Question

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

If the difference of their perimeters is $24m$, find the sides of the two squares.

Answer 1

Step 1: Let the sides of the first and second square be $X$ and $Y$

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

side

Area of the first square $=\left(X\right)²$

Area of the second square $=\left(Y\right)²$

According to question, $\left(X\right)²+\left(Y\right)²=468m²$ ——(1).

Perimeter of first square $=4\times X$ and Perimeter of second square $=4\times Y$

Step 2: According to question

$4X–4Y=24$ ——–(2)

From equation (2) we get,

$\begin{array}{rcl}4X–4Y& =& 24\\ 4(X-Y)& =& 24\\ X–Y& =& \frac{24}{4}\\ X–Y& =& 6\\ X& =& 6+Y...........\left(3\right)\end{array}$

Step 3: Putting the value of $X$ in equation (1)

$\begin{array}{rcl}\left(X\right)²+\left(Y\right)²& =& 468\\ (6+Y)²+\left(Y\right)²& =& 468\\ \left(6\right)²+\left(Y\right)²+2\times 6\times Y+\left(Y\right)²& =& 468\\ 36+Y²+12Y+Y²& =& 468\\ 2Y²+12Y–468+36& =& 0\\ 2Y²+12Y-432& =& 0\\ 2(Y²+6Y–216)& =& 0\\ Y²+6Y–216& =& 0\\ Y²+18Y–12Y-216& =& 0\\ Y(Y+18)–12(Y+18)& =& 0\\ (Y+18)(Y-12)& =& 0\\ (Y+18)& =& 0\\ Or(Y-12)& =& 0\\ Y& =& -18\\ orY& =& 12\end{array}$

Step 4: Putting $Y=12$ in equation (3)

$$\begin{gathered} X=6+Y \\ X=6+12 \\ X=18 \end{gathered}$$

Side of first square $=X=18m$

Side of second square $=Y=12m$.

Hence, the sides of the two squares are $18m$ and $12m$.