Desperately need help with maths question
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# Desperately need help with maths question

[From: ] [author: ] [Date: 11-05-24] [Hit: ]
Now you are given L and want to solve for W.For instance, with L = 6, this gives W = 2*6 / (6-2) = 12/4 = 3.w = 2 * l / (l - 2 ) unit ans-You must have been given more info than that........
I desperately need a formula to figure out the width of a rectangle when only given the length.

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The perimeter of a rectangle is related to its width and length by
p = 2*W + 2*L

The area of a rectangle is just
A = W*L

If you want the area to be the same (numerically) as the perimeter, you just throw the two equations together by requiring p=A, i.e.
2*W + 2*L = W*L

Now you are given L and want to solve for W. This is pretty easy:
2*W + 2*L = W*L :: subtract W*L from both sides
2*W - W*L + 2*L = 0 :: subtract 2*L from both sides
2*W - W*L = -2*L :: factor out a W from the left
W*(2-L) = -2*L :: divide both sides by (2-L)
W = -2*L / (2-L) :: get rid of the negative sign by flipping the bottom's terms
W = 2*L / (L-2)

For instance, with L = 6, this gives W = 2*6 / (6-2) = 12/4 = 3.

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area = perimeter of rectangle
l * w = 2(l + w )
l * w - 2w = 2* l
w (l - 2 ) = 2 * l
w = 2 * l / (l - 2 ) unit ans

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You must have been given more info than that... the area or perimeter of the rectangle was surely given to you?
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