Question

The sum of the area of two square is 640m2.If the difference in perimeter be 64m find the two sides...​

Answer 1

Let the side of one square be x and the other square be y.

Formula of perimeter of a square:

4×side

So,the perimeters of the both the squares 4x and 4y.

According to the given condition,

4x-4y=64

=>4(x-y)=64

=>x-y=16

=>x=16+y

=>y=x-16...............[1]

Now,

Given that the areas of the squares have a difference of 640m^2.

Area of a square:

side×side=side^2

Hence,

$x {}^{2} - y {}^{2} = 640$

Using [1],

$\implies \: x {}^{2} - (x - 16) {}^{2} = 640 \\ \\ \implies \: x {}^{2} - (x {}^{2} + 256 - 32x) = 640 \\ \\ \implies \: x {}^{2} - x {}^{2} - 256 + 32x = 640 \\ \\ \implies \: 32x = 896 \\ \\ \implies \: x = \frac{896}{32} \\ \\ \implies \: \huge\boxed {x = 28}$

Putting x=32 in [1],

$y = x - 16 \\ \\ \implies \: y = 28- 16 \\ \\ \implies \: \huge \boxed{y = 12}$

Hence,(x,y)=(28,12)