Each side of square $A B C D$ is 8 . A circle is drawn through $A$ and $D$ so that it is tangent to $B C$. What is the radius of this circle?
(A) 4
(B) 5
(C) 6
(D) $4 \sqrt{2}$
(E) 5.25

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-150.jpg?height=342&width=349&top_left_y=409&top_left_x=1365)

Part C: Each question is worth 8 credits.
Each side of square $A B C D$ is 8 . A circle is drawn through $A$ and $D$ so that it is tangent to $B C$. What is the radius of this circle?
(A) 4
(B) 5
(C) 6
(D) $4 \sqrt{2}$
(E) 5.25

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-283.jpg?height=377&width=395&top_left_y=587&top_left_x=1361)

## Solution

Let $r$ represent 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 4 , since it is half of a side of square $A B C D$. The length of $O A$ is $r$, and the length of $P O$ is $P N-O N=8-r$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-283.jpg?height=380&width=458&top_left_y=1062&top_left_x=1365)

Using the Pythagorean Theorem we get

$$
\begin{aligned}
r^{2} & =4^{2}+(8-r)^{2} \\
r^{2} & =16+64-16 r+r^{2} \\
16 r & =80 \\
r & =5
\end{aligned}
$$

Thus, the radius of the circle is 5 .

ANSWER: (B)

## PART C: