3A. Two faces of a three-sided pyramid are equilateral triangles with side $a$ cm. The planes of these triangles are perpendicular to each other. Calculate the area and volume of the pyramid.
Solution. Let the triangular pyramid $ABCD$ be given, where the sides $ABC$ and $ABD$ are equilateral triangles with side $a \, \text{cm}$, i.e.,

$$
\overline{AC}=\overline{BC}=\overline{AB}=\overline{AD}=\overline{BD}=a \, \text{cm}.
$$

Let $CN$ and $DN$ be the altitudes in triangles $ABC$ and $ABD$ respectively, to the common side $AB$, meaning $DN \perp AB$. From the problem's condition, we also have $CN \perp DN$ (see the diagram). If we take one of the equilateral triangles as the base of the pyramid, for example, triangle $ABC$, then the height of the pyramid will be $DN$, so the volume of the pyramid is

$$
V=\frac{1}{3} \cdot P_{\triangle ABC} \cdot \overline{DN}=\frac{1}{3} \cdot \frac{a^{2} \sqrt{3}}{4} \cdot \frac{a \sqrt{3}}{2}=\frac{a^{3}}{8} \, \text{cm}^{3}
$$

Triangle $CND$ is an isosceles right triangle with a right angle at vertex $N$ and $\overline{CN}=\overline{DN}=\frac{a \sqrt{3}}{2} \, \text{cm}$. Then,

$$
\overline{CD}=\sqrt{\left(\frac{a \sqrt{3}}{2}\right)^{2}+\left(\frac{a \sqrt{3}}{2}\right)^{2}}=\frac{a \sqrt{3}}{2} \sqrt{2}=\frac{a \sqrt{6}}{2} \, \text{cm}
$$

Triangles $CDA$ and $CDB$ are congruent isosceles triangles with base $\frac{a \sqrt{6}}{2} \, \text{cm}$ and leg $a \, \text{cm}$. Then, their area is

$$
P_{\triangle CDA}=P_{\triangle CDB}=\frac{1}{2} \cdot \frac{a \sqrt{6}}{2} \cdot \sqrt{a^{2}-\left(\frac{a \sqrt{6}}{4}\right)^{2}}=\frac{1}{2} \cdot \frac{a \sqrt{6}}{2} \cdot \frac{a \sqrt{10}}{4}=\frac{a^{2} \sqrt{15}}{8} \, \text{cm}^{2}
$$

and finally, the surface area of the pyramid is

$$
P=2 \cdot \frac{a^{2} \sqrt{3}}{4}+2 \cdot \frac{a^{2} \sqrt{15}}{8}=\frac{a^{2} \sqrt{3}(2+\sqrt{5})}{4} \, \text{cm}^{2}
$$