### Segment of a Circle:

** Definition: **A region bounded by a chord and a corresponding arc lying between the chord’s endpoints is known as segment of a circle. It is to be noted that the segments do not contain the center point.

The segment of a circle divides it into two region namely major segment and minor segment. The segment having larger area is known as the **major segment** and the segment having smaller area is known as** minor segment**.

In fig. 1, ADB is the major segment and ABC is the minor segment. If nothing is stated, segment means minor segment. In order to calculate the area of segment of a circle, one should know how to calculate the area of sector of circle.

### Area of Segment of Circle:

In fig. 2, if ∠AOB=θ (in degrees), then the area of the sector AOBC (\(A_{sector ~AOBC}\)) is given by the formula

\(A_{sector ~AOBC}\) = \(\frac{θ}{360°}~\times~ πr^2\)

Let the area of \(∆AOB\) be \(A_{∆AOB}\). So, area of the segment \(ABC(A_{segment~ ABC})\) is given by

\(A_{segment~ ABC}\) = \(A_{sector~ AOBC} – A_{∆AOB}\)

\(A_{segment ~ABC}\) = \(\frac{θ}{360°}~\times~πr^2 – A_{∆AOB}\)

The area of \(∆AOB\) can be calculated in two steps, As shown in fig. 3,

Calculate the height of \(∆AOB\) i.e. \(AP\) using Pythagoras theorem as:

\(AP\) = \(\sqrt{r^{2} – \left ( \frac{AB}{2} \right )^{2}}\), if length of \(AB\) is given

or, \(AP\) = \(r~cos~\frac{θ}{2}\), if θ is given (in degrees)

Calculate the area of ∆AOB using the formula

\(A_{area ~∆AOB}\) = \(\frac{1}{2}×base×height\) = \(\frac{1}{2}×AB×AP\)

Substituting the values in area of segment formula, the area can be calculated.

**Theorems on Segment of a Circle: **

**Theorem 1: **

**Alternate Segment Theorem: **

The alternate angle is the angle made in the other sector from a chord.

Consider the alternate angle ACD (as x) is equal to the angle ABC shown on the other side of the chord, where DC is the tangent to the circle.

Triangle ABC has points A,B and C on the circumference of a circle with centre O. Join points OA and OC to form triangle AOC.

Let \(\angle ACD = x\), and \(\angle OCA = y\).

We know that tangent to a circle is at a right angle to the radius of a circle,

therefore \(x + y = 90^{\circ}\) ——————————(i)

Bisect the triangle AOC from the point O, then the triangle formed is a right-angled triangle at E. Let the bisected angle be z,

therefore \(\angle AOE = \angle COE = z\)

Sum of angles of a triangle is \(180^{\circ}\).

In \(\bigtriangleup COE\), \(y + z + 90 = 180\)

or \(y + z = 90^{\circ}\) —————————(ii)

Equating equation (i) and (ii), we have

x = z

We know, \(\angle ABC = \angle ACD = x\)

Therefore, \(\angle AOC = 2z\)

and \(\angle ABC = x\)

which implies \(\angle AOC = 2 \angle ABC\)

**Theorem 2:**

**Angle in the same segment are equal:**

Consider triangle ABC and ADC having \(\angle ABC \;\; and \;\; \angle ADC\) in the major segment of a circle.

**To prove: \(\angle ABC = \angle ADC\)**

**Construction: **Join O to A and C.

**Proof: **

Let \(\angle AOC = x\)

From **Theorem 1,** we have

\(x = 2.\angle ABC\) —————–(i)

Also, \(x = 2.\angle ADC\) ———————-(ii)

From (i) and (ii), we have

\(\angle ABC = \angle ADC\)

Lets Work Out:
Area of the sector AOB (blue region + green region) = \(\frac{\theta }{360} \times \pi r^{2} = \frac{60}{360} \times \pi \times 6^{2}\) = \(6π~cm^2\) \(Area \; of ∆AOB\) = \(\frac{1}{2}~×~OC~×~AB\) where \(OC\) = \(6~cos~30°\) = \(6 \times \frac{\sqrt{3}}{2} = 3\sqrt{3}\) cm and \(AB\) = \(2BC\) = \(2~×~6~sin~30°\) = \(2~×~6~×~\frac{1}{2}\) = \(6~cm\) Substituting the values, Area of ∆AOB = \(\frac{1}{2} \times 3\sqrt{3} \times 6 = 9 \sqrt{3} \; cm^{2}\) So, area of segment AB = \(6π~cm^2 – 9√3~cm^2\) = \(3(2π – 3√3)cm^2\) |

To solve more questions on segment of circle, download BYJU’S – The Learning App.