Task B-3.6. The base of a triangular pyramid is a right-angled triangle with legs of length 12 cm and 35 cm. All lateral faces form an angle of $60^{\circ}$ with the base plane. Determine the surface area and volume of the pyramid.
## Solution.

![](https://cdn.mathpix.com/cropped/2024_05_30_247a158c2fc9e6452974g-14.jpg?height=794&width=792&top_left_y=535&top_left_x=642)

For the sketch, it should be clear which angle is $60^{\circ}$, $r$, the height of the lateral face, and the height of the pyramid—whether inside the pyramid or just in the triangle. It is not necessary to have the base specifically drawn.

(2 points)

$$
\begin{aligned}
& c=\sqrt{12^{2}+35^{2}}=37 \text{ cm} \\
& c=A M+B M=B Q+A P=(a-r)+(b-r) \\
& r=\frac{a+b-c}{2}=5 \text{ cm}
\end{aligned}
$$

Note: $r$ can also be calculated using the area $r \cdot s=\frac{a b}{2}$.

Since all lateral faces form an angle of $60^{\circ}$ with the base plane, it follows that the heights $v_{p}$ of all lateral faces are the same. From the triangle $S D V$, we get

$$
v_{p}=2 \cdot 5=10 \text{ cm}
$$

and the height of the entire pyramid is

$$
h=\frac{10 \sqrt{3}}{2}=5 \sqrt{3} \text{ cm}
$$

The surface area is

$$
O=B+P l=\frac{12 \cdot 35}{2}+\frac{10 \cdot(35+12+37)}{2}=210+420=630 \text{ cm}^2
$$

The volume is:

$$
V=\frac{B \cdot h}{3}=\frac{210 \cdot 5 \sqrt{3}}{3}=350 \sqrt{3} \text{ cm}^3
$$