Three faces of a cuboid

The diagonal of three faces of a cuboid are 13,√281, and 20 units. Then the total surface area of the cuboid is.

Correct answer:

S =  664

Step-by-step explanation:

$$\begin{gathered} d_1=13 \\ {d}_2=\sqrt{281}=\sqrt{281}\doteq 16.7631 \\ {d}_3=20 \\ {d}_1^2=a^2+b^2 \\ {d}_2^2=b^2+c^2 \\ {d}_3^2=a^2+c^2 \\ {a}^2=d_1^2-b^2=d_1^2-d_2^2+c^2=d_1^2-d_2^2+d_3^2-a^2 \\ 2a^2=d_1^2-d_2^2+d_3^2 \\ a=\sqrt{{\displaystyle \frac{d_1^2-d_2^2+d_3^2}{2}}}=\sqrt{{\displaystyle \frac{13^2-16.7631^2+20^2}{2}}}=12 \\ \dots \\ b=\sqrt{{\displaystyle \frac{d_1^2+d_2^2-d_3^2}{2}}}=\sqrt{{\displaystyle \frac{13^2+16.7631^2-20^2}{2}}}=5 \\ \dots \\ c=\sqrt{{\displaystyle \frac{-d_1^2+d_2^2+d_3^2}{2}}}=\sqrt{{\displaystyle \frac{-13^2+16.7631^2+20^2}{2}}}=16 \\ S=2\cdot (a\cdot b+b\cdot c+c\cdot a)=2\cdot (12\cdot 5+5\cdot 16+16\cdot 12)=664 \end{gathered}$$

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