How do you find the altitude in a pyramid? (SAT math question)

Question

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?

Answer 1

e is the hypotenus of the triangle VXY where X & Y are the base center and a base corner. Given m, you can find the length of XY.

Answer 2

Hint: Connect the point where $h$ meets the base to any vertex on the square. Do you see a right angled triangle, with one leg $h$, the hypotenuse $e$ and the other leg?

Explanation: Let us call the point $A$ where $h$ meets the base, and $B$ any one vertex of the square, and the apex of the pyramid $C$. I am assuming that it is a straight pyramid, that its apex lies straight above the center of the base [without this information we cannot solve this problem].

Since $A$ is the center of square, its distance from an vertex of the square, that is $AB$, must be half an diagonal. You can find using Pythagorean Theorem that the length of the diagonal is $\sqrt 2m$. Hence $AB$ = $\frac{\sqrt 2m}{2}$.

Now come back to the pyramid. You can visualize that $\angle CAB = 90^{\circ}$. Thus we may again apply the Pythagorean theorem here, to get:

$$CB^2-AB^2=AC^2$$ $$e^2 - (\frac{\sqrt 2m}{2})^2 =h^2$$ $$m^2 - \frac{m^2}{2}=h^2$$

In the last step we used $e=m$. Can you solve it from here?

Answer 3

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

$$\sqrt{e^2-\left(\frac{\sqrt2}2m\right)^2}=\sqrt{m^2-\frac{m^2}2}=\frac m{\sqrt2}$$