4. For a truncated triangular pyramid, the sides of the smaller base are $7 \mathrm{~cm}$, $5 \mathrm{~cm}$, and $3 \mathrm{~cm}$, and the lateral edges form an angle of $45^{\circ}$ with the plane of the larger base. Calculate the volume of the truncated pyramid if its height is $\frac{2}{3}$ of the height of the corresponding pyramid.
Solution. Let $A B C A_{1} B_{1} C_{1}$ be the truncated pyramid, and $A B C S$ the corresponding pyramid (see diagram). From the fact that the lateral edges of the larger base form an angle of $45^{\circ}$, it follows that $\overline{A A_{1}}=\overline{B B_{1}}=\overline{C C_{1}}$, so the projection of $S$ onto the plane $A B C$ is the center of the circumscribed circle around triangle $A B C$. Now, it is easily seen that $H=2 R_{1}$, where $H$ is the height of the truncated pyramid, and $R_{1}$ is the radius of the circumscribed circle around triangle $A_{1} B_{1} C_{1}$.

![](https://cdn.mathpix.com/cropped/2024_06_05_a9dc3c06af03b935fe82g-4.jpg?height=560&width=473&top_left_y=398&top_left_x=994)

According to Heron's formula, for the area $P_{1}$ of triangle $A_{1} B_{1} C_{1}$ we have:

$$
P_{1}=\sqrt{s_{1}\left(s_{1}-a_{1}\right)\left(s_{1}-b_{1}\right)\left(s_{1}-c_{1}\right)}=\frac{15 \sqrt{3}}{4} \mathrm{~cm}^{2}
$$

so

$$
R_{1}=\frac{a_{1} b_{1} c_{1}}{4 P_{1}}=\frac{7 \sqrt{3}}{3} \mathrm{~cm}, \quad H=\frac{14 \sqrt{3}}{3} \mathrm{~cm}
$$

The height of the corresponding pyramid is $H_{1}=\frac{3}{2} H=\frac{14 \sqrt{3}}{2} \mathrm{~cm}$, so for the area $P$ of triangle $A B C$ we will have:

$$
P: P_{1}=\left(\frac{14 \sqrt{3}}{2}\right)^{2}:\left(\frac{14 \sqrt{3}}{2}-\frac{14 \sqrt{3}}{3}\right)^{2}
$$

i.e., $P=9 P_{1}$.

Finally, for the volume $V$ of the truncated pyramid, we will have:

$$
V=\frac{H}{3}\left(P+P_{1}+\sqrt{P P_{1}}\right)=\frac{13}{3} H \cdot P_{1}=227.5 \mathrm{~cm}^{3} .
$$

## Third Year