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.

Given:

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.

To do:

We have to find the area 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 and breadth reduced each by 2 units, the area is reduced by 28 square units.

New length $=l+2$

New breadth $=b-2$

The area formed by the new rectangle $=(l+2)(b-2)$

According to the question,

$(l+2)(b-2)=lb-28$

$lb-2l+2b-4=lb-28$

$2l-2b=28-4$

$2(l-b)=24$

$l-b=12$.....(i)

In the second case, the length is reduced by 1 unit and the breadth increased by 2 units, the area increases by 33 square units.

New length $=l-1$

New breadth $=b+2$

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

According to the question,

$(l-1)(b+2)=lb+33$

$lb+2l-b-2=lb+33$

$2l-b=33+2$

$2l-b=35$.....(ii)

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

$2l-b-(l-b)=35-12$

$2l-l-b+b=23$

$l=23$ units

$23-b=12$    (From (i))

$b=23-12$

$b=11$ units

The area of the original rectangle $=lb$

$=23\times11$

$=253$ square units.

The area of the rectangle is $253$ square units.

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Updated on: 10-Oct-2022

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