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If the length of a rectangle is $4$ $4$ cm. more than its width and the perimeter of the rectangle is $28$ $28$ cm., what is the area of the rectangle?

1 Answer

Alan P. · Shwetank Mauria

Feb 22, 2017

$Area = 45 {cm}^{2}$ $"Area" = 45 " cm"^2$

Explanation:

Let the length be $L$ $L$
and the width be $W$ $W$ (with all measurements in cm.)

We are told that $L = W + 4$ $L=W+4$
and
that the perimeter is $28$ $28$

Since the perimeter of a rectangle is $2 (L + W)$ $2(L+W)$
we have
$XXX 2 (L + W) = 28$ $color(white)("XXX")2(L+W)=28$

$XXX 2 (W + 4 + W) = 28$ $color(white)("XXX")2(W+4+W)=28$

$XXX 2 (2 W + 4) = 28$ $color(white)("XXX")2(2W+4)=28$

$XXX (2 W + 4) = 14$ $color(white)("XXX")(2W+4)=14$

$XXX 2 W = 10$ $color(white)("XXX")2W=10$

$XXX W = 5$ $color(white)("XXX")W=5$

and since $L = W + 4$ $L=W+4$
$XXX L = 9$ $color(white)("XXX")L=9$

Finally
$XXX Area = L \times W = 9 \times 5 = 45$ $color(white)("XXX")"Area" = L xx W = 9 xx 5 =45$ (sq.cm.)

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