Question

Sum of the areas of two squares is 640 m². If the difference of their perimeters is 64
find the sides of the two squares.
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Answer 1

Answer:

ANSWER

Let the sides of the two squares be of a and b.

The according to the problem

a

2

+b

2

=640.........(1).

Again 4(a−b)=64

or, a−b=16.......(2).

Using (2) in (1) we get,

a

2

+(−16+a)

2

=640

or, 2a

2

−32a+256=640

or, 2a

2

−32a−384=0

or, a

2

−16a−192=0

or, (a−24)(a+8)=0

Since a is the side of a square then a=24⇒b=8.

So the sides of the squares be 24 m and 8 m.