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15y ago

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How do you find the area from the perimeter and width of a rectangle?

Let P be perimeter, A is area, L is length, and W is width.A = L x W, P = 2L + 2W; you know P and W, so L = (P - 2W)/2and A = W x (P - 2W)/2


W plus 65 equals 1000 w equals what?

Well, darling, if W plus 65 equals 1000, then W must be 935. It's simple math, honey. Just subtract 65 from 1000 and you've got your answer.


What is W equals P half minus L solve for L?

L = P/2 - W.


How do you determine length and width if given the area and perimeter?

If length = L and width = W then area A = L*W and perimeter P = 2(L+W). So, from the area equation, L = A/W Substituting this in the equation for P gives P=2(A/W + W) = 2A/W + 2W Then multiplying through by W gives PW = 2A + 2W2 or, in standrad form, 2W2 - PW + 2A = 0 Then W = 1/4*[P +/- sqrt(P2 - 16A)] The two solutions of this quadratic will be W and L.


What is the perimeter of a rectangle with an area of 100 ft?

First of all, an area of 100 feet is not possible since area is measured in square feet - not in feet. But suppose we overlook that fundamental error. The area of a rectangle is not sufficient to determine its perimeter. It can have any value greater than 40 ft. Let L be ANY length ≥ 10 ft and let W = 100/L ft. Then, a rectangle with length L and width W has an area of L*W = L*100/L = 100 square feet. But since there are infinitely many choices for L, there are infinitely many perimeters (P). For example, L = 10 => W = 10 => P = 40 L = 100 => W = 1 => P = 202 L = 1000=> W = 0.1 => P = 2000.2 L = 10000=> W = 0.01 => P = 20000.02 You should see from that, there is no limit to how large L can be and so no limit on how large P can be.