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## Circumference of a Semicircle | AMC 8, 2014 | Problem 25

Try this beautiful problem from AMC-8-2014 (Geometry) based on Circumference of a Semicircle

## Circumference of a Semicircle- AMC 8, 2014 – Problem 25

On A straight one-mile stretch of highway, 40 feet wide, is closed. Robert rides his bike on a path composed of semicircles as shown. If he rides at 5miles per hour, how many hours will it take to cover the one-mile stretch?

• $\frac{\pi}{11}$
• $\frac{\pi}{10}$
• $\frac{\pi}{5}$

### Key Concepts

Geometry

Semicircle

Distance

Answer:$\frac{\pi}{10}$

AMC-8, 2014 problem 25

Challenges and Thrills of Pre College Mathematics

## Try with Hints

Find the circumference of a semi-circle

Can you now finish the problem ……….

If Robert rides in a straight line, it will take him $\frac {1}{5}$ hours

can you finish the problem……..

If Robert rides in a straight line, it will take him $\frac {1}{5}$ hours. When riding in semicircles, let the radius of the semicircle r, then the circumference of a semicircle is ${\pi r}$. The ratio of the circumference of the semicircle to its diameter is $\frac {\pi}{2}$. so the time Robert takes is  $\frac{1}{5} \times \frac{\pi}{2}$. which is equal to $\frac{\pi}{10}$

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## Rolling ball Problem | Semicircle |AMC 8- 2013 -|Problem 25

Try this beautiful problem from Geometry based on a rolling ball on a semicircular track.

## A Ball rolling Problem from AMC-8, 2013

A ball with diameter 4 inches starts at point A to roll along the track shown. The track is comprised of 3  semicircular arcs whose radii are $R_1=100$ inches ,$R_2=60$ inches ,and $R_3=80$ inches respectively. The ball always remains in contact with the track and does not slip. What is the distance the center of the ball travels over the course from  A to B?

• $235 \pi$
• $238\pi$
• $240 \pi$

### Key Concepts

Geometry

circumference of a semicircle

Circle

Answer:$238 \pi$

AMC-8, 2013 problem 25

Pre College Mathematics

## Try with Hints

Find the circumference of semicircle….

Can you now finish the problem ……….

Find the total distance by the ball….

can you finish the problem……..

The radius of the ball is 2 inches. If you think about the ball rolling or draw a path for the ball (see figure below), you see that in A and C it loses $2\pi \times \frac{2}{2}=2\pi$  inches each, and it gains $2\pi$ inches on B .

So, the departure from the length of the track means that the answer is

$\frac{200+120+160}{2} \times \pi$ + (-2-2+2) $\times \pi$=240$\pi$ -2$\pi$=238$\pi$