find area of region

[galactus]

These pictorial problems are fun. Try this one if you like.

Below is a diagram of a small circle inside a larger one. The origin is at the center if the larger circle.

Given the dimensions shown, find the area of the white region.

There is 9 units from the left edge of the big circle to the left edge of the smaller one.

There is 5 units from the top and bottom of the smaller circle to the top and bottom of the smaller one.

 

[mmm4444bot]

There are 5 units vertically from the top and bottom of the smaller circle to the top and bottom of the larger circle, respectively.

This is how I read it. Are the edits in red correct ?

I misunderstand something, here.

Let D = diameter of the big circle

The diameter of the small circle is D - 9 horizontally, but D - 10 vertically.

 

[galactus]

Yes, that's it. I suppose I could have worded i better. There is 5 units vertically from the top and bottom and 9 units horizontally from the left.

The small circle is tangent to the larger one at the right edge.

 

[galactus]

Here is how I tackled it. It is not that bad.

Note that the radii of the bigger and smaller circles can be equated as such:

where R=radius of big circle and r=radius of small circle.

\(\displaystyle 2r=2R-9\rightarrow r=\frac{2R-9}{2}\)........[1]

Using circle equation formula:

\(\displaystyle r^{2}=(r-(R-9))^{2}+(R-5)^{2}\)

Sub in [1] and we get \(\displaystyle R-25=0\)

Therefore, \(\displaystyle R=25\)

The area of the entire large circle is \(\displaystyle 625{\pi}\)

The radius of the smaller black circle is then \(\displaystyle r=20.5\)

Giving an area of \(\displaystyle \frac{1681{\pi}}{4}\)

Subtract the two to get the area of the white region.

\(\displaystyle 625{\pi}-\frac{1681{\pi}}{4}=\boxed{\frac{819{\pi}}{4}\approx 643.24}\)

I do not remember where I saw this, but I thought it was a cool little problem.