Let $t$ be TNYWR.

In the diagram, the vertices of rectangle $A B C D$ lie on a circle. Diagonal $A C$ is a diameter of the circle and has length $t$. If $C D=2 A D$, find the area of the shaded region, and write your answer in the form $a \pi-\frac{b}{c}$ with $a, b, c$ positive integers and with $b$ and $c$ having no common positive divisor larger than 1 .

![](https://cdn.mathpix.com/cropped/2024_04_17_385ef00c66fdf6396726g-3.jpg?height=572&width=502&top_left_y=1313&top_left_x=1386)
Since the diameter of the circle has length $t$, then the radius of the circle is $\frac{t}{2}$ and so the area of the circle is $\pi\left(\frac{t}{2}\right)^{2}=\frac{\pi t^{2}}{4}$.

Since $C D=2 A D$, then the area of rectangle $A B C D$ is $(A D)(C D)=2 A D^{2}$.

Since $A B C D$ is a rectangle, then $\angle A D C=90^{\circ}$.

By the Pythagorean Theorem in $\triangle A D C$, we have $A D^{2}+C D^{2}=A C^{2}$.

Thus, $A D^{2}+(2 A D)^{2}=t^{2}$ and so $5 A D^{2}=t^{2}$, or $2 A D^{2}=\frac{2 t^{2}}{5}$. (Recall that this is the area of rectangle $A B C D$.)

The area of the shaded region is the area of the circle minus the area of the rectangle, which equals $\frac{\pi t^{2}}{4}-\frac{2 t^{2}}{5}$.

Since the answer to (b) is 6 , then $t=6$ and so the shaded area is $\frac{\pi(36)}{4}-\frac{2(36)}{5}=9 \pi-\frac{72}{5}$. ANSWER: $144,6,9 \pi-\frac{72}{5}$