How to Find the Area of a Circular Segment

A circular segment is the region a chord cuts off from a circle. Subtracting the triangle from the sector gives the area radius² ÷ 2 × (angle in radians − sin angle). The chord, the arc length and the height of the segment are also shown.

Explanation

A circular segment is what a single chord cuts off from a circle: the shape of a bow. Take the sector and remove the triangle formed by the center and the chord, and what is left is the segment.

S=r22(θ−sin⁡θ)S = \dfrac{r^2}{2}(\theta - \sin\theta)

Note that θ\theta is in radians. An angle aa in degrees becomes θ=a×π÷180\theta = a \times \pi \div 180. Enter degrees here and the calculator converts them for you.

The chord cc, the arc ℓ\ell and the height hh, measured from the middle of the chord to the arc, follow from the same rr and θ\theta.

c=2rsin⁡θ2ℓ=rθh=r(1−cos⁡θ2)c = 2r\sin\dfrac{\theta}{2} \qquad \ell = r\theta \qquad h = r\left(1 - \cos\dfrac{\theta}{2}\right)

Example

With the defaults, a radius of r=6r = 6 and a central angle of 60∘60^\circ, which is θ=π3≈1.0472\theta = \dfrac{\pi}{3} \approx 1.0472 radians.

S=622(1.0472−sin⁡1.0472)=18×(1.0472−0.8660)≈3.2611S = \dfrac{6^2}{2}(1.0472 - \sin 1.0472) = 18 \times (1.0472 - 0.8660) \approx 3.2611

The chord is 2×6×sin⁡30∘=62 \times 6 \times \sin 30^\circ = 6, the arc is 6×1.0472≈6.28326 \times 1.0472 \approx 6.2832 and the height is 6×(1−cos⁡30∘)≈0.80386 \times (1 - \cos 30^\circ) \approx 0.8038.

Where it is used

The liquid in a tank lying on its side, the plank that can be cut from a log, the cross-section of an arch: wherever a straight line cuts across a circle.

Watch out

Do not confuse the segment with the sector. The sector is the slice cut from the center, with area r2θ2\dfrac{r^2\theta}{2}; the segment is that minus the triangle r2sin⁡θ2\dfrac{r^2\sin\theta}{2}. At 180∘180^\circ the segment is a half circle, and at 360∘360^\circ it is the whole circle.