Question

Sum of the areas of two square is $468m^2$. If the difference of their perimeter is $24m$, find the sides of the two squares.

Answer 1

Let side of square $1$ be $x$metres.

Perimeter of square$1=4\times$side$=4x$

Given: Difference of perimeter of squares$=24$

$4x-$Perimeter of square $2=24$

$\Rightarrow$Perimeter of square $2=4x-24$

$\Rightarrow 4\times$ side of square $2=4x-24$

side of square $2=\frac{4(x-6)}{4}=x-6$

Hence, side of square$1$ is $x$ and side of square$2$ is $x-6$

Given: Sum of area of square$=468m^2$

$\Rightarrow$ Area of square$1+$Area of square$2=468m^2$

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

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

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

or $x^2-6x-216=0$ is of the form $a{x}^2+bx+c=0$ where $a=1,b=-6,c=-216$

$D=b^2-4ac={(-6)}^2-4\times 1\times -216=900$

So, the roots are $x=\frac{-b\pm \sqrt{D}}{2a}=\frac{-(-6)\pm \sqrt{900}}{2}=\frac{6\pm 30}{2}=18,-12$

Since $x$ is the side of the square and $x$cannot be negative.

So, $x=18$ is the solution.

$\therefore$ Side of square$1=x=18$m

Side of square $2=x-6=18-6=12$m