = 640 ............ (2)

Putting the value of *x* in equation (2) from equation (1)

x^{2} + (x - 16)2 = 640

x^{2} + x^{2} - 32x + 256 = 640

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

2x^{2} - 32x - 384 = 0

2(x^{2} - 16x - 192) = 0

x^{2} - 16x - 192 = 0

x^{2} - 24x + 8x - 192 = 0

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

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

x - 24 = 0

x = 24

or

x + 8 = 0

x = -8

Sides of the square never are negative.

Therefore, putting the value of *x* in equation (1)

y = x - 16 = 24 - 16 = 8

Hence, sides of the square be 24m and 8m respectively.