The area of a rectangle gets reduced by 9 square units if its length is reduced by 5 minutes and breadth is increased by 3 units if we increase the length by 3 units if we increase a length by three years and the breadth by 2 units the area increases 67 square then find the dimension of the rectangle. (By elimination method)

Academic Mathematics NCERT Class 10

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,

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

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

$3l-5b=15-9$

$3l-5b=6$

$3l=6+5b$

$l=\frac{6+5b}{3}$.....(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)

Substituting the value of $l=\frac{6+5b}{3}$ in (ii), we get,

$2(\frac{6+5b}{3})+3b=61$

Multiplying by $3$ on both sides, we get,

$3\times2(\frac{6+5b}{3})+3\times3b=3\times61$

$2(6+5b)+9b=183$

$12+10b+9b=183$

$19b=183-12$

$19b=171$

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

$b=9$

Substituting $b=9$ in equation (i), we get,

$l=\frac{6+5\times9}{3}$

$l=\frac{6+45}{3}$

$l=\frac{51}{3}$

$l=17$

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

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Updated on: 10-Oct-2022

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