Question

Sum of the areas of two squares is 468 m${}^2$. If the difference of their perimeters is 24 m, formulate the quadratic equation to find the sides of the two squares.

Answer 1

Given that,

The sum of the areas of two squares is $468m^2$.

The difference of their perimeters is $24m$

Let the length of each side of one of the squares be $xm$.

Then, its perimeter will be $4xm$.

It is given that the difference of the perimeters of two squares is $24m$
Perimeter of second square$=(24+4x)m$

Hence, length of each side of the second square $=\frac{24+4x}{4}m$

$=(6+x)m$

We know that, area of a square $=side\times side$

Hence, area of the first square $=x^2{m}^2$

And, area of the second square $=(6+x{)}^2{m}^2$

Sum of their areas is $468m^2$

$\therefore x^2+{(6+x)}^2=468$

$\Rightarrow x^2+(36+12x+x^2)=468[\because (a+b{)}^2=a^2+2ab+b^2]$

$\Rightarrow 2x^2+12x-432=0$

$\Rightarrow x^2+6x-216=0$

Hence, $x^2+6x-216=0$ is the required quadratic equation.