# Area of Semi-Circle region from Chord

• Aug 8th 2007, 02:19 PM
Chris Rickets
Area of Semi-Circle region from Chord
http://img261.imageshack.us/img261/7381/problemgq0.png

Hello,
I would like to know how I am supposed to approach this problem.
A small semi-circle is inserted into a bigger one.
The only measurement that I have been given is that chord WX = 36mm, and is parallel to diameter YZ.
I need to find the area of the grey area, and I am completely stuck for any idea on how to tackle this with only the one measurement.

p.s. The diameter of the smaller semi-circle is not necessarily half the diameter of the larger one.

Thanks if anyone can help.
• Aug 8th 2007, 02:35 PM
ThePerfectHacker
This shape is called the Arbelos (or Shoemaker's Knife) it has been discussed in the Book of Lemmas by Archimedes Proposition 14.

NOTE: You need to find a link called "Book of Lemmas", it is not possible to post this as a redirect.
• Aug 8th 2007, 02:53 PM
Chris Rickets
Thanks, but I am still none the wiser as to how to solve the question at hand.

I found proposition 14 in the Book of Lemmas, but I couldn't figure out how to utilise any of the information.

Also, it seems that all of the Arbelos' have two semi-circles inscribed and not just one as in my problem.

Thanks in advance for any further help!
• Aug 8th 2007, 03:57 PM
Soroban
Hello, Chris Rickets!

Quote:

http://img261.imageshack.us/img261/7381/problemgq0.png

A small semi-circle is inserted into a bigger one.
The only measurement is that chord WX = 36mm, and is parallel to diameter YZ.
Find the area of the grey area.

Let $\displaystyle r$ = radius of small circle.
Let $\displaystyle R$ = radius of large circle.

Let $\displaystyle O$ be the center of the large circle.
Then: .$\displaystyle OW \,=\, R$

From $\displaystyle O$ erect a perpendicular, cutting $\displaystyle WX$ at $\displaystyle P$.
\. . Note that: $\displaystyle OP \,= \,r$

In right triangle $\displaystyle WPO$, we have: .$\displaystyle OP^2 + WP^2 \:=\:OW^2$
. . So: .$\displaystyle r^2 + WP^2 \:=\:R^2\quad\Rightarrow\quad WP \:=\:\sqrt{R^2-r^2}$

Since $\displaystyle WZ = 36,\:WP = 18\!:\;\sqrt{R^2-r^2} \:=\:18\quad\Rightarrow\quad R^2-r^2\:=\:324$ .[1]

The area of the large semicircle is: .$\displaystyle \frac{1}{2}\pi R^2$
The area of the small semicircle is: .$\displaystyle \frac{1}{2}\pi r^2$

Then the grey area is: .$\displaystyle A\:=\:\frac{1}{2}\pi R^2 - \frac{1}{2}\pi r^2 \:=\:\frac{1}{2}\pi\left(R^2-r^2\right)$ .[2]

Substitute [1] into [2]: .$\displaystyle A \:=\:\frac{1}{2}\pi(324) \:=\:\boxed{162\pi\text{ mm}^2}$

• Aug 8th 2007, 04:13 PM
Chris Rickets
Thank you so much!
I just couldn't get my head around it!