#### Question

Sum of the area of two squares is 400 cm^{2}. If the difference of their perimeters is 16 cm, find the sides of two squares.

#### Solution

Let the side of the smaller square be *x *cm.

Perimeter of any square = (4 × side of the square)* *cm.

It is given that the difference of the perimeters of two squares is 16 cm.

Then side of the bigger square = \[\frac{16 + 4x}{4} = \left( 4 + x \right)\] cm.

According to the question,

\[x^2 + \left( 4 + x \right)^2 = 400\]

\[ \Rightarrow x^2 + 16 + x^2 + 8x = 400\]

\[ \Rightarrow 2 x^2 + 8x - 384 = 0\]

\[ \Rightarrow x^2 + 4x - 192 = 0\]

\[ \Rightarrow x^2 + 16x - 12x - 192 = 0\]

\[ \Rightarrow x(x + 16) - 12(x + 16) = 0\]

\[ \Rightarrow (x - 12)(x + 16) = 0\]

\[ \Rightarrow x - 12 = 0 \text { or } x + 16 = 0\]

\[ \Rightarrow x = 12 \text { or } x = - 16\]

Since, side of the square cannot be negative

Thus, the side of the smaller square is 12 cm.

and the side of the bigger square is (4 + 12) = 16 cm.