The pyramid shown above has altitude h and a square base of side m. The four edges that meet at V, the vertex of the pyramid, each have length e. If e = m, what is the value of h in terms of m?
- A) $\frac{m}{\sqrt2}$
- B) $\frac{m\sqrt3}{2}$
- C) $m$
- D) $\frac{2m}{\sqrt3}$
- E) $m\sqrt2$
I know I need to use special triangles to solve this problem and tried using the 90, 60, 30 triangle rule but ended up with $\frac{m\sqrt3}{2}$. And it's wrong! The correct answer is A. Can you walk me through to how to get to that answer?
Best Answer
Since you've already tried here's the answer:
The base's diagonal's length is $\;\sqrt 2\,m\;$, as in any square (root of two times the sides' length). Assuming this is a straight pyramid, its apex $\;V\;$ is directly over the base's center, i.e.: the line through $\;V\;$ and perpendicular to the base's plane intersects this plane at the base's diagonal's intersection.
Thus, we have that the distance from any of the base's vertices (say, vertex $\;T\;$) to the center of the base (say, point $\;O\;$) is $\;\frac{\sqrt2}2m\;$ , and we thus have a straight triangle $\;\Delta VOT\;$, whose vertical leg (the pyramid's height) is, by Pythagoras Theorem
$$\sqrt{e^2-\left(\frac{\sqrt2}2m\right)^2}=\sqrt{m^2-\frac{m^2}2}=\frac m{\sqrt2}$$