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# The length of a rectangle exceeds its width by 6 m. If its perimeter is 44 m, find its dimensions.

Given:

The length of a rectangle exceeds its width by $6\ m$.

Its perimeter is $44\ m$.

To do:

We have to find its dimensions.

Solution:

Let the width of the rectangle be $x$.

This implies,

The length of the rectangle $=(x+6)\ m$

We know that,

Perimeter of a rectangle $= 2( l + b)$

Therefore,

$2(x+x+6) = 44$

$4x + 12 = 44$

$4x = 44-12$

$x = \frac{32}{4}$

$x=8\ m$

This implies,

Width $=x = 8\ m$

Length $= x+6=8+6 = 14\ m$

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