The width of a rectangle is 6 less than twice its length. If the area of the rectangle is 102 cm 2 , what is the length of the diagonal?

Question

The width of a rectangle is 6 less than twice its length. If the area of the rectangle is 102 , what is the length of the diagonal?

Answer 1

W=2L-6

Area = 102 =LW = L(2L-6) = 2L^2 -6L

2L^2 -6L -102 divide by 2

L^2 -3L -51 = 0 use the quadratic formula x=-b/2a + or - (1/2a)sqr(b^2-4ac) a=2, b=-3 c=-51, x=L

L= 3/2 + or -(1/2)sqr (9+4(51)) = 1.5 + 7.3 = about 8.8 cm (ignore the negative square root, as L>0)

W= 17.6-6 = 11.6 cm

diagonal = d

d^2 + 8.8^2 + 11.6^2 = 77.44 + 34.56 = 212

d=sqr212 = about 14.6 cm = the diagonal