17. The shape shown is made by removing four equilateral triangles with side-length 1 from a regular octagon with side-length 1 .
What is the area of the shape?
A $2-2 \sqrt{2}+\sqrt{3}$
B $2+2 \sqrt{2}-\sqrt{3}$
C $2+2 \sqrt{2}+\sqrt{3}$
D $3-2 \sqrt{2}-\sqrt{3}$
E $2-2 \sqrt{2}-\sqrt{3}$
Solution
B
As the diagram on the right shows, a regular octagon with side length 1 may be obtained by cutting four triangular corners from the square.
Each of these corners is an isosceles right-angled triangles with a hypotenuse of length 1. It follows, by Pythagoras' Theorem that the other two sides of these triangles have length $1 / \sqrt{2}$.
It follows that the side length of the square is given by $1 / \sqrt{2}+1+1 / \sqrt{2}=1+\sqrt{2}$. Therefore the area of the square is $(1+\sqrt{2})^{2}=1+2 \sqrt{2}+2=3+2 \sqrt{2}$.
The four triangular corners fit together to make a square of side length 1 and hence area 1 .

Therefore the area of the octagon is $(3+2 \sqrt{2})-1=2+2 \sqrt{2}$.
From this area we need to subtract the area of the four equilateral triangles that are removed from the octagon to make the shaded shape.
We let $P Q R$ be an equilateral triangle with side length 1 , and let $P K$ be the perpendicular from $P$ to $Q R$ as shown.
It can be checked that $K$ is the midpoint of $Q R$ (see Problem 17.1) and hence that $Q K$ has length $\frac{1}{2}$. By Pythagoras' Theorem applied to the right-angled triangle $P Q K$ we have $Q K^{2}+P K^{2}=P Q^{2}$. Therefore $P K^{2}=P Q^{2}-Q K^{2}=1-\frac{1}{2}^{2}=\frac{3}{4}$.

Therefore $P K=\frac{1}{2} \sqrt{3}$.
We can now deduce that the area of the triangle $P Q R$ is $\frac{1}{2}(Q R \times P K)=\frac{1}{2}\left(1 \times \frac{1}{2} \sqrt{3}\right)=\frac{1}{4} \sqrt{3}$.
It now follows that the area of the shape is
$$
2+2 \sqrt{2}-4 \times \frac{1}{4} \sqrt{3}=2+2 \sqrt{2}-\sqrt{3} \text {. }
$$

FOR INVESTIGATION
17.1 Prove that the triangles $P K Q$ and $P K R$ are congruent. Deduce that the length of $Q K$ is $\frac{1}{2}$.
17.2 Use the $\frac{1}{2} a b \sin C$ formula for the area of a triangle to confirm that the area of an equilateral triangle with side length 1 is $\frac{1}{4} \sqrt{3}$.