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2 August, 08:38

Henry knows that the area of a rectangle is 30 square inches the perimeter is 12 inches if the length is 1 inch longer than the width what are the length and width Henry's rectangle explain how you know

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  1. 2 August, 08:52
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    What you typed is impossible, the greatest possible area for a four sided polygon is a square. If the perimeter is 12, the sides of the square would be 3 inches. The area of that square would be 9 in^2.

    Work through this and see that it is impossible.

    P=2 (W+L), we are told that P=12 so

    12=2 (W+L)

    6 = (W+L) and we are told that L=W+1 so

    6=2W+1

    5=2W

    W=2.5, and then L=3.5 so the Area would be:

    A=2.5 (3.5) = 8.75in^2 NOT 30 in^2

    Okay now that that crashed and burned, let's ignore the perimeter statement and just go with L=W+1

    Then A=LW and using L=W+1 in the area equation we get:

    A=W^2+W and since A=30 we have:

    30=W^2+W

    W^2+W-30=0

    (W+6) (W-5) = 0 since W>0 to have any meaning ...

    W=5 and L=6

    Okay that's fine, the area fits the length and width parameters ...

    Now notice that P=2 (5+6) = 22in NOT 12 in

    :D So you mistyped?
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