16. $G$ and $H$ are midpoints of two adjacent edges of a cube. A trapezium-shaped cross-section is formed cutting through $G$, $H$ and two further vertices, as shown. The edge-length of the cube is $2 \sqrt{2}$.
What is the area of the trapezium?
A 9
B 8
C $4 \sqrt{5}$
D $4 \sqrt{3}$
E $4 \sqrt{2}$
SolUTION
A
We let $P, Q, S, U, V$ and $W$ be vertices of the cube, as shown in the diagram, and $K, L$ be the feet of the perpendiculars from $G, H$, respectively, to $U W$.

We apply Pythagoras' Theorem to the right-angled triangle $U V W$. This gives $U W^{2}=U V^{2}+W V^{2}=$ $(2 \sqrt{2})^{2}+(2 \sqrt{2})^{2}=8+8=16$. Hence $U W=4$.
$G$ is the midpoint of the edge $P Q$. Hence $P G=\sqrt{2}$. Similarly, $P H=\sqrt{2}$. Therefore, applying Pythagoras' Theorem to the right-angled triangle $G P H$ gives $G H^{2}=$ $P G^{2}+P H^{2}=(\sqrt{2})^{2}+(\sqrt{2})^{2}=2+2=4$. Hence $G H=2$.

Also, applying Pythagoras' Theorem to the right-angled triangle $G Q U$ gives $G U^{2}=G Q^{2}+U Q^{2}=$ $(\sqrt{2})^{2}+(2 \sqrt{2})^{2}=2+8=10$. Hence $G U=\sqrt{10}$.

Since $G K L H$ is a rectangle, $K L=G H=2$. Therefore $U K+W L=U W-K L=4-2=2$. By symmetry, $U K=W L$. Hence $U K=W L=1$.
Applying Pythagoras' Theorem to the right-angled triangle $G K U$, gives $G K^{2}=G U^{2}-U K^{2}=$ $(\sqrt{10})^{2}-1^{2}=10-1=9$. Hence $G K=3$.
We now use the formula $\frac{1}{2}(a+b) h$ for the area of a trapezium whose parallel sides have lengths $a$ and $b$ and which has height $h$. [You are asked to prove this formula in Problem 16.1.]

It follows from this formula that
area of the trapezium $G U W H=\frac{1}{2}(U W+G H) G K=\frac{1}{2}(4+2) \times 3=9$.