Question

The side of one square is $3cm$ longer than the twice of the second square the difference in their areas is $144c{m}^2.$ Find the sides of the two squares.

Answer 1

Let $x=$ the length of the side of the first square

And, $y=$ the length of the side of the second square

the side of one square is $3cm$ longer than twice the side of the second square

$x=3+2y\dots \dots \left(i\right)$

the difference in their areas is $144c{m}^2$

$x^2-y^2=144$ [$\because$ Area of square $=(Side{)}^2$]

on putting the value of $x$ from $eq\left(i\right)$ in the above equation, we get

$(3+2y{)}^2-y^2=144$

$\Rightarrow 9+4y^2+2\times 3\times 2y-y^2=144$

$\Rightarrow 3y^2+12y+9=144$

$\Rightarrow y^2+4y+3=\frac{144}{3}$

$\Rightarrow y^2+4y+3=48$

$\Rightarrow y^2+4y+3-48=0$

$\Rightarrow y^2+9y-5y-45=0$

$\Rightarrow y(y+9)-5(y+9)=0$

$\Rightarrow (y+9)(y-5)=0$

$\Rightarrow (y+9)=0or,(y-5)=0$

$\therefore y=5$ (since the length of the side cannot be a negative number)

so, $y=5$

$x=3+2\times 5=13$

Hence, the sides of the two squares are $5cm$ and $13cm.$