A rectangle is one inch longer than it is wide. Its diagonal is five inches. What’s the width of the rectangle?

A rectangle is one inch longer than it is wide. Its diagonal is five inches. What’s the width of the rectangle?

Detailed Explanation

The diagonal formula for a rectangle is \(D^2 = L^2 + W^2\). In this case, D = 5, and L = W + 1. Substituting the known values into the formula results in \(5^2 = (W+1)^2 + W^2\)

\(25 = (W+1)(W+1) + W^2\)

\(25 = W^2 + 2W +1 + W^2\)

\(25 = 2W^2 + 2W +1\)

(Note: This equation is a quadratic equation and can be solved by setting it equal to zero and factoring.)

\(0 = 2W^2 + 2W - 24\)

\(1/2(0) = 1/2(2W^2 + 2W - 24)\)

\(0 = W^2 + W - 12\)

\(0 = (W - 3) × (W + 4)\)

\(W - 3 = 0\) or \(W + 4 = 0\)

\(W = 3\) or \(W = -4\) (not a possible solution)

\(25 = (W+1)(W+1) + W^2\)

\(25 = W^2 + 2W +1 + W^2\)

\(25 = 2W^2 + 2W +1\)

(Note: This equation is a quadratic equation and can be solved by setting it equal to zero and factoring.)

\(0 = 2W^2 + 2W - 24\)

\(1/2(0) = 1/2(2W^2 + 2W - 24)\)

\(0 = W^2 + W - 12\)

\(0 = (W - 3) × (W + 4)\)

\(W - 3 = 0\) or \(W + 4 = 0\)

\(W = 3\) or \(W = -4\) (not a possible solution)