For a certain rectangular prism, the area of the rectangular diagonal section perpendicular to two parallel edges can be one of three possibilities: $t_{1}=60, t_{2}=4 \sqrt{153}, t_{3}=12 \sqrt{10}$. Calculate the surface area and volume of the rectangular prism.
![](https://cdn.mathpix.com/cropped/2024_05_02_b09b31c77f8a95a7e67dg-1.jpg?height=271&width=480&top_left_y=122&top_left_x=817)

Solution. Using the notation in the diagram. According to the Pythagorean theorem:

$$
e=\sqrt{a^{2}+b^{2}}, \quad f=\sqrt{b^{2}+c^{2}}, \quad g=\sqrt{a^{2}+c^{2}}
$$

We know that

$$
t_{1}=c \sqrt{a^{2}+b^{2}}=60, \quad t_{2}=a \sqrt{b^{2}+c^{2}}=4 \sqrt{153}, \quad t_{3}=b \sqrt{a^{2}+c^{2}}=12 \sqrt{10}
$$

We get the following equations:

$$
\begin{array}{lll}
c^{2}\left(a^{2}+b^{2}\right)=3600, & \text { i.e., } & a^{2} c^{2}+b^{2} c^{2}=3600 \\
a^{2}\left(b^{2}+c^{2}\right)=2448, & \text { i.e., } & a^{2} b^{2}+a^{2} c^{2}=2448 \\
b^{2}\left(a^{2}+c^{2}\right)=1440, & \text { i.e., } & a^{2} b^{2}+b^{2} c^{2}=1440
\end{array}
$$

Adding the three equations and dividing by two: $a^{2} b^{2}+a^{2} c^{2}+b^{2} c^{2}=3744$. Since we know the sum of any two terms on the left side, we get:

$$
a^{2} b^{2}=144, \quad b^{2} c^{2}=1296, \quad a^{2} c^{2}=2304
$$

Since $a, b$, and $c$ are positive, we have $a b=12, b c=36, a c=48$.

Based on this, the surface area of the rectangular prism is:

$$
A=2(a b+a c+b c)=2(12+36+48)=192
$$

Since

$$
V^{2}=a^{2} b^{2} c^{2}=a b \cdot a c \cdot b c=12 \cdot 48 \cdot 36=20736
$$

the volume of the rectangular prism is: $V=144$.