In the diagram, $P Q R S T$ is a pentagon with $P Q=8$, $Q R=2, R S=13, S T=13$, and $T P=8$. Also, $\angle T P Q=\angle P Q R=90^{\circ}$. What is the area of pentagon $P Q R S T$ ?
(A) 76
(B) 84
(C) 92
(D) 100
(E) 108

![](https://cdn.mathpix.com/cropped/2024_04_20_6ed09463f225f8ba1f07g-034.jpg?height=632&width=439&top_left_y=1343&top_left_x=1296)

## Part C: Each correct answer is worth 8.
We drop a perpendicular from $R$ to $X$ on $P T$, join $R$ to $T$, and drop a perpendicular from $S$ to $Y$ on $R T$.

![](https://cdn.mathpix.com/cropped/2024_04_20_faa9db06d997f2a5b9b1g-059.jpg?height=634&width=455&top_left_y=1060&top_left_x=954)

Since quadrilateral $P Q R X$ has three right angles (at $P, Q$ and $X$ ), then it must have four right angles and so is a rectangle. Its area is $8 \times 2=16$.

Next, $\triangle R X T$ is right-angled at $X$.

Since $P Q R X$ is a rectangle, then $X R=P Q=8$ and $P X=Q R=2$.

Since $P X=2$, then $X T=P T-P X=8-2=6$.

Thus, the area of $\triangle R X T$ is $\frac{1}{2} \times X T \times X R=\frac{1}{2} \times 6 \times 8=24$.

By the Pythagorean Theorem, $T R=\sqrt{X T^{2}+X R^{2}}=\sqrt{6^{2}+8^{2}}=\sqrt{36+64}=\sqrt{100}=10$, since $T R>0$.

Since $\triangle T S R$ is isosceles with $S T=S R$ and $S Y$ is perpendicular to $T R$, then $Y$ is the midpoint of $T R$.

Since $T Y=Y R=\frac{1}{2} T R$, then $T Y=Y R=5$.

By the Pythagorean Theorem, $S Y=\sqrt{S T^{2}-T Y^{2}}=\sqrt{13^{2}-5^{2}}=\sqrt{169-25}=\sqrt{144}=12$, since $S Y>0$.

Therefore, the area of $\triangle S T R$ is $\frac{1}{2} \times T R \times S Y=\frac{1}{2} \times 10 \times 12=60$.

Finally, the area of pentagon $P Q R S T$ is the sum of the areas of the pieces, or $60+24+16=100$.