Solve the problem.
-The length of the diagonal of a rectangle is given by $D = \sqrt { L ^ { 2 } + W ^ { 2 } }$ where $L$ and $W$ are the length and width of the rectangle. What is the length of the diagonal, D, of a rectangle that is 68 inches long and 41 inches wide? Round your answer to the nearest tenth of an inch, if necessary.
A) $52.8 \mathrm { in }$.
B) $54.2 \mathrm { in }$.
C) $10.4 \mathrm { in }$.
D) $79.4 \mathrm { in }$.

Question

Quizplus · Accepted Answer

Using the given formula $D = \sqrt { L ^ { 2 } + W ^ { 2 } }$, with $L = 68$ inches and $W = 41$ inches, we get $D = \sqrt { 68^2 + 41^2 } = \sqrt { 4624 + 1681 } = \sqrt { 6305 }$ inches, which rounds to approximately $79.4$ inches.

Solve the Problem D=L2+W2D = \sqrt { L ^ { 2 } + W ^ { 2 } }D=L2+W2​

Solve the Problem $D = \sqrt { L ^ { 2 } + W ^ { 2 } }$