For the cube $A B C D E F G H$, whose edge measures $6 \, \text{cm}$, the point $M$ is the midpoint of $\overline{E F}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_96f63abf6bdef97495eag-15.jpg?height=690&width=790&top_left_y=1299&top_left_x=710)

a) Determine the area of triangle $A M H$.

b) Determine the volume of the pyramid $A M H E$. (The volume of a pyramid can be calculated by one-third of the product of the base area and the height relative to this base).

c) Calculate the measure of the height relative to the base $A M H$.
Solution

a) $\overline{A H}$ is the diagonal of a face, that is, $A H=6 \sqrt{2} \mathrm{~cm} . \overline{M H}$ and $\overline{A M}$ are hypotenuses of triangles whose legs measure $6 \mathrm{~cm}$ and $3 \mathrm{~cm}$, that is, $M H=A M=3 \sqrt{5} \mathrm{~cm}$. Drawing the height $M M^{\prime}$ of the triangle $A M H$, relative to the side $\overline{A H}$, we can find it by applying the Pythagorean Theorem, that is, $M M^{\prime}=3 \sqrt{3} \mathrm{~cm}$. Finally, the area of the triangle $A M H$ is $\frac{6 \sqrt{2} \cdot 3 \sqrt{3}}{2}=9 \sqrt{6} \mathrm{~cm}^{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_01_96f63abf6bdef97495eag-16.jpg?height=439&width=901&top_left_y=740&top_left_x=456)

b) Using as the base of the pyramid $A M H E$ the triangle $E H M$, whose area is $\frac{6 \cdot 3}{2}=9 \mathrm{~cm}^{2}$, its height will be $\overline{A E}$ and its volume will be $\frac{9 \cdot 6}{3}=18 \mathrm{~cm}^{3}$.

c) For the calculation of the measure of the height relative to the base $A M H$ of the pyramid $A M H E$, we will calculate the volume of this pyramid in two ways: the first with base $A M H$ and height $h$, which is what we are looking for, and the second with base $M H E$.

$$
\begin{aligned}
\frac{9 \sqrt{6} \cdot h}{3} & =18 \\
h & =\frac{6}{\sqrt{6}} \\
h & =\sqrt{6} \mathrm{~cm}
\end{aligned}
$$