The length of a rectangle exceeds its width by 3 m. If the width is increased by 4 m and the length is decreased by 6 m, the area is decreased by 22 sq. m. Find the dimensions of the rectangle.

Given :

The length of a rectangle exceeds its width by 3m.

If the width is increased by 4 m and the length is decreased by 6 m, the area is decreased by 22 sq. m.

To do :

We have to find the dimensions of the rectangle.

Solution :

Let the width of the rectangle be 'w'm.

The length of the rectangle $= (w+3)m$

We know that,

The area of a rectangle of length l and breadth b is $l \times b$.

Therefore,

Area of the given rectangle $= w \times (w+3) m^2$.

If the width is increased by 4m and the length is decreased by 6m, the area is decreased by 22 sq.m.

The length of the new rectangle $= (w+3)-6 m = (w+3-6)m =(w-3)m$

The breadth of the new rectangle $= (w+4) m$

The area of the new rectangle $= (w-3) \times (w+4) sq.m.$

Therefore,

$w \times (w+3) - (w-3) \times (w+4) sq.m. = 22 sq.m.$

$w^2+3w-(w^2+4w-3w-12) sq.m. = 22 sq.m.$

$w^2+3w-(w^2+w-12) sq.m. = 22 sq.m.$

$w^2+3w-w^2-w+12 sq.m. = 22 sq.m.$

$2w+12 = 22$

$2w = 22-12$

$2w = 10$

$w = \frac{10}{2}$

$w = 5 m$

The breadth of the original rectangle $= 5 m$.

The length of the original rectangle $= (5+3) m = 8m$.

Therefore, the length and breadth of the rectangle are 5 m and 8 m respectively.

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