Solve the problem.
-The length of the diagonal of a box is given by $\mathrm { D } = \sqrt { \mathrm { L } ^ { 2 } + \mathrm { W } ^ { 2 } + \mathrm { H } ^ { 2 } }$ where $\mathrm { L } , \mathrm { W }$, and $\mathrm { H }$ are the length, width, and height of the box. Find the length of the diagonal, D, of a box that is $4 \mathrm { ft }$ long, $3 \mathrm { ft }$ high, and $2 \mathrm { ft }$ wide. Give the exact value.
A) $9 \mathrm { ft }$
B) $2 \sqrt { 6 } \mathrm { ft }$
C) $\sqrt { 29 } \mathrm { ft }$
D) $24 \mathrm { ft }$

Question

Quizplus · Accepted Answer

By substituting L = 4 ft, W = 2 ft, and H = 3 ft into the formula $D = \sqrt{L^2 + W^2 + H^2}$, we get $D = \sqrt{4^2 + 2^2 + 3^2} = \sqrt{16 + 4 + 9} = \sqrt{29}$ ft.

Solve the Problem D=L2+W2+H2\mathrm { D } = \sqrt { \mathrm { L } ^ { 2 } + \mathrm { W } ^ { 2 } + \mathrm { H } ^ { 2 } }D=L2+W2+H2​

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