INTRODUCING 5 - days-a-week problem solving session for Math Olympiad and ISI Entrance. Learn More 

February 16, 2020

Perimeter of a circle : AMC 8 2013 Problem 25

What is the area and perimeter of a circle?

A circle is a curve which maintains same distance from a fixed point called center.

The perimeter of a circle is the length of the curve and area of a circle is portion of a plane bounded by the curve.

Try the problem

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?

$\textbf{(A)}\ 238\pi \qquad \textbf{(B)}\ 240\pi \qquad \textbf{(C)}\ 260\pi \qquad \textbf{(D)}\ 280\pi \qquad \textbf{(E)}\ 500\pi$

AMC 8 2013 Problem 25

Geometry : Perimeter of a circle

7 out of 10

Mathematical Circles.

Knowledge Graph

Perimeter of a circle- knowledge graph

Use some hints

First I want to give you the formula required.

You can clearly notice that we have to find the perimeters of all of the semicircles

The perimeter of a circle of radius $r$ unit can be obtained by the formula $2\pi r$. Then can you find perimeter of the semicircles ?!!!

So using the formula, the perimeters of

Semicircle 1 =$\frac{2\pi\times 100}{2}$ inches.

Semicircle 2 =$\frac{2\pi\times 60}{2}$ inches.

Semicircle 3 =$\frac{2\pi\times 80}{2}$ inches.

So the total path covered by the ball is

$\pi(100+60+80)=240\pi$ inches.

Is it the final answer??? Or have we ignored something ?

OK !!! please notice that they have asked for the distance covered by the center of the ball.

And the ball is of radius \(2\) inches.

So for the \(1^{st}\) and \(3^{rd}\) semicircle : The center will roll along a semicircular path of radius \(R_1-2\) and \(R_3-2\).

See this image :

And for the \(2^{nd}\) semicircle : The center will roll along a semicircular path of radius \(R_2+2\).

See the image below :

So the length of the path covered by the center of the ball is

\([\pi(100-2)+\pi(60+2)+\pi(80-2)] \quad \text{inches} \\=\pi(98+62+78) \quad \text{inches}\\=238\pi \quad \text{inches}\).

Subscribe to Cheenta at Youtube

Leave a Reply

This site uses Akismet to reduce spam. Learn how your comment data is processed.

Cheenta. Passion for Mathematics

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.