The width of a rectangle is 8 less than twice its length If the area of the rectangle is 28 cm2 what is the length of the diagonal?

Suppose the length is L cmThen the width is 2L - 8 cm

Therefore, the area is L*W = L*(2L - 8) = 2L^2 - 8L sq cm.

Then 2L^2 - 8L = 28

or 2L^2 - 8L - 28 = 0

that is L^2 - 4L - 14 = 0

therefore L = 2+3*sqrt(2) [the other root of the quadratic equation is negative and so cannot be a length].

and then W = 6*sqrt(2) - 4

By Pythagoras,

L^2 = 4 + 9*2 + 12*sqrt(2) = 22 + 12*sqrt(2)

W^2 = 36*2 + 16 - 48*sqrt(2) = 88 - 48*sqrt(2)

and then D^2 = L^2 + W^2 = 110 - 36*sqrt(2) = 59.0883 approx

and so D = 7.6869 cm, approx.