Problema Solution
The width of a rectangle is 8 less than twice its length. If the area of the rectangle is 129 "cm"^2, what is the length of the diagonal?
Answer provided by our tutors
let
l = the length og the rectangle
w = the width of the rectangle
the width of a rectangle is 8 less than twice its length
w = 2l - 8
the area of the rectangle is 129 "cm"^2
w*l = 129
plug w = 2l - 8 into the last equation
(2l - 8)l = 129
by solving the quadratic equation we find
l = (8 + (1096)^0.5)/4
l = 2 + 68.5^0.5
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w = 2l - 8 = 2( 2 + 68.5^0.5) - 8 = -4 + 2*68.5^0.5
we need to find the diagonal d
using Pythagorean Theorem
d^2 = w^2 + l^2
d^2 = (2 + 68.5^0.5)^2 + (-4 + 2*68.5^0.5)^2
d = 16.22 cm
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