# If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units.

Question:

If in a rectangle, the length is increased and breadth reduced each by 2 units, the area is reduced by 28 square units. If, however the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units. Find the area of the rectangle.

Solution:

Let the length and breadth of the rectangle be  and  units respectively

Then, area of rectangle  square units

If length is increased and breadth reduced each by  units, then the area is reduced by square units

\begin{aligned} &(x+2)(y-2)=x y-28 \\ \Rightarrow & x y-2 x+2 y-4=x y-28 \\ \Rightarrow &-2 x+2 y-4+28=0 \\ \Rightarrow &-2 x+2 y+24=0 \\ \Rightarrow & 2 x-2 y-24=0 \end{aligned}

Therefore,

Then the length is reduced by  unit and breadth is increased by units then the area is increased by  square units

$(x-1)(y+2)=x y+33$

$\Rightarrow x y+2 x-y-2=x y+33$

$\Rightarrow 2 x-y-2-33=0$

$\Rightarrow 2 x-y-35=0$

Therefore, $2 x-y-35=0 \quad \ldots \ldots(i i)$

Thus we get the following system of linear equation

$2 x-2 y-24=0$

$2 x-y-35=0$

By using cross multiplication, we have

$\frac{x}{(-2 \times-35)-(-1 \times-24)}=\frac{y}{(2 \times-35)-(2 \times-24)}=\frac{1}{(2 \times-1)-(2 \times-2)}$

$\frac{x}{70-24}=\frac{-y}{-70+48}=\frac{1}{-2+4}$

$\frac{x}{46}=\frac{-y}{-22}=\frac{1}{2}$

$x=\frac{46}{2}$

$x=23$

and

$y=\frac{22}{2}$

$y=11$

The length of rectangle is units.

The breadth of rectangle is units.

$=x \times y$
$=23 \times 11$
$=253$ square units