SOLUTION: I need help with this question. The length of a rectangle is 4cm more that 2 times its width. If the area of the rectangle is 61cm^2, find the dimensions of the rectangle to t

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Question 110343: I need help with this question.

The length of a rectangle is 4cm more that 2 times its width. If the area of the rectangle is 61cm^2, find the dimensions of the rectangle to the nearest thousandth.

Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!
the area of a rectangle is:
A+=+L%2AW
given:
A+=+61cm%5E2
and the length=L of a rectangle is 4cm+more that 2+times its width=W, and we can write it this way:
L+=+2W+%2B+4cm............we will substitute this in formula for area
in order to find the dimensions of the rectangle
A+=+L%2AW
61cm%5E2+=+%282W+%2B+4cm%29%2AW
61cm%5E2+=+2W%5E2+%2B+4Wcm.........move 61cm%5E2 to the right
0+=+2W%5E2+%2B+4Wcm+-+61cm%5E2
or
+2W%5E2+%2B+4Wcm+-+61cm%5E2+=+0.............now we can use quadratic formula to solve for W

W%5B1%2C2%5D=%28-b+%2B-+sqrt+%28b%5E2+-4%2Aa%2Ac+%29%29+%2F+%282%2Aa%29
W%5B1%2C2%5D=%28-4+%2B-+sqrt+%284%5E2+-4%2A2%2A%28-61%29+%29%29+%2F+%282%2A2%29
W%5B1%2C2%5D=%28-4+%2B-+sqrt+%2816+%2B+488+%29%29%2F4
W%5B1%2C2%5D=%28-4+%2B-+sqrt+%28504+%29%29%2F4
W%5B1%2C2%5D=%28-4+%2B-+%2822.449%29%29%2F4
then:
W%5B1%5D=%28-4+%2B+%2822.449%29%29%2F4
W%5B1%5D=%2818.449%29%2F4
W%5B1%5D=+4.612cm

W%5B2%5D=%28-4+-+%2822.449%29%29%2F4..........we don't need negative value, lenght is positive value

so W+=+4.612cm
now find L

L+=+2W+%2B+4cm
L+=+2%284.612%29cm+%2B+4cm
L+=%289.224%29cm+%2B+4cm
L+=13.224cm