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### Circle and Rectangle Area Problem

```Date: 11/03/2003 at 22:09:15
From: Victor
Subject: Intersection area of a circle and a rectangle

Assume an arbitrary circle and an arbitrary rectangle.  Put one corner
of the rectangle exactly in the center of the circle.  Assume that the
circle radius is larger than the small side of the rectangle but
smaller than the large side of the rectangle.  What is the area of the
intersection of the circle and the rectangle?

```

```
Date: 11/04/2003 at 05:10:01
From: Doctor Jeremiah
Subject: Re: Intersection area of a circle and a rectangle

Hi Victor,

Interesting question!  The area covered by the rectangle is 1/4 of the
circle minus 1/2 of a chord of the circle:

+++++
+++            +++
+++                     +++
+                               +
+                                   +
+                                     +
+                                       +
+                                         +
+                                         +
+                                           +
+                     +---------R-----------+
+                     |                     +
+                    |                    +
+                    H                    +
+                   |                   +
+                  |                  +
+ - - - - - - - - +-----------------+
+               |   1/2 chord   +
+++         R-H         +++
+++     |     +++
+++++

So if we can calculate the area of the chord we'll be in good shape.
First notice that a sector of the circle is two triangles plus the
full chord:

+++++
+++            +++
+++                     +++
+                               +
+                                   +
+                                     +
+                                       +
+                                         +
+                                         +
+                                           +
+                     +                     +
+                  + a|a +                  +
+              +     |     +              +
+           R        H        R           +
+       +           |           +       +
+   +              |              +   +
+-----------------+-----------------+
+                               +
+++     full chord      +++
+++           +++
+++++

If we can calculate what fraction of the circle the sector makes up we
can subtract the area of the triangle and that will leave us with the
area of the chord.

The angle I have labeled a is important because once we know it we can
calculate the area of the sector as a fraction of the area of the
whole circle.

We need trigonometry to calculate angles.  We know H (the short side
of the rectangle) and R (the radius of the circle) so a would be:

cos(a) = H/R   or   a = arccos(H/R)

Now, the angle of the full sector is 2 times a because there are two
triangles:

Angle_of_sector = 2a = 2 arccos(H/R)

Sector's_fraction_of_circle = 2a / 360

Sector's_fraction_of_circle = 2 arccos(H/R) / 360

So the sector as a fraction of the total area of the circle is
2 arccos(H/R) / 360 and since the area of the circle is Pi R^2 the
area of the sector is:

Sector_area = Area_of_circle x Sector's_fraction_of_circle

Sector_area = Pi R^2 (2 arccos(H/R) / 360)

Which is:

Sector_area = 2 Pi R^2 arccos(H/R) / 360

The area of the triangle can be found once we know the length of the
third side:

+
+  |  +
+     |     +
R        H        R
+           |           +
+              |              +
+---------L-------+------L----------+

The value of L can be found using the Pythagorean Theorem:

L^2 = R^2 - H^2   or   L = square_root(R^2 - H^2)

The area of each triangle therefore is:

Triangle_area = LH/2 = square_root(R^2 - H^2) H/2

But there are two triangles so the area of both triangles is:

Double_triangle_area = 2 square_root(R^2 - H^2) H/2

Which is:

Double_triangle_area = square_root(R^2 - H^2) H

Now, the area of the full chord is the area of the sector minus the
area of both triangles:

Chord_area = Sector_area - Double_triangle_area

Chord_area = 2 Pi R^2 arccos(H/R) / 360
- square_root(R^2 - H^2) H

And the area of half the chord is:

Half_chord_area = Pi R^2 arccos(H/R) / 360
- square_root(R^2 - H^2) H/2

So, back to the original statement:

The area covered by the rectangle is 1/4 of the circle minus 1/2 of
a chord of the circle.

So the area covered by the rectangle is:

Intersection_area = Pi R^2 / 4 - Half chord_area

Intersection_area = Pi R^2 / 4
- (Pi R^2 arccos(H/R) / 360 - square_root(R^2 - H^2) H/2)

Which is:

Intersection_area = Pi R^2 / 4
- Pi R^2 arccos(H/R) / 360
+ square_root(R^2 - H^2) H/2

Or even better:

Intersection_area = Pi R^2 (90-arccos(H/R)) / 360
+ square_root(R^2 - H^2) H/2

Does this make sense?  Feel free to write back if you have questions
on anything I did.

- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Trigonometry

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