In the diagram, points $P, Q$ and $R$ lie on a circle with centre $O$ and radius 12, and point $S$ lies on $O R$. If $\angle P O R=135^{\circ}$, the area of trapezoid $O P Q S$ is closest to
(A) 216
(B) 144
(C) 108
(D) 112.5
(E) 114.6

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-071.jpg?height=320&width=488&top_left_y=886&top_left_x=1293)
Join $O$ to $Q$ and draw a perpendicular from $O$ to $T$ on $P Q$. Since the radius of the circle is 12 , then $O P=O Q=12$. Consider $\triangle O T P$ and $\triangle O T Q$. Each is right-angled, they share side $O T$, and they have hypotenuses $(O P$ and $O Q)$ of equal length. Therefore, these triangles are congruent. Consider quadrilateral $T Q S O$. Since the quadrilateral has three right angles, then it must be a rectangle so its fourth

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-137.jpg?height=315&width=499&top_left_y=1000&top_left_x=1431)
angle, $\angle T O S$, is $90^{\circ}$.

Thus, $\angle T O P=135^{\circ}-90^{\circ}=45^{\circ}$.

Since the angles in $\triangle O T P$ add to $180^{\circ}$, then $\angle O P T=180^{\circ}-90^{\circ}-45^{\circ}=45^{\circ}$.

Therefore, $\triangle O T P$ is isosceles and right-angled with hypotenuse 12 .

Since $\triangle O T Q$ is congruent to $\triangle O T P$, it is also isosceles and right-angled with hypotenuse 12 . Since $\angle T O P=\angle T O Q=45^{\circ}$, then $\angle Q O S=135^{\circ}-45^{\circ}-45^{\circ}=45^{\circ}$, which tells us that $\triangle O Q S$, which is right-angled, has one $45^{\circ}$ angle and so must have a second. Therefore, $\triangle O Q S$ is also isosceles and right-angled, and also has hypotenuse $O Q=12$.

So $\triangle O Q S$ is congruent to $\triangle O T Q$.

Therefore, the area of trapezoid $O P Q S$ equals three times the area of an isosceles right-angled triangle with hypotenuse 12 .

We calculate the area of $\triangle O T P$, which is one of these triangles.

Suppose that $O T=T P=a$.

Since $\triangle O T P$ is right-angled and isosceles, then $O P=\sqrt{2} a$.

(We can see this by using the Pythagorean Theorem to obtain

$$
O P=\sqrt{O T^{2}+T P^{2}}=\sqrt{a^{2}+a^{2}}=\sqrt{2 a^{2}}=\sqrt{2} a
$$

since $a>0$.)

Since $O P=12$, then $\sqrt{2} a=12$ or $a=\frac{12}{\sqrt{2}}$.

The area of $\triangle O T P$ is $\frac{1}{2}(O T)(T P)=\frac{1}{2} a^{2}=\frac{1}{2}\left(\frac{12}{\sqrt{2}}\right)^{2}=\frac{1}{2}\left(\frac{144}{2}\right)=36$.

Thus, the area of trapezoid $O P Q S$ is $3 \times 36=108$.