8. In the diagram, the outer hexagon is regular and has an area of 216 . What is the shaded area?
A 108
В 96
С 90
D 84
E 72
Solution
$\mathbf{E}$
Method 1
In the diagram on the right we have added three lines to the diagram given in the question.

The outer hexagon is now divided into 12 congruent equilateral triangles and 6 congruent triangles with angles $120^{\circ}, 30^{\circ}$ and $30^{\circ}$.

We leave it to the reader to check that each of the triangles with angles $120^{\circ}, 30^{\circ}$, and $30^{\circ}$ has the same area as each of the equilateral triangles (see Problem 8.1).

It follows that the outer hexagon is divided into 18 triangles all with the same area.
The shaded area is made up of 6 of these triangles. Hence its area is $\frac{6}{18}$, that is, $\frac{1}{3}$ of the area of the outer hexagon. Therefore the area of the shaded hexagon is $\frac{1}{3} \times 216=72$.

METHOD 2
We let $s$ be the side length of the outer hexagon, and $P, Q$ and $R$ be three adjacent vertices of this hexagon, as shown.

We let $L$ be the midpoint of $P R$.
We leave it to the reader to check that $P L Q$ is a right-angled triangle in which $\angle P Q L=60^{\circ}$ and $P L=\frac{\sqrt{3}}{2} s$ (see Problem 8.2). It follows that $P R=\sqrt{3} s$.
The inner hexagon is regular. The distance between its parallel sides is $s$. The outer hexagon is regular. The distance between its parallel sides is $\sqrt{3} s$. The ratio of the areas of similar figures is the same as the ratio of the squares of their corresponding lengths. Therefore
area of inner hexagon : area of outer hexagon $=s^{2}:(\sqrt{3} s)^{2}=s^{2}: 3 s^{2}=1: 3$.

It follows that the area of the shaded hexagon is $\frac{1}{3} \times 216=72$.
For investigation
8.1 Show that in the diagram of Method 1, all the 18 triangles into which the outer hexagon is divided have the same area.
8.2 Show that in the diagram of Method 2, the triangle $P L Q$ is right-angled, $\angle P Q L=60^{\circ}$ and $P L=\frac{\sqrt{3}}{2} s$