Sometimes, in a circle, we may be required study a particular portion which is bounded only its chord and the intercepted arc. The center of the circle is not considered and may not even be available or seen. For example, a shape may have a limited circular curve joined by a line segment at the end points of the curve. Such an area is called as the segment of a circle. In real life, a number of shapes can be seen as a segment of a circle and hence there is a requirement for a study about segment of a circle.
In this article let us take a closer look on the same.

## Segment of a Circle Definition

As explained earlier, a segment of a circle is a part of the area of the circle enclosed by part of its circumference and a chord. Look at the diagram shown below.

In the above diagram, the shaded is a portion of the circle bounded by the arc APB and the chord AB. The arc AQP is the remaining portion of the circle (or deemed to be a circle) with center at O. The shaded area is called the segment of the circle.

It may be noted that the center or radius of the circle are not the parts of the segment of the circle.
Thus a segment of a circle is formally defined as the part of the area of a circle bounded by a chord and the intercepted a as a corollary we can say that if the chord is a diameter, then the segment of a circle is a semicircle.

## Area of a Segment of a Circle

The area of a segment can be calculated as explained below.

The shaded area represents the area of the segment of the circle shown. It can be computed in an indirect way that, it is the difference between the area of the sector OAPB of the circle and the area of the triangle OAB.

From the center of the circle draw a perpendicular OR on AB and extend that to P, at the circumference.
As per the property of circles,  AR = RB = $\frac{c}{2}$ and angle AOP = angle BOP = $\frac{\theta}{2}$

Area of the sector OAPBO =  ($\frac{\theta}{2 \pi}$)(πr2) = [($\frac{(\theta r^2)}{2}$] = ($\frac{1}{2}$)(θr2))

Area of the triangle OAB.= ($\frac{1}{2}$)(AB)(OR) = ($\frac{1}{2}$)(2AR)(OR)

= ($\frac{1}{2}$)[2rsin($\frac{\theta}{2}$)][rcos($\frac{\theta}{2}$)]

= ($\frac{1}{2}$)[2r2)rsin($\frac{\theta}{2}$)cos($\frac{\theta}{2}$)]

= ($\frac{1}{2}$)[r2)sin(θ)]

Therefore, area of the segment of the circle As is,

As = ($\frac{1}{2}$)(θr2)) –  ($\frac{1}{2}$)[r2)sin(θ)] = ($\frac{r^2}{2}$)[(θ) – sin(θ)]

## Segment of a Circle Formula

A circle basically is governed just by its radius. But in a segment of a circle a number of parameters are involved.
The length of the chord of the segment, its distance from the center of the circle, the length of arc of the segment of the circle which depends on its central angle, the height of the segment of the circle. The area of the circle depends on all these parameters. Let us derive the formulas for the measures of all these parameters in terms of the radius and the angle subtended by the arc at the center.

Refer to the following diagram.

The shaded area is the segment of the circle shown above. It is surrounded by the arc APB and below the chord AB of the circle.
First let us derive the formula for the length of the arc of the segment of the circle. Let ‘r’ be the radius of the circle. The arc APB is a part of the circumference of the circle and directly proportional to the angle (in radians) subtended by it at the center. Let that be θ (in radians) be that angle. That is, if ‘l’ is the length of the arc,
($\frac{l}{\theta}$) = ($\frac{C}{2 \pi}$) = ($\frac{2πr}{2π}$) = r, since C = circumference of the circle = 2πr
or, l = r $\theta$
Next let us try to establish a formula for ‘c’, the length of the chord.
From the center of the circle draw a perpendicular OR on AB and extend that to P, at the circumference.
As per the property of circles,  AR = RB = $\frac{c}{2}$ and angle AOP = angle BOP = $\frac{\theta}{2}$
Consider the triangle AOR or BOR.
($\frac{c}{2r}$) = sin ($\frac{\theta}{2}$)  or, c = 2r $\times$ sin ($\frac{\theta}{2}$)

We can derive another formula  for ‘c’.  Applying cosine law for the triangle OAB,
AB2 = OA2 + OB2 – 2 (OA)(OB)cos θ
or, c2 = r2 + r2 – 2 (r)(r)cos θ = r2 + r2 – 2 r2cos θ = 2 r2 – 2 r2cos θ = r2(2 - 2 cos θ)
or, c = r√(2 – 2 cos θ)
Thus, c = 2r * sin ($\frac{\theta}{2}$) = r√(2 – 2 cos θ)

Let us now derive a formula for ‘h’, the height of the segment of a circle.
Consider again the triangle AOR or BOR.
OR (r) = cos ($\frac{\theta}{2}$), or  d = r $\times$ cos ($\frac{\theta}{2}$)
Now, h = RP = OP – OR = r - r $\times$ cos ($\frac{\theta}{2}$) = r[1 - cos ($\frac{\theta}{2}$)]
The height ‘h’ can also be expressed in terms of ‘r’ and ‘c’.
In the triangle AOR or BOR, d2 = r2 – ($\frac{c^2}{4}$)
or, d = √[ r2 – ($\frac{c^2}{4}$)]
so, h = r – d = r - √[ r2 – ($\frac{c^2}{4}$)]
Thus, h = r[1 - cos ($\frac{\theta}{2}$)] = r - √[ r2 – ($\frac{c^2}{4}$)]

Next, let us determine the angle θ.
(d) = r $\times$ cos ($\frac{\theta}{2}$),
or, ($\frac{\theta}{2}$) = cos-1($\frac{d}{2}$)
or, θ = 2cos-1($\frac{d}{r}$)

Sometimes, a segment of a circle alone may be available or described. In such a case we may need to find the radius of the circular arc. We can do that by a formula that is derived below.
The obvious formula is r = d + h, but we can also derive a formula in terms of ‘c’ and ‘h’. It will be more useful because ‘d’ may not be known in most cases when ‘r’ is not known.
Again in the triangle AOR or BOR,
r2 – d2 = ($\frac{c^2}{4}$)

or,  r2 – (r – h)2 = ($\frac{c^2}{4}$)

r2 – (r2 - 2r h + h2) = ($\frac{c^2}{4}$)

2rh - h2 = ($\frac{c^2}{4}$)

2r - h = ($\frac{c^2}{4h}$)

2r = h + ($\frac{c^2}{4h}$) = [$\frac{4h^2 + c^2}{4h}$]

or, r = [$\frac{4h^2 + c^2}{8h}$]

## Segment of a Circle Examples

### Solved Examples

Question 1: A segment of a circle is formed by a chord and arc of a circle of 20in. radius. If the central angle of the arc is 60o, find the length of the chord.
Solution:

The chord length ‘c’ of a segment of a circle is given by the formula,
c = 2r sin($\frac{\theta}{2}$), where θ is the central angle of the arc of the segment and ‘r’ is the radius.
Given that, θ = 60o and the radius = $\frac{20}{2}$ = 10 in. Therefore,
c = 2(10)sin ($\frac{60^o}{2}$) = 20 sin (30o) = 20($\frac{1}{2}$) = 10 in.

Question 2: In the segment of the circle described in the previous problem find the height of the segment.
Solution:

The height ‘h’ of a segment of a circle is given by the formula,
h = r[1 – cos($\frac{θ}{2}$)], where θ is the central angle of the arc of the segment and ‘r’ is the radius.
Given that, θ = 60o and the radius = $\frac{20}{2}$ = 10 in. Therefore,
h = r[1 – cos($\frac{\theta}{2}$)] = 10[1 – cos(30o)] = 10[1 – ($\frac{\sqrt{3}}{2}$] in. = (5)[2 - √3] in. ≈ 1.34 in.

Question 3: In a segment of a circle, having a radius of 6 in., the arc subtends an angle of 45o at the center. Find the length of the arc.
Solution:

The length ‘l’ of an arc of the segment of a circle is,
l = r $\times$ θ, where ‘θ’ is the central angle in radians and ‘r’ is the radius of the segment of the circle.
Given, θ = 45o = $\frac{\pi}{4}$ radians and r = 6 in.
Therefore, l = (10) $\times$ ($\frac{\pi}{4}$) ≈ 7.85 in.

Question 4: The diameter of a circular garden is 40 m. At 10 m from the center of the garden there is a pathway across the circle and a lawn for the rest of the area. Find the angle subtended by the pathway at the center of the circle.
Solution:

The pathway described in the problem is a chord and the lawn is the segment of the circle. The angle subtended by the pathway can be found from the formula,
θ = 2cos-1($\frac{d}{r}$), where, ‘d’ is the distance of the pathway from the center and ‘r’ is the diameter of the circular park.
Given, d = 10 m and r = $\frac{40}{2}$ = 20 m.
Therefore, θ = 2cos-1($\frac{d}{r}$) = 2cos-1($\frac{10 m}{20 m}$) = 2cos-1(0.5) = 2 (60o) = 120o

Question 5: A cam plate, in the shape of a segment of a circle has a width of 20cm and a height of 5 cm.
Solution:

If the cam plate is to be cut from a circular plate, what is the radius of the circular plate?
The radius of the required circular plate is the radius ‘r’ of the segment of the circle having a height ’h’ of 5cm and chord length ‘c’ of 20 cm. It can be found by using the formula,
r = [$\frac{4h^2 + c^2}{8h}$]
or, r = [$\frac{4{(5)}^2 + {(20)}^2}{4(5)}$] cm = $\frac{[100 + 400]}{(20)}$ cm = 25 cm.
Thus, the radius of the required circular plate is 25 cm.

Question 6: The shape of the garden area of a hotel is a circle of radius of 40 ft. At 25 feet from the center one side, a swimming pool is built for the remaining area. What is the area of the swimming pool?
Solution:

The swimming pool described in the problem is a segment of a circle. The angle θ subtended at the center of the garden can be found from,
θ = 2cos-1($\frac{d}{r}$), where ‘d’ is the distance of the chord of the segment = 25 feet and r = radius  of the garden = 40 ft.
So, θ = 2cos-1($\frac{25 ft}{40 ft}$) = 2cos-1($\frac{5}{8}$) ≈ 1.32 radians
The area As of the swimming pool is given by the formula
As = ($\frac{r^2}{2}$)[(θ) – sin(θ)]
= ($\frac{40^2}{2}$)[(1.32) – sin(1.32)] ≈ ($\frac{1600}{2}$)[(1.32) – (0.97)] = 280 sq. ft
So the area of the swimming pool is about 280 sq. ft.