4B. The base of a right prism is a right triangle, with legs in the ratio $24: 7$, the hypotenuse of the base is in the ratio to the height of the prism as $5: 2$, and the lateral surface area of the prism is $140 m^{2}$. Find the volume of the prism.
Solution. From $a: b=24: 7$ and $c: H=5: 2$ we have that $b=\frac{7}{24} a$ and $c=\frac{5}{2} H$. From the Pythagorean theorem for a right-angled triangle we have

$$
\begin{aligned}
& a^{2}+b^{2}=c^{2} \Rightarrow a^{2}+\frac{49}{576} a^{2}=\frac{25}{4} H^{2} \Rightarrow \\
& \quad \Rightarrow \frac{625}{576} a^{2}=\frac{25}{4} H^{2} \Rightarrow a^{2}=\frac{144}{25} H^{2} \Rightarrow a=\frac{12}{5} H
\end{aligned}
$$

By substituting into $b=\frac{7}{24} a$, we get

$$
b=\frac{7}{24} \cdot \frac{12}{5} H=\frac{7}{10} H
$$

We have that $M=140 m^{2}$, and on the other hand

$$
M=(a+b+c) \cdot H
$$

![](https://cdn.mathpix.com/cropped/2024_06_05_5cbf21d72fe8685988dag-03.jpg?height=300&width=437&top_left_y=1148&top_left_x=1029)

so

$$
\begin{aligned}
& 140 m^{2}=(a+b+c) \cdot H=\left(\frac{12}{5} H+\frac{7}{10} H+\frac{5}{2} H\right) \cdot H=\frac{56}{10} H^{2} \\
& H^{2}=140 \cdot \frac{10}{56} m^{2}=25 m^{2}
\end{aligned}
$$

Thus $H=5 m$, from which we get

$$
a=\frac{12}{5} H=12 m \text { and } b=\frac{7}{10} H=3.5 m \text {. }
$$

Then, for the volume of the prism we have

$$
V=B \cdot H=\frac{a \cdot b}{2} \cdot H=\frac{12 \cdot 3.5}{2} \cdot 5=105 m^{3}
$$

## 2nd Year