Question

the sum of the area of two squares is 640 meter square if the difference in their perimeters be 64m find the sides of the two squares

Answer 1

Let, the side of two two squares are x and y.

Given that : The difference in their perimeters is 64 m

According to the question :

$4x - 4y = 64 \\ \\ = > x - y = 16......(1)$

Also, given that : the sum of their area is 640 m²

Again, according to the question :

${x}^{2} + {y}^{2} = 640 \\ \\ = > {(x - y)}^{2} + 2xy = 640 \\ \\ = > {(16)}^{2} + 2xy = 640 \\ \\ = > 2xy = 640 - 256 \\ \\ = > 2xy = 384 \\ \\ = > xy = 192....(1)$

From equation (2)

$x = \frac{192}{y}$

On putting the value of x in equation (1)

$\frac{192}{y} - y = 16 \\ \\ = > {y}^{2} + 16y - 192 = 0 \\ \\ = > {y}^{2} + 24y - 8y - 192 = 0 \\ \\ = > y(y + 24) - 8(y + 24) = 0 \\ \\ = > (y - 8)(y + 24) = 0$

Required value of y = 8 and -24

The value of y = -24 doesn't exist.

So, the value of y will be 8

On putting the value of y in equation (1)

$x = 16 + 8 \\ \\ = > x = 24$

So, the sides will be 24 and 8