Macedonian Mathematical Competitions

Let $ABCD$ be a pyramid, where $ABC$ and $ABD$ are equilateral triangles with length $a\,\mathrm{cm}$, i.e. $\overline{AC} = \overline{BC} = \overline{AB} = \overline{AD} = \overline{BD} = a\,\mathrm{cm}$. Let $CN$ and $DN$ be the heights in the triangles $ABC$ and $ABD$ respectively. Therefore $CN \perp AB$ and

If the triangle $ABC$ is the pyramid base, then the height of the pyramid is $DN$, so the volume is $V = \frac{1}{3} \cdot P_{\Delta ABC} \cdot \overline{DN} = \frac{1}{3} \cdot \frac{a^2 \sqrt{3}}{4} \cdot \frac{a\sqrt{3}}{2} = \frac{a^3}{8}\,\mathrm{cm}^3$,
because $CN \perp DN$. $CND$ is isosceles right triangle and $\overline{CN} = \overline{DN} = \frac{a\sqrt{3}}{2}\,\mathrm{cm}$.
Therefore $\overline{CD} = \sqrt{\left(\frac{a\sqrt{3}}{2}\right)^2 + \left(\frac{a\sqrt{3}}{2}\right)^2} = \frac{a\sqrt{3}}{2} \sqrt{2} = \frac{a\sqrt{6}}{2}\,\mathrm{cm}$. The triangles $CDA$ and $CDB$
are congruent, so $P_{\Delta CDA} = P_{\Delta CDB} = \frac{1}{2} \cdot \frac{a\sqrt{6}}{2} \cdot \sqrt{a^2 - \left(\frac{a\sqrt{6}}{4}\right)^2} = \frac{1}{2} \cdot \frac{a\sqrt{6}}{2} \cdot \frac{a\sqrt{10}}{4} = \frac{a^2\sqrt{15}}{8}\,\mathrm{cm}^2$.
Finally the area of the pyramid is $P = 2 \cdot \frac{a^2\sqrt{3}}{4} + 2 \cdot \frac{a^2\sqrt{5}}{8} = \frac{a^2\sqrt{3}(2+\sqrt{5})}{4}\,\mathrm{cm}^2$.