In the previous section we saw theorem 27.5 and 27.6. We also saw some solved examples. In this section we will learn about Segments.
• Consider the chord AB in fig.27.36(a) below. It divides the circle into two arcs. Arc AXB and AYB.
• Consider the arc AXB and chord AB. We get an area enclosed between them. This area is a segment of the circle.
• There is another enclosed area also present. It is between the chord AB and arc AYB. It is also a segment.
• So we can define a segment of a circle as follows:
■ A segment of a circle is the area enclosed between a chord and it’s arc.
• The difference between a segment and a sector can be seen from figs.27.36 b and c. We have earlier seen the details about sectors in chapter 21.8.
■ Just as every arc have an alternate arc, every segment will have an alternate segment. In the fig.27.36(b), segment AYB is the alternate segment of segment AXB and vice versa.
∠AQB will be equal to half the central angle of arc AXB
∠ARB will be equal to half the central angle of arc AXB
3. We find that all three angles are equal to half the central angle of arc AXB
• That means, all the three angles are equal
■ In fact, what ever number of angles we draw in the sector AYB, they will all have the same value
1. Now consider fig(b).
2. Some angles are drawn inside the sector AXB. Let us find the values of those angles:
Using theorem 27.4,
∠AKB will be equal to half the central angle of arc AYB
∠ALB will be equal to half the central angle of arc AYB
∠AMB will be equal to half the central angle of arc AYB
3. We find that all three angles are equal to half the central angle of arc AYB
• That means, all the three angles are equal
■ In fact, what ever number of angles we draw in the sector AXB, they will all have the same value
1. Consider a segment and its chord AB
2. Mark any number of points on the arc of the segment
3. Join all those points to A and B
4. All the angles so formed will be equal
5. They are all equal to 'half of the central angle of the alternate arc'
■ So we can say this:
• Every segment will have a particular value of angle.
• We will call it the unique angle of a segment
• The unique angle of any segment is equal to 'half of the central angle of the alternate arc'
6. Let us find the sum of c⁄2 and d⁄2:
We get: c⁄2 + d⁄2 = (c+d)⁄2
7. But (c+d) is the total angle at centre O. The total angle at any point is 360o.
That means, (c + d) = 360o.
8. So we can write: c⁄2 + d⁄2 = (c+d)⁄2 = 360⁄2 = 180o
• That means the two angles are supplementary
1. Consider any segment. It will have an unique angle
2. The alternate segment will also have it's own unique angle
3. The sum of the two unique angles will always be 180o
4. That is., the unique angles of alternate segments are always supplementary.
So far, we know the following two points:
• Every segment has it's own unique angle
• This unique angle is equal to 'half of the central angle of the alternate arc'
■ That means, to find the unique angle, we must first know the 'central angle of the alternate arc'.
Would it not be better, if we do not need to know the 'central angle of the alternate arc'?
Let us try:
1. Consider the segment AYB in fig.27.39(a) below. Mark a point P any where on the arc AYB
2. We have seen that, wherever we mark P, the angle APB will be the same, and it is the unique angle of the segment AYB. Let us denote this unique angle as uo.
3. In fig(b), the central angle of 'it's own arc', the arc AYB is marked as do.
• We want the relation between u and d
4. Let the central angle of the alternate arc AXB be co. Then, by converse of theorem 27.4, we get:
c = 2u. This is shown in fig(c).
5. Now, c and d are angles around a point. So there sum will be 360o
• Thus we get: c+d = 360 ⇒ d = 360 – c ⇒ d = 360 – 2u
• This is the required relation.
■ That is., if we know the unique angle (uo) of a segment, the 'central angle of the arc of that segment' will be equal to (360 -2u)o
2. We have seen that, wherever we mark M, the angle AMB will be the same, and it is the unique angle of the segment AMB. Let us denote this unique angle as vo.
3. In fig(b), the central angle of 'it's own arc', the arc AXB is marked as co.
• We want the relation between v and c
4. Let the central angle of the alternate arc AYB be co. Then, by converse of theorem 27.4, we get:
d = 2v. This is shown in fig(c).
5. Now, c and d are angles around a point. So there sum will be 360o
• Thus we get: c+d = 360 ⇒ c = 360 – d ⇒ c = 360 – 2v
• This is the required relation.
■ That is., if we know the unique angle (vo) of a segment, the 'central angle of the arc of that segment' will be equal to (360 -2v)o
• Consider the chord AB in fig.27.36(a) below. It divides the circle into two arcs. Arc AXB and AYB.
 |
| Fig.27.36 |
• There is another enclosed area also present. It is between the chord AB and arc AYB. It is also a segment.
• So we can define a segment of a circle as follows:
■ A segment of a circle is the area enclosed between a chord and it’s arc.
• The difference between a segment and a sector can be seen from figs.27.36 b and c. We have earlier seen the details about sectors in chapter 21.8.
■ Just as every arc have an alternate arc, every segment will have an alternate segment. In the fig.27.36(b), segment AYB is the alternate segment of segment AXB and vice versa.
Now we will learn about the angles inside a segment.
1. Consider fig.27.37(a) below. The chord AB makes two arcs AXB and AYB.
 |
| Fig.27.37 |
2. Some angles are drawn inside the sector AYB. Let us find the values of those angles:
Using theorem 27.4,
∠APB will be equal to half the central angle of arc AXB∠AQB will be equal to half the central angle of arc AXB
∠ARB will be equal to half the central angle of arc AXB
3. We find that all three angles are equal to half the central angle of arc AXB
• That means, all the three angles are equal
■ In fact, what ever number of angles we draw in the sector AYB, they will all have the same value
1. Now consider fig(b).
2. Some angles are drawn inside the sector AXB. Let us find the values of those angles:
Using theorem 27.4,
∠AKB will be equal to half the central angle of arc AYB
∠ALB will be equal to half the central angle of arc AYB
∠AMB will be equal to half the central angle of arc AYB
3. We find that all three angles are equal to half the central angle of arc AYB
• That means, all the three angles are equal
■ In fact, what ever number of angles we draw in the sector AXB, they will all have the same value
Based on the above results, we can write the following theorem:
Theorem 27.7:1. Consider a segment and its chord AB
2. Mark any number of points on the arc of the segment
3. Join all those points to A and B
4. All the angles so formed will be equal
5. They are all equal to 'half of the central angle of the alternate arc'
■ So we can say this:
• Every segment will have a particular value of angle.
• We will call it the unique angle of a segment
• The unique angle of any segment is equal to 'half of the central angle of the alternate arc'
Now we will see the relationship between two unique angles
1. Consider segment AXB in fig.27.38(a) below:
 |
| Fig.27.38 |
2. An ∠AMB is drawn inside the segment. We know that, where ever be the position of M along the arc AXB, The ∠AMB will be equal to the unique angle of the segment AXB.
3. Also, by theorem 27.4, ∠AMB will all be equal to half the central angle of the alternate arc AYB.
4. That is., the unique angle of segment AXB will be equal to d⁄2 as shown in fig(b)
5. In a similar way, the unique angle of segment AYB will be equal to c⁄2 6. Let us find the sum of c⁄2 and d⁄2:
We get: c⁄2 + d⁄2 = (c+d)⁄2
7. But (c+d) is the total angle at centre O. The total angle at any point is 360o.
That means, (c + d) = 360o.
8. So we can write: c⁄2 + d⁄2 = (c+d)⁄2 = 360⁄2 = 180o
• That means the two angles are supplementary
We can write the above result in the form of a theorem:
Theorem.27.8:1. Consider any segment. It will have an unique angle
2. The alternate segment will also have it's own unique angle
3. The sum of the two unique angles will always be 180o
4. That is., the unique angles of alternate segments are always supplementary.
• Every segment has it's own unique angle
• This unique angle is equal to 'half of the central angle of the alternate arc'
■ That means, to find the unique angle, we must first know the 'central angle of the alternate arc'.
Would it not be better, if we do not need to know the 'central angle of the alternate arc'?
Let us try:
1. Consider the segment AYB in fig.27.39(a) below. Mark a point P any where on the arc AYB
 |
| Fig.27.39 |
3. In fig(b), the central angle of 'it's own arc', the arc AYB is marked as do.
• We want the relation between u and d
4. Let the central angle of the alternate arc AXB be co. Then, by converse of theorem 27.4, we get:
c = 2u. This is shown in fig(c).
5. Now, c and d are angles around a point. So there sum will be 360o
• Thus we get: c+d = 360 ⇒ d = 360 – c ⇒ d = 360 – 2u
• This is the required relation.
■ That is., if we know the unique angle (uo) of a segment, the 'central angle of the arc of that segment' will be equal to (360 -2u)o
Let us see if this is true for the other arc also:
1. Consider the segment AXB in fig.27.40(a) below. Mark a point M any where on the arc AXB
 |
| Fig.27.40 |
3. In fig(b), the central angle of 'it's own arc', the arc AXB is marked as co.
• We want the relation between v and c
4. Let the central angle of the alternate arc AYB be co. Then, by converse of theorem 27.4, we get:
d = 2v. This is shown in fig(c).
5. Now, c and d are angles around a point. So there sum will be 360o
• Thus we get: c+d = 360 ⇒ c = 360 – d ⇒ c = 360 – 2v
• This is the required relation.
■ That is., if we know the unique angle (vo) of a segment, the 'central angle of the arc of that segment' will be equal to (360 -2v)o
• So we proved the relation in both cases
Now we will see a solved example
Solved example 27.14
In the fig.27.41(a) below, ΔABC is equilateral and O is it's circumcentre.
Prove that, the length of AD is equal to the radius of the circle.
 |
| Fig.27.41 |
Solution:
• In the given problem, ABC is an equilateral triangle. And O is the circumcentre of that triangle
• O is joined to vertex C. So OC is a radius
• This OC is extended until it meets the circle at D. So CD is a diameter and OD is a radius
• Point D is then joined to A.
• We have to prove that, AD is equal to the radius of the circle
1. Consider the three segments: ADB, BYC and CXA. [fig.27.41(b)]
• Since ΔABC is equilateral, AB = BC = CA
• That means, the three segments are equal
• So they have the same central angle
2. Join A and B to O. (Note that C is already joined to O)
• OA, OB and OC shows the central angles of the three segments
• It is clear that each of the three segments have a 1⁄3 share at the centre
• So ∠AOB = ∠BOC = ∠COA = 1⁄3 × 360 = 120o.
3. Now consider fig(c)
• B is joined to D by a red dotted line. This gives us ∠ABD
• This ∠ABD is the unique angle of segment ABD. The segment ABD has a central angle of 120o.
• So 120 = [360 - (2 ×∠ABD)] ⇒ 2×∠ABD = 360-120 ⇒ 2×∠ABD = 240 ⇒ ∠ABD = 120o.
4. AB is a chord and OD is a radius. So by theorem 17.1, OD is the perpendicular bisector of chord AB.
So we get: ∠BDO = ∠ADO = 1⁄2×120 = 60o.
5. Now consider ΔOAD
• We have OA = OD (∵ radii of same circle)
• So ΔOAD is isosceles. The base angles are equal.
• From (4) we already have one base angle ∠ADO = 60o
• So the other base angle ∠DAO = 60o
• If these two angles are 60o each, the third angle ∠DOA is also 60o. (∵ 180 -60 -60 = 60)
• So we find that all the three angles in OAD is 60o
• It is an equilateral triangle. All three sides are equal
• Thus we get AD = AO
• In the given problem, ABC is an equilateral triangle. And O is the circumcentre of that triangle
• O is joined to vertex C. So OC is a radius
• This OC is extended until it meets the circle at D. So CD is a diameter and OD is a radius
• Point D is then joined to A.
• We have to prove that, AD is equal to the radius of the circle
1. Consider the three segments: ADB, BYC and CXA. [fig.27.41(b)]
• Since ΔABC is equilateral, AB = BC = CA
• That means, the three segments are equal
• So they have the same central angle
2. Join A and B to O. (Note that C is already joined to O)
• OA, OB and OC shows the central angles of the three segments
• It is clear that each of the three segments have a 1⁄3 share at the centre
• So ∠AOB = ∠BOC = ∠COA = 1⁄3 × 360 = 120o.
3. Now consider fig(c)
• B is joined to D by a red dotted line. This gives us ∠ABD
• This ∠ABD is the unique angle of segment ABD. The segment ABD has a central angle of 120o.
• So 120 = [360 - (2 ×∠ABD)] ⇒ 2×∠ABD = 360-120 ⇒ 2×∠ABD = 240 ⇒ ∠ABD = 120o.
4. AB is a chord and OD is a radius. So by theorem 17.1, OD is the perpendicular bisector of chord AB.
So we get: ∠BDO = ∠ADO = 1⁄2×120 = 60o.
5. Now consider ΔOAD
• We have OA = OD (∵ radii of same circle)
• So ΔOAD is isosceles. The base angles are equal.
• From (4) we already have one base angle ∠ADO = 60o
• So the other base angle ∠DAO = 60o
• If these two angles are 60o each, the third angle ∠DOA is also 60o. (∵ 180 -60 -60 = 60)
• So we find that all the three angles in OAD is 60o
• It is an equilateral triangle. All three sides are equal
• Thus we get AD = AO
Some additional solved examples can be seen here.
In the next section, we will see Cyclic quadrilaterals.
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