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?
$\endgroup$3 Answers
$\begingroup$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
$$\sqrt{e^2-\left(\frac{\sqrt2}2m\right)^2}=\sqrt{m^2-\frac{m^2}2}=\frac m{\sqrt2}$$
$\endgroup$ $\begingroup$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.
$\endgroup$ $\begingroup$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?
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