Square $A B C D$ has sides of length 14. A circle is drawn through $A$ and $D$ so that it is tangent to $B C$, as shown. What is the radius of the circle?
(A) 8.5
(B) 8.75
(C) 9
(D) 9.25
(E) 9.5

![](https://cdn.mathpix.com/cropped/2024_04_20_ac36362783317e0251fdg-149.jpg?height=336&width=349&top_left_y=569&top_left_x=1365)
Square $A B C D$ has sides of length 14. A circle is drawn through $A$ and $D$ so that it is tangent to $B C$, as shown. What is the radius of the circle?
(A) 8.5
(B) 8.75
(C) 9
(D) 9.25
(E) 9.5

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-306.jpg?height=374&width=393&top_left_y=1293&top_left_x=1300)

## Solution

Let $r$ represent the length of the radius and let $O$ represent the centre of the circle. Draw diameter $M N$ that bisects chord $A D$ perpendicularly at $P$. Join $O A$.

$\triangle O A P$ is a right-angled triangle with $\angle A P O=90^{\circ}$. The length of $A P$ is 7 , since it is half of a side of the square. The length of $O A$ is $r$, and the length of $P O$ is $P N-O N=14-r$.

![](https://cdn.mathpix.com/cropped/2024_04_20_33a8a6559d675f59baa2g-306.jpg?height=379&width=463&top_left_y=1762&top_left_x=1292)

Using the Pythagorean Theorem we get

$$
\begin{aligned}
r^{2} & =7^{2}+(14-r)^{2} \\
r^{2} & =49+196-28 r+r^{2} \\
28 r & =245 \\
r & =8.75
\end{aligned}
$$

Thus, the radius of the circle is 8.75 .

ANSWER: (B)