1. ## Arc Segment Area

Hi everybody
I am trying to the prove the formula of Arc Segment Area

but I am ending up with the Arc Area formula

Can you pleas tell me what wrong with my work ( I attached my work )

2. the segment of an arc is that area between a chord and the circle (your picture shows half a segment)

area of sector - area of the isosceles triangle formed by the chord and two radii = area of the segment

$\displaystyle \frac{1}{2}r^2 \theta - \frac{1}{2} r^2 \sin{\theta} = \frac{1}{2}r^2(\theta - \sin{\theta})$

3. Can you explain more what do you mean, I don't get it .
Why I end up with the Arc Area formula ?

4. If $\displaystyle \theta$ is the central angle of an arc, in radians, then $\displaystyle \frac{\theta}{2\pi}$ is the fraction of the circle contained in that arc. If the circle has radius r then it has area $\displaystyle \pi r^2$ and so the area of the portion of the circle contained in the arc is $\displaystyle \frac{\theta}{2\pi}\left(\pi r^2\right)= \frac{\theta r^2}{2}$.

The isosceles triangle formed by the two radii and the chord can be divided into two right triangles each of angle $\displaystyle \frac{\theta}{2}$ and hypotenuse of length r. The "opposite side" to $\displaystyle \frac{\theta}{2}$ has length $\displaystyle r sin(\theta/2)$ and is half the base of the iscosceles triangle. The "near side" to $\displaystyle \theta/2$ has length $\displaystyle r cos(\theta/2)$ and is the altitude of the isosceles triangle. The area of the triangle, "(1/2)base times height" is $\displaystyle (1/2)(2r sin(\theta/2))(r cos(\theta/2))= r^2 sin(\theta/2) cos(\theta/2)$. Using the trig identity, $\displaystyle sin(2x)= 2 sin(x)cos(x)$ so $\displaystyle sin(\theta/2)cos(\theta/2)= (1/2)sin(2(\theta/2))= (1/2) sin(\theta)$ and the area of the isosceles triangle is $\displaystyle \frac{1}{2} r^2 sin(\theta)$.

The area of the region you are seeking, between the chord and the circle is the difference between the area of the entire region between the radii and that triangle: $\displaystyle \frac{\theta r^2}{2}- \frac{1}{2} r^2 sin(\theta)= \frac{1}{2}r^2(\theta- sin(\theta))$.

By the way, you have two small pictures in your solution, showing an arc segment with a perpendicular drawn from one end of the arc to the opposite radius and on one of them the right triangle formed is shaded and labeled "A". Is that what you were calculating? That is NOT the "arc segment". The region you want is the region between the chord between the two ends of the radii and the circle.

5. crystal clear