A wheel of radius 8 rolls along the diameter of a semicircle of radius 25 until it bumps into this semicircle. What is the length of the portion of the diameter that cannot be touched by the wheel?
(A) 8
(B) 12
(C) 15
(D) 17
(E) 20

![](https://cdn.mathpix.com/cropped/2024_04_20_e85ef69091c57cb633f3g-150.jpg?height=249&width=460&top_left_y=1174&top_left_x=1315)
A wheel of radius 8 rolls along the diameter of a semicircle of radius 25 until it bumps into this semicircle. What is the length of the portion of the diameter that cannot be touched by the wheel?
(A) 8
(B) 12
(C) 15
(D) 17
(E) 20

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-284.jpg?height=274&width=525&top_left_y=378&top_left_x=1361)

## Solution

Draw $\triangle O B C$, where $O$ is the centre of the large circle, $B$ is the centre of the wheel, and $C$ is the point of tangency of the wheel and the diameter of the semicircle. Since $B C$ is a radius of the wheel, $\angle O B C=90^{\circ}$ and $\triangle O B C$ is rightangled at $C$.

![](https://cdn.mathpix.com/cropped/2024_04_20_688a3b42f69739f2e244g-284.jpg?height=331&width=531&top_left_y=829&top_left_x=1361)

Extend $O B$ to meet the semicircle at $D$. Then $B D=B C=8$, since they are both radii of the wheel, and $O B=25-8=17$.

Use the Pythagorean Theorem in $\triangle O B C$ to find $O C$.

$$
\begin{aligned}
O C^{2} & =17^{2}-8^{2} \\
O C^{2} & =225 \\
O C & =15
\end{aligned}
$$

Then $A C=25-15=10$. The length of the portion of the diameter that cannot be touched by the wheel is a length equivalent to $2 A C$ or 20 .

ANSWER: (E)