In the diagram, $A B C D E F$ is a regular hexagon with side length 4 and centre $O$. The line segment perpendicular to $O A$ and passing through $A$ meets $O B$ extended at $P$. What is the area of $\triangle O A P ?$

![](https://cdn.mathpix.com/cropped/2024_04_17_040cabdd5aeac1142eb4g-2.jpg?height=455&width=377&top_left_y=1800&top_left_x=1248)
Join $F O$ and $O C$.

Since $A B C D E F$ is a regular hexagon with side length 4 , then $F A=A B=B C=4$.

Since $A B C D E F$ is a regular hexagon with centre $O$, then $F O=A O=B O=C O$.

Therefore, $\triangle F O A, \triangle A O B$ and $\triangle B O C$ are congruent triangles, by symmetry, since their corresponding side lengths are equal.

Also, each of $\angle F O A, \angle A O B$ and $\angle B O C$ must be $\frac{1}{6}$ of the total angle around $O$.

Thus, $\angle F O A=\angle A O B=\angle B O C=\frac{1}{6}\left(360^{\circ}\right)=60^{\circ}$.

Since $\triangle A O B$ is isosceles with $A O=B O$ and $\angle A O B=60^{\circ}$, then $\angle O A B=\angle O B A$, so $\angle O A B=\frac{1}{2}\left(180^{\circ}-\angle A O B\right)=\frac{1}{2}\left(180^{\circ}-60^{\circ}\right)=60^{\circ}$.

Thus, $\triangle A O B$ is equilateral, and so $A O=A B=4$.

![](https://cdn.mathpix.com/cropped/2024_04_17_4b96281abafa035ed6b6g-03.jpg?height=461&width=379&top_left_y=152&top_left_x=927)

Now $\triangle O A P$ has $\angle O A P=90^{\circ}$ and $\angle A O P=\angle A O B=60^{\circ}$. This means that $\triangle O A P$ is a $30^{\circ}-60^{\circ}-90^{\circ}$ triangle.

By the ratio of sides in a $30^{\circ}-60^{\circ}-90^{\circ}, A P=\sqrt{3}(A O)=4 \sqrt{3}$.

Therefore, the area of $\triangle O A P$ is $\frac{1}{2}(O A)(A P)=\frac{1}{2}(4)(4 \sqrt{3})=8 \sqrt{3}$.

ANSWER: $8 \sqrt{3}$