Pls solve Q4

Dear Student,
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,

$$\begin{gathered} x^2+{\left(4+x\right)}^2=400 \\ \Rightarrow x^2+16+x^2+8x=400 \\ \Rightarrow 2x^2+8x-384=0 \\ \Rightarrow x^2+4x-192=0 \\ \Rightarrow x^2+16x-12x-192=0 \\ \Rightarrow x(x+16)-12(x+16)=0 \\ \Rightarrow (x-12)(x+16)=0 \\ \Rightarrow x-12=0 \mathrm{or} x+16=0 \\ \Rightarrow x=12 \mathrm{or} x=-16 \end{gathered}$$

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.

regards