Question

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

Answer 1

Step-by-step explanation:

Let both squares sides are x and y respectively.

Let area of both squares are x^2 and y^2.

Perimeter of both squares are 4x and 4y.

A/Q,

x^2 + y^2 = 640

and, 4(x-y) = 64 => (x-y) = 16....(1)

Now, Square both side,

(x-y) ^2 = 256......(2)

=> x^2 + y^2 - 2xy = 256

=> 640 - 2 xy = 256

=> 2xy = 640 - 256 = 384

=> xy = 192 => y = 192/x .... (3)

Now, from eqn (1) &(2) ,we have,

x - 192/x = 16

=> x^2 - 16x - 192 = 0

=> x = [ 16 + √ (256 + 768) ] / 2 = (16+32) / 2 = 24 m

Or

x = (16-32)/2 = -8m

but side can't be negative. So x = 24 m

then, y = 192/24 = 8m