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)
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