25. The diagram shows a regular nonagon $N$. Moving clockwise around $N$, at each vertex a line segment is drawn perpendicular to the preceding edge. This produces a smaller nonagon $S$, shown shaded.
What fraction of the area of $N$ is the area of $S$ ?
A $\frac{1-\cos 40^{\circ}}{1+\cos 40^{\circ}}$
B $\frac{\cos 40^{\circ}}{1+\cos 40^{\circ}}$
C $\frac{\sin 40^{\circ}}{1+\sin 40^{\circ}}$
D $\frac{1-\sin 40^{\circ}}{1+\sin 40^{\circ}}$
$\mathrm{E} \frac{1}{9}$
SOLUTION
A

We let the side length of the larger regular nonagon $N$ be $x$ and the side length of the smaller regular nonagon $S$ be $y$. Because $S$ and $N$ are similar figures, their areas are in the ratio $y^{2}: x^{2}$. Therefore the required fraction, giving the area of $S$ in terms of the area of $N$, is $\frac{y^{2}}{x^{2}}$.
The area between the nonagons is divided into nine triangles. Two of these adjacent triangles $J K L$ and $P L M$ are shown in the figure on the right.

The triangles $J K L$ and $P L M$ have right angles at $K$ and $L$, respectively. The sides $K L$ and $L M$ each have length $x$ as they are sides of the regular nonagon $N$. The angles at $J$ and $P$ in these triangles are exterior angles of $S$ and therefore they are each $\frac{1}{9} \times 360^{\circ}$, that is, $40^{\circ}$.
In the right-angled triangles $J K L$ and $P L M$ the angles are equal and $K L=L M$. Therefore the triangles are congruent. Hence $J K=P L$. We let $h$ be the common length of $J K$ and $P L$.
In the right-angled triangle $J K L$, the hypotenuse $J L$ has length $y+h$. Therefore, applying Pythagoras' Theorem to this triangle, we have $x^{2}+h^{2}=(y+h)^{2}$. On expansion, this gives $x^{2}+h^{2}=y^{2}+2 y h+h^{2}$, and hence $x^{2}=y^{2}+2 h y$. It follows that $\frac{x^{2}}{y^{2}}=\frac{y+2 h}{y}$ and hence
$$
\frac{y^{2}}{x^{2}}=\frac{y}{y+2 h} .
$$

From the triangle $J K L$ we have $\frac{h}{y+h}=\cos 40^{\circ}$. Hence $y+h=\frac{h}{\cos 40^{\circ}}$ and therefore
$$
y=\frac{h}{\cos 40^{\circ}}-h \text { and } \quad y+2 h=\frac{h}{\cos 40^{\circ}}+h .
$$

Substituting from (2) into the right-hand side of equation (1), we may now deduce that the area of $S$, as a fraction of the area of $N$, is
$$
\frac{y^{2}}{x^{2}}=\frac{\frac{h}{\cos 40^{\circ}}-h}{\frac{h}{\cos 40^{\circ}}+h}=\frac{h-h \cos 40^{\circ}}{h+h \cos 40^{\circ}}=\frac{1-\cos 40^{\circ}}{1+\cos 40^{\circ}} \text {. }
$$