Question

The length of a rectangle is greater than the breadth by 3 cm. If the length increased is by 9 cm and the breadth is reduced by 3 cm, the area remains the same Find the area(in $c{m}^2$) of the rectangle.

Answer 1

Let the breadth of the rectangle be $Xcm$

Then, the length of the rectangle be $(x+3)cm$

$\therefore$ Area of the rectangle $=\left(x+3\right)\times xc{m}^2-----\left(1\right)$

Given that

Length is increased by $9=x+3+9=x+12cm$

and, breadth is reduced by $3=x-3cm$

Now,

Area of new rectangle $\left(x+12\right)\times \left(x-3\right)c{m}^2-----\left(2\right)$

But in the given question we have given that the both area is same

then we can get

$\begin{array}{c}\left(x+3\right)\times x=\left(x+12\right)\times \left(x-3\right)\\ \Rightarrow x^2+3x=x^2+12x-3x-36\\ \Rightarrow 6x=36\\ \therefore x=6\\ Now,fromequation\left(1\right)\\ Areaoftherec\tan gle=\left(x+3\right)\times x\\ =\left(6+3\right)6\\ =54c{m}^2.\end{array}$