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 result:

S = 664

Solution:

d_{1} = 13 d_{2} = \sqrt{281} = \sqrt{281} ≐ 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} 2 a^{2} = d_{1}^{2} - d_{2}^{2} + d_{3}^{2} a = \sqrt{\frac{d_{1}^{2} - d_{2}^{2} + d_{3}^{2}}{2}} = \sqrt{\frac{1 3^{2} - 16.763 1^{2} + 2 0^{2}}{2}} = 12 \dots b = \sqrt{\frac{d_{1}^{2} + d_{2}^{2} - d_{3}^{2}}{2}} = \sqrt{\frac{1 3^{2} + 16.763 1^{2} - 2 0^{2}}{2}} = 5 \dots c = \sqrt{\frac{- d_{1}^{2} + d_{2}^{2} + d_{3}^{2}}{2}} = \sqrt{\frac{- 1 3^{2} + 16.763 1^{2} + 2 0^{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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