25. A drinks carton is formed by arranging four congruent triangles as shown. $Q P=R S=4 \mathrm{~cm}$ and $P R=$ $P S=Q R=Q S=10 \mathrm{~cm}$.
What is the volume, in $\mathrm{cm}^{3}$, of the carton?
A $\frac{16}{3} \sqrt{23}$
B $\frac{4}{3} \sqrt{2}$
C $\frac{128}{25} \sqrt{6}$
D $\frac{13}{2} \sqrt{23}$
E $\frac{8}{3} \sqrt{6}$
Solution
A
The lengths in this question are given in terms of centimetres. However, for convenience, we will ignore these units in the calculations until we reach the final answer.

The drinks carton is in the shape of a pyramid. We therefore use the fact that the volume of a pyramid is given by the formula
$$
\text { volume }=\frac{1}{3}(\text { area of base } \times \text { height }) .
$$

We let $M$ be the midpoint of $P Q$ and let $L$ be the midpoint of $R S$.

The carton may be thought of as made up of two pyramids, one with the triangle $R M S$ as its base and $P$ as its apex, and the other with $R M S$ as its base and $Q$ as its apex.

Since $P S=Q S$, and $M$ is the midpoint of $Q P$, the triangles $P M S$ and $Q M S$ are congruent.

It follows that the angles $\angle P M S$ and $\angle Q M S$ are equal. Because they are angles on a line, it follows that they are both right angles.
Since $\angle P M S$ is a right angle, $P M$ is the height of the pyramid with base $R M S$ and apex $P$. Therefore the volume of this pyramid is $\frac{1}{3}($ area of $R M S \times P M)$.
Similarly the volume of the pyramid with base $R M S$ and apex $Q$ is $\frac{1}{3}$ (area of $R M S \times M Q$ ). We let $V$ be the volume of the carton.

We now have that
$$
\begin{aligned}
V & =\frac{1}{3}(\text { area of } R M S \times P M)+\frac{1}{3}(\text { area of } R M S \times M Q) \\
& =\frac{1}{3}(\text { area of } R M S \times(P M+M Q)) \\
& =\frac{1}{3}(\text { area of } R M S \times P Q) .
\end{aligned}
$$

We therefore now calculate the area of the triangle $R M S$.
Since the triangles $P R Q$ and $P S Q$ are congruent it may be shown that $R M=S M$ (see Problem 25.3).

It follows that the triangles $R M L$ and $S M L$ are congruent, and hence that $M L$ is at right angles to $R S$.

Therefore
$$
\text { area of } R M S=\frac{1}{2}(R S \times M L) \text {. }
$$

By Pythagoras theorem applied to the right-angled triangle $S M P$, we have $M S^{2}=P S^{2}-P M^{2}=$ $10^{2}-2^{2}=100-4=96$. Therefore, by Pythagoras' Theorem applied to the right-angled triangle $S L M$, we have $M L^{2}=M S^{2}-S L^{2}=96-2^{2}=92$. Hence $M L=\sqrt{92}=2 \sqrt{23}$.

We can now deduce, by (1) and (2) that
$$
\begin{aligned}
V & =\frac{1}{3}\left(\frac{1}{2}(R S \times M L) \times P Q\right) \\
& =\frac{1}{6}(R S \times M L \times P Q) \\
& =\frac{1}{6}(4 \times 2 \sqrt{23} \times 4) \\
& =\frac{16}{3} \sqrt{23} .
\end{aligned}
$$

Hence the volume of the carton is $\frac{16}{3} \sqrt{23} \mathrm{~cm}^{3}$.
FOR INVESTIGATION
25.1 How can it be proved that the volume of a pyramid is given by the formula
$$
\left.\frac{1}{3} \text { (area of base } \times \text { height }\right) \text { ? }
$$
25.2 Another method for solving this problem is to regard the carton as a pyramid with the triangle $P Q S$ as its base and $R$ as its apex. Then the volume of the carton is given
$$
\frac{1}{3}(\text { area of } P Q S \times R K),
$$
where $K$ is the foot of the perpendicular from $R$ to the triangle $P Q S$.
Use this method to find the volume of the carton.
25.3 Explain how $R M=S M$ follows from the fact that the triangles $P R Q$ and $P S Q$ are congruent.