# The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units. If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units. Find the dimensions of the rectangle.

Given:

The area of a rectangle gets reduced by 9 square units, if its length is reduced by 5 units and breadth is increased by 3 units.

If we increase the length by 3 units and the breadth by 2 units, the area increases by 67 square units.

To do:

We have to find the dimensions of the rectangle.

Solution:

Let the original length of the rectangle be $l$ and the breadth be $b$.

Area of the original rectangle $=lb$.

In the first case, the length is reduced by 5 units and breadth is increased by 3 units, the area of the rectangle is reduced by 9 square units.

New length $=l-5$

New breadth $=b+3$

The area formed by the new rectangle $=(l-5)(b+3)$ sq. units

According to the question,

$(l-5)(b+3)=lb-9$

$lb-5b+3l-15=lb-9$

$3l-5b=15-9$

$3l-5b=6$.....(i)

In the second case, the length is increased by 3 units and breadth is increased by 2 units, the area is increased by 67 sq. units.

New length $=l+3$

New breadth $=b+2$

The area formed by the new rectangle $=(l+3)(b+2)$ sq. units

According to the question,

$(l+3)(b+2)=lb+67$

$lb+2l+3b+6=lb+67$

$2l+3b=67-6$

$2l+3b=61$.....(ii)

Subtracting $2\times(i)$ from $3\times(ii)$, we get,

$3(2l+3b)-2(3l-5b)=3(61)-2(6)$

$6l-6l+9b+10b=183-12$

$19b=171$

$b=\frac{171}{19}$

$b=9$

$2l+3(9)=61$ (From (ii))

$2l=61-27$

$2l=34$

$l=\frac{34}{2}$

$l=17$

The dimensions of the rectangle are $17$ units (length) and $9$ units (breadth).

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