A2. The lengths of the diagonals of the faces of a cuboid are $|X Y|=8 \mathrm{~cm}$, $|Y Z|=9 \mathrm{~cm}$, and $|Z X|=\sqrt{55} \mathrm{~cm}$ (see figure). How many centimeters is the length of the space diagonal of this cuboid?
(A) $\sqrt{90}$
(B) 10
(C) $\sqrt{120}$
(D) $\sqrt{200}$
(E) 20

![](https://cdn.mathpix.com/cropped/2024_06_07_716acb2f660115fb0b44g-10.jpg?height=249&width=354&top_left_y=1269&top_left_x=1565)
A2. Let the length, width, and height of the cuboid be denoted by $a, b$, and $c$. Then, by the Pythagorean theorem, we have $a^{2}+b^{2}=|X Y|^{2}=64, b^{2}+c^{2}=|Z X|^{2}=55$ and $c^{2}+a^{2}=|Y Z|^{2}=81$. Adding these equations, we get $2\left(a^{2}+b^{2}+c^{2}\right)=200$. The length of the space diagonal of the cuboid is $\sqrt{a^{2}+b^{2}+c^{2}}=\sqrt{\frac{200}{2}}=10 \mathrm{~cm}$.