# Area of a Circle on Each Side of a Chord

#### EvanJ

Half of the area of a circle will be on each side of the diameter, but the areas on each side of a chord will not be equal. Given the equation of a circle and the two points the chord intersects, is it possible to calculate the area on each side of the chord? Call the points of intersection A and B. I would think that all chords of length AB that go through the same circle would make the same small area and the same large area.

Forum Staff
EvanJ

#### EvanJ

Take a circle with radius of 1 and chord length of 1.5. If I used the formula correctly, the smaller area is about 0.833, which is about 26.5 percent of the total area of pi. It looks visually possible. Is that correct?

#### skipjack

Forum Staff
It's too high. Can you post your work so that any mistakes can be identified?

#### EvanJ

I used the formula with the central angle in degrees that didn't display when I copied it here.

theta = 2arcsin (c/2R)

c = 1.5 and R = 1, so C/2R = 0.75
theta = 2*48.59 = about 97.18 degrees

A = (1^2)/2 (97.18pi/180) - sin(97.18pi/180))
A = 1/2 (1.696 - sin 1.696)
A = 1/2 (1.696 - 0.296) = 1/2 (1.666) = 0.833

I prefer degrees to radians, but I feel like I'm converting to radians and then taking sin of 1.696, which is wrong. However, if I change that part to sin of 97.18 degrees, I get an answer that's way too low:

A = (1^2)/2 (97.18pi/180) - sin(97.18pi/180))
A = 1/2 (1.696 - sin 97.18)
A = 1/2 (1.696 - 0.992) = 1/2 (0.704) = 0.352

#### skipjack

Forum Staff
You possibly tried to find sin(1.696°), but that's 0.0296 approximately, not 0.296. Both are incorrect.

You had already found that $\theta$ = about 97.18°, and sine of that is about 0.992, as you later found.

Alternatively, sine of 1.696 radians gives the same value.

Your calculation of 0.352 is correct.

#### EvanJ

A chord of length 1.5 is three-quarters of the diameter. The area is about 11.2 percent of the circle's area, which is pi. I'm not saying you're wrong, but I'm surprised the area is that small. Since half the area is on each side of the diameter, that would mean the area in between the chord and the diameter is about 38.8 percent of the circle's area.

#### skipjack

Forum Staff
Here's a diagram.

EvanJ

In the proof :

Could U plz tell where does the very last part comes from ?

#### skeeter

Math Team
In the proof :
View attachment 10836
Could U plz tell where does the very last part comes from ?
from the diagram on the proof page, $\cos\left(\dfrac{\theta}{2}\right) = \dfrac{d}{R}$

double angle identity ... $\cos{\theta} = 2\cos^2\left(\dfrac{\theta}{2}\right) - 1$

$2R\sqrt{2-2\cos{\theta}} = 2R\sqrt{2 - 2\left[2\cos^2\left(\dfrac{\theta}{2}\right) - 1\right]} = 2R\sqrt{2 - 4\cos^2\left(\dfrac{\theta}{2}\right) + 2} = 2R\sqrt{4 - 4\cos^2\left(\dfrac{\theta}{2}\right)} = 2R\sqrt{4\left[1 - \cos^2\left(\dfrac{\theta}{2}\right)\right]} = 4R\sqrt{1 - \left(\dfrac{d}{R}\right)^2}$