# 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.

Result

S =  664

#### Solution:

$d_{1}=13 \ \\ d_{2}=\sqrt{ 281 }=\sqrt{ 281 } \doteq 16.7631 \ \\ d_{3}=20 \ \\ \ \\ d_{1}^2=a^2+b^2 \ \\ d_{2}^2=b^2+c^2 \ \\ d_{3}^2=a^2+c^2 \ \\ \ \\ a^2=d_{1}^2 - b^2=d_{1}^2-d_{2}^2+c^2=d_{1}^2-d_{2}^2+d_{3}^2-a^2 \ \\ 2a^2=d_{1}^2-d_{2}^2+d_{3}^2 \ \\ \ \\ a=\sqrt{ \dfrac{ d_{1}^2-d_{2}^2+d_{3}^2 }{ 2 } }=\sqrt{ \dfrac{ 13^2-16.7631^2+20^2 }{ 2 } }=12 \ \\ \ \\ ... \ \\ b=\sqrt{ \dfrac{ d_{1}^2+d_{2}^2-d_{3}^2 }{ 2 } }=\sqrt{ \dfrac{ 13^2+16.7631^2-20^2 }{ 2 } }=5 \ \\ \ \\ ... \ \\ c=\sqrt{ \dfrac{ -d_{1}^2+d_{2}^2+d_{3}^2 }{ 2 } }=\sqrt{ \dfrac{ -13^2+16.7631^2+20^2 }{ 2 } }=16 \ \\ \ \\ S=2 \cdot \ (a \cdot \ b+b \cdot \ c+c \cdot \ a)=2 \cdot \ (12 \cdot \ 5+5 \cdot \ 16+16 \cdot \ 12)=664$

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