22. The diagram shows a regular octagon and a square formed by drawing four diagonals of the octagon. The edges of the square have length 1 .
What is the area of the octagon?
A $\frac{\sqrt{6}}{2}$
B $\frac{4}{3}$
C $\frac{7}{5}$
D $\sqrt{2}$
E $\frac{3}{2}$
Solution
D

Let $O$ be the centre of the regular octagon, and let $P, Q$ and $R$ be adjacent vertices of the octagon as shown in the figure on the right. Let $K$ be the point where $O Q$ meets $P R$.
Let $x$ be distance of $O$ from the vertices of the octagon.
Since the edges of the square have length $1, P R=1$. By Pythagoras' Theorem applied to the right-angled triangle $P R O$, we have $x^{2}+x^{2}=1^{2}$. Therefore $x^{2}=\frac{1}{2}$ and hence $x=\frac{1}{\sqrt{2}}$.
The triangle $R O Q$ has a base $O Q$ of length $x$, that is $\frac{1}{\sqrt{2}}$, and height $R K$ of length $\frac{1}{2}$. Therefore the area of the triangle $R O Q$ is $\frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{1}{2}$, which equals $\frac{1}{4 \sqrt{2}}$.
The octagon is made up of 8 triangles each congruent to triangle $R O Q$.
Therefore the area of the octagon is given by $8 \times \frac{1}{4 \sqrt{2}}=\sqrt{2}$.