Find the area of the rectangle whose length is 2 more that twice the width and the perimeter is 22

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First, we have to solve for the length and the width.  We can do this by setting up the following equations.

l = 2w + 2

2l + 2w = 22

2 (2w +2) + 2w = 22

4w + 4 + 2w = 22

6w + 4 = 22

6w = 18

w = 3

If w = 3, then the length of the rectangle is 2 more than twice that.  That means that length = 2*3 + 2 or l = 6 + 2 or l = 8

So the width is 3 and the length is 8.  To get the area, we multiply these and we get 24.

So the area is 24.

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Area of the rectangle is:

a = length * width = L*w

But we know that:

L = 2+ 2w

Also:

2L + 2W = 22

Now substitute with L value:

==> 2L + 2W = 22

==> 2(2+2w) + 2w = 22

==> 4 + 4w + 2w = 22

==> 6w = 18

==> w = 3

==> L = 2+ 2*3 = 8

==> a = L*w = 3*8 = 24

==> the area of the rectangle is 24 square unit.

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