Question

Sum of area of two squares is $468m^2$. If the difference of their permeter is 24 m, and the sides of the two squares.

Answer 1

Sum of the areas of two squares $=468m^2$

Let $a$ and $b$ be the sides of the two squares.

$\Rightarrow a^2+b^2=468$ .....$\left(1\right)$

Also given that,

the difference of their perimeters $=24m$

$\Rightarrow 4a-4b=24$

$\Rightarrow a-b=6$

$\Rightarrow a=b+6$ .......$\left(2\right)$

We need to find the sides of the two squares.

Substituting the value of $a$ from equation $\left(2\right)$ in equation $\left(1\right)$, we have,

${(b+6)}^2+b^2=468$

$\Rightarrow b^2+12b+36+b^2=468$

$\Rightarrow 2b^2+12b+36-468=0$

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

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

$\Rightarrow b^2+18b-12b-216=0$

$\Rightarrow b(b+18)-12(b+18)=0$

$\Rightarrow (b+18)(b-12)=0$

$\Rightarrow b=-18$ or $12$

Since side cannot be negative.Hence $b=12m$

$\therefore a=b+6=12+6=18m$