# Area of a part of circle

1. Jun 28, 2012

### kaizokonpaku

1. The problem statement, all variables and given/known data

the figure is shown on the attachment. Find the area of the smallest part of the circle.

2. Relevant equations

area of circle, sector, segment

3. The attempt at a solution

I cannot use the said equations since the part of the circle is not with reference to the center. I cannot think also of any formulas for this kind of problem.

#### Attached Files:

• ###### circle.jpg
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2. Jun 28, 2012

### Saitama

Consider that part as a part of another circle whose radius is 7.

3. Jun 28, 2012

### kaizokonpaku

4. Jun 28, 2012

### Infinitum

Are you sure?

5. Jun 28, 2012

### Saitama

Can you show us your steps and calculations?

Not quite, but let me check it out once more.

Last edited: Jun 28, 2012
6. Jun 28, 2012

### kaizokonpaku

If that is another circle, then the area would be a quarter circle
A = (pi)(7^2) / 4 = 38.48. however, the given answer is 31.

7. Jun 28, 2012

### Staff: Mentor

Your first step should be to demonstrate that the corner of that square lies at the centre of the large circle.

Last edited: Jun 28, 2012
8. Jun 28, 2012

### kaizokonpaku

I do not get the first step.

9. Jun 28, 2012

### Staff: Mentor

Lest anyone be misled into thinking this apparently is an easy problem you can solve in your head ....

You do need some trig and a calculator, and the answer is not an integer. I get 31.0126

10. Jun 28, 2012

### Curious3141

This can be done the hard way, or a very easy way.

Hard way: use trig to find certain angles, then prove that a geometric property of a circle (angle subtended by a chord at circumference = half angle subtended at centre). I was mucking around with this for a few minutes (and actually did it) before inspiration struck and I found the easy way.

Easy way: Reflect the figure vertically and superimpose the mirror images (getting two vertically stacked 5X5 squares). Now reflect that figure laterally, and superimpose the mirror images (to get four 5X5 squares). You will find that, with the dimensions given, one vertex of each of those squares will exactly touch the others, proving that that is the centre.

Now, after you've done that, the easiest way to find the required area is to visualise it as the sum of areas of a segment and a right triangle. Determining the angle subtended by the segment requires a little trig, but nothing too difficult.

11. Jun 28, 2012

### Infinitum

I did this. Couldn't post about it because I had to go out The answer doesn't come out to be exact, as NascentOxygen pointed out.

Interesting method, Curious(10^3)Pi!

Last edited: Jun 28, 2012
12. Jun 28, 2012

### Curious3141

Thanks, but I only found it because I'm always trying to find the lazy way out. :tongue:

13. Jun 29, 2012

### kaizokonpaku

I solved it and I got it using the easy way. Thanks for the help :)