Regular triangular pyramid

Text

A regular pyramid has an equilateral-triangle base of side $12$ cm and height $8$ cm. Find the volume, the pyramid’s slant height (apothem) and the lateral surface. (Take $\sqrt3\approx1.732$.)

Solution

Base area: $\mathcal A_b=\dfrac{\sqrt3}{4}\cdot144=36\sqrt3\approx62.35$ cm$^2$. Volume: $V=\frac13\cdot36\sqrt3\cdot8=96\sqrt3\approx166.28\ \text{cm}^3.$ Base apothem (inradius of the triangle): $\dfrac{\ell}{2\sqrt3}=\dfrac{12}{2\sqrt3}=2\sqrt3$. Pyramid apothem: $a=\sqrt{h^2+(2\sqrt3)^2}=\sqrt{64+12}=\sqrt{76}=2\sqrt{19}\approx8.72\ \text{cm}.$ Lateral surface: $S_\ell=\dfrac12\cdot(3\cdot12)\cdot2\sqrt{19}=36\sqrt{19}\approx156.92$ cm$^2$. $\boxed{V=96\sqrt3\approx166.28\ \text{cm}^3;\ a=2\sqrt{19}\approx8.72\ \text{cm};\ S_\ell=36\sqrt{19}\approx156.92\ \text{cm}^2}$