Area of a Circular Segment

A triangle's area is half its base times its perpendicular height. With sides $a$ and $b$ enclosing angle $C$, the height is $b\sin C$, so its area is $\frac12ab\sin C$.

For an obtuse angle (between 90° and 180°), the height meets the extended base. The right triangle uses angle $\pi-C$. Its sine equals $\sin C$, since these two angles give the same height on a circle, so the area rule still works.

Our two sides are both radii, so $A_{\triangle}=\frac12r^2\sin\theta$. The sector uses $\theta$; the triangle uses $\sin\theta$. They are different quantities.

For a minor segment, use its central angle $0<\theta<\pi$ in radians:

$A_{\text{segment}}=\frac12r^2(\theta-\sin\theta)$.

You can use this combined form or calculate the two areas separately. Both show the same subtraction.

The major segment is everything else in the circle: subtract the minor segment from $\pi r^2$. If the chord is a diameter, the two segments are equal semicircles. At $\theta=\pi$, the radius triangle has zero area.

Keep the angle fixed and double the radius. The arc should double, but both areas should multiply by four. Then keep the radius fixed and move the angle towards a half-turn; the radius triangle becomes flatter.