Question

Find the lateral area the regular pyramid.
L. A. =

Answer 1

Check the picture below.

since the pyramid is a hexagonal one, so the base of it is a regular hexagon in this case, that means that we can split up the base into six equilateral triangles, keeping an eye on the equiateral triangle in the picture, let's find it's altitude of "y", which is the perpendicular distance from the center to a side.

`\textit{height of an equilateral triangle}\\\\ h=\cfrac{s√(3)}{2}~~ \begin{cases} s=\stackrel{length~of}{a~side}\\[-0.5em] \hrulefill\\ s=6 \end{cases}\implies h=\cfrac{6√(3)}{2}\implies h=3√(3)\implies h=√(27)=y`

now, since we know what "y" is, let's use that green triangle on the right, to find the altitude of each triangular face in the pyramid, namely "z".

`\begin{array}{llll} \textit{using the pythagorean theorem} \\\\ c^2=a^2+o^2 \end{array} \qquad \begin{cases} c=\stackrel{hypotenuse}{z}\\ a=\stackrel{adjacent}{√(27)}\\ o=\stackrel{opposite}{8} \end{cases} \\\\\\ (z)^2= (√(27))^2 + (8)^2\implies z^2=27+64\implies z^2=91\implies z=√(91) \\\\[-0.35em] ~\dotfill`

`\stackrel{ \textit{area of six triangles} }{6\left[\cfrac{1}{2}(\underset{b}{6})(\underset{h}{√(91)}) \right]}\implies 18√(91) ~~\approx ~~ \text{\LARGE 171.71}\impliedby \textit{lateral area}`