In a rectangle, if the length is increased by 3 metres and breadth is decreased by 4 metres, the area of the rectangle is reduced by 67 square metres. If length is reduced by 1 metre and breadth is increased by 4 metres, the area is increased by 89 sq. metres. Find the dimensions of the rectangle.

Given:

In a rectangle, if the length is increased by 3 metres and breadth is decreased by 4 metres, the area of the rectangle is reduced by 67 square metres. If length is reduced by 1 metre and breadth is increased by 4 metres, the area is increased by 89 sq. metres.

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 increased by 3 metres and breadth is decreased by 4 metres, the area of the rectangle is reduced by 67 square metres. 

New length $=l+3$ m

New breadth $=b-4$ m

The area formed by the new rectangle $=(l+3)(b-4)\ m^2$

According to the question,

$(l+3)(b-4)=lb-67$

$lb-4l+3b-12=lb-67$

$4l-3b=67-12$

$4l-3b=55$.....(i)

In the second case, the length is reduced by 1 metre and breadth is increased by 4 metres, the area is increased by 89 sq. metres.

New length $=l-1$ m

New breadth $=b+4$ m

The area formed by the new rectangle $=(l-1)(b+4)\ m^2$

According to the question,

$(l-1)(b+4)=lb+89$

$lb+4l-b-4=lb+89$

$4l-b=89+4$

$4l-b=93$.....(ii)

Subtracting (i) from (ii), we get,

$4l-b-(4l-3b)=93-55$

$4l-4l-b+3b=38$

$2b=38$

$b=19\ m$

$4l-19=93$    (From (ii))

$4l=93+19$

$4l=112$

$l=\frac{112}{4}$

$l=28\ m$

The dimensions of the rectangle are $28\ m$ (length) and $19\ m$ (breadth). 

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Updated on: 10-Oct-2022

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