Question

The side of a square exceeds the side of the another square by $4cm$ and the sum of the areas of the two squares is $400$ sq.m. Find the dimensions of the squares.

Answer 1

Let $S_1$ and $S_2$ be two squares. Let the side of the square $S_2$ be $x$ cm in length.

Then, the side of square $S_1$ is $(x+4)$cm

Area of square $S_1={(x+4)}^2$
and Area of square $S_2=x^2$

It is given that,
(Area of square $S_1)$+(Area of square $S_2)=400{cm}^2$

$\Rightarrow {(x+4)}^2+x^2=400$

$\Rightarrow (x^2+8x+16)+x=400\Rightarrow 2x^2+8x-384=0\Rightarrow x^2+4x-192=0$

$\Rightarrow x^2+16x-12x-192=0\Rightarrow (x+16)(x-12)=0\Rightarrow x=12orx=-16$

As the length of the side of a square cannot be negative. Therefore, $x=12$

Side of square $S_1=x+4=12+4=16cm$

Side of square $S_2=12cm$