A circle is inscribed in trapezoid $P Q R S$.

If $P S=Q R=25 \mathrm{~cm}, P Q=18 \mathrm{~cm}$ and $S R=32 \mathrm{~cm}$, what is

the length of the diameter of the circle?
(A) 14
(B) 25
(C) 24
(D) $\sqrt{544}$
(E) $\sqrt{674}$

![](https://cdn.mathpix.com/cropped/2024_04_20_6027bc27089ed4fc493cg-105.jpg?height=235&width=220&top_left_y=1113&top_left_x=1446)
A circle is inscribed in trapezoid $P Q R S$.

If $P S=Q R=25 \mathrm{~cm}, P Q=18 \mathrm{~cm}$ and $S R=32 \mathrm{~cm}$, what is the length of the diameter of the circle?
(A) 14
(B) 25
(C) 24
(D) $\sqrt{544}$
(E) $\sqrt{674}$

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-474.jpg?height=393&width=331&top_left_y=286&top_left_x=1382)

ANSWER: (B)

## Solution

We start by drawing perpendiculars from $P$ and $Q$ to meet $S R$ at $X$ and $Y$ respectively.

By symmetry, we see that $X Y=P Q=18$. We also note that $S X=Y R$ which means that $S X=Y R=\frac{32-18}{2}=7$.

By applying the Pythagorean Theorem in $\triangle P X S$ we find,

$$
\begin{aligned}
(P X)^{2}+7^{2} & =25^{2} \\
(P X)^{2} & =576 \\
P X & =24
\end{aligned}
$$

The diameter of the circle is thus $24 \mathrm{~cm}$.
![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-474.jpg?height=740&width=328&top_left_y=846&top_left_x=1366)

![](https://cdn.mathpix.com/cropped/2024_04_20_748fdafcde2a47412608g-474.jpg?height=433&width=225&top_left_y=1754&top_left_x=1427)

ANSWER: (C)