# Thread: Area of Semi-Circle region from Chord

1. ## Area of Semi-Circle region from Chord

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.

2. 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.

3. 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!

4. Hello, Chris Rickets!

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

5. Thank you so much!
I just couldn't get my head around it!