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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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