Arectangle is nine inches longer than it is wide, and its diagonal is 10 inches longer than its width. what is the width of the quadrilateral?

L = length

w = width

r = diagonal, or hypotenuse

In order to solve this, you must use the Pythagorean theorem (a^2 + b^2 = c^2)

L = w + 9 = a

w = ? = b

r = 10 + w = c

Solve like so:

a^2 + b^2 = c^2

(w + 9)^2 + w^2 = (10 + w)^2

(w + 9)(w + 9) + w^2 = (10 + w)(10 +w)

w^2 + 18w + 81 + w^2 = w^2 + 20w + 100

2w^2 + 18w + 81 = w^2 + 20w + 100

w^2 + 18w + 81 = 20w + 100

w^2 + 81 = 2w + 100

w^2 - 19 = 2w

2 = w - 19/w

w = 2 + 19/w

w ends up being equal to 5.47 as a decimal answer. The work to solve this equation further would require me to spend another couple hours to present. So if you wish to see how this is done, go to quickmath.com.

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