# Math Help - Circle and square problem

1. ## Circle and square problem

There is a circle with a radius of 4cm and a square that is 5 * 5.

2 corners (diagonally opposite) of the square are touching the circumference of the circle. Calculate the area overlap of the square and circle.

At first when I read the problem I thought it was pretty simple. But I then realised it wasn't - there's 2 segments that stick out that you have to subtract (segment AC and BD).

I just can't seem to find the area of the 2 segments that stick out. I can solve the rest. Can anyone show me or hint to me what I'm not seeing?

2. Hi educated.

Are you describing a square inscribed in a circle ?

3. Hello, Educated!

There is a circle with a radius of 4cm and a square that is 5-by-5.

2 diagonally opposite corners of the square are on the circumference of the circle.

Calculate the area overlap of the square and circle.

The diagram looks like this:

Code:
              * * *
*          /*
*           /   *
*          4/     *
/
*     S ♥--/--------♥ P
*       | o         *
*       |           *
|           |
*     5|          *|5
*     |         * |
*   |       *   |
♥-*-*-------♥ Q
R     5

Vertices $P$ and $R$ are on the circle.

. . And, no . . . I haven't solved it yet.

4. I don't see any neat way to do this. I tried putting the vertices of the square ABCD at the points $(0,\pm5/\sqrt2),\ (\pm5/\sqrt2,0)$. Then the centre S of the circle has to be at the point $(0,\sqrt{7/2})$. The overlap of the square and circle is then the sector DSB of the circle, with the two little protruding sections from D to Q and from P to B subtracted, together with the triangles ASB and ASD.

If the angle DSB is $\theta$ then $\tan(\theta/2) = 5/\sqrt7$, from which you can find the area of the sector DSB. To subtract the two little protruding sections, you need to find the angle $\phi$ in the attached picture. Using the inner product for the vectors $\vec{SB}$ and $\vec{SP}$, I found that $\cos\phi = 5\sqrt7/16$ (correctness not guaranteed! – I didn't check the calculation). Once you know $\phi$, you can find the area of the triangle SBP and subtract it from the area of the sector SBP to get the area of the sliver between P and B.

Edit. I meant to add that P is the point $\bigl(\frac{\sqrt7}{\sqrt2},\frac{-5+\sqrt7}{\sqrt2}\bigr)$.