# Area of Segment of Circle

## Proof of Area of Segment of Circle

• The segment of a circle is the portion of the circle enclosed by an arc and the chord of the circle.
• The sector of circle containing large arc of the circle is the major segment of circle and the sector of the circle containing small arc of the circle is the minor segment of the circle.
• The following figure shows the major and minor segment of circle.

Note:

The semicircle is the largest segment enclosed by the diameter of circle and the arc of circle.

### Derivation for the area of segment of circle:

• Let us consider the circle with center O and radius r. the chord PQ of circle forms the segment PSQP of circle.
• Let ϴ be the angle subtended by radii OP and OQ at the center.
• Now, to find the area of segment of circle we subtract the area of triangle from the area of the sector of the circle.

• Hence, area of the segment of the circle = area of sector OPSQO – area of triangle OPQ
• We know that, the area of the sector of the circle if ϴ is in radian is given by, Area of sector OPSQO = (ϴ/ 2π) *π*r2 = r2*ϴ/2

#### Now, to find the area of the triangle OPQ:

To find area of triangle OPQ we draw a perpendicular OM on PQ as shown in following figure.

In ΔPOM, <AOM = ϴ/ 2, OP = r

And in ΔBOM, <BOM = ϴ/ 2, OQ = r

Also, PM = MQ

Thus, in ΔPOM,

• Sin (ϴ/2) = PM/ PO = PM/ r

Hence, PM = r*sin (ϴ/2)

• And cos (ϴ/2) = OM/ OP = OM/r

Hence, OM = r*cos(ϴ/2)

Now, area of triangle OPQ = ½*base*height = ½*2PM*OM

= ½*2* r*sin (ϴ/2) * r*cos(ϴ/2)

= r2*sin(ϴ/2) *Cos(ϴ/2)

We know that, sin(2ϴ) = 2sinϴ*cosϴ

Hence, area of triangle OPQ = r2*(sinϴ/2)

Now, area of segment PSQMP = area of sector OPSQO – area of triangle OPQ

= r2*ϴ/2 – r2*(sinϴ/2)

= r2/2*[ϴ – sinϴ]

Hence, the area of segment of a circle when ϴ is in radian is given by,

Area of segment of a circle = r2/2*[ϴ – sinϴ]

Hence proved.

Updated: September 13, 2021 — 8:57 pm