 Since arc AB and BC are equal.
So, ∠AOB = ∠BOC = 50°
Now,
∠AOC = ∠AOB + ∠BOC = 50° + 50° = 100°
As arc AB, arc BC and arc CD so,
∠AOB = ∠BOC = ∠COD = 50°
∠AOD = ∠AOB + ∠BOC + ∠COD = 50° + 50° + 50° = 150°
Now, ∠BOD = ∠BOC + ∠COD
∠BOD = 50° + 50°
∠BOD = 100°

The triangle thus formed, ΔAOC is an isosceles triangle with OA = OC as they are radii of the same circle.
Thus ∠OAC = ∠OCA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠AOC + ∠OAC + ∠OCA = 180°
2∠OAC + 100° = 180°     as, ∠OAC = ∠OCA
2∠OAC = 180° – 100°
2∠OAC = 80°
∠OAC = 40°

as ∠OCA = ∠OAC So,
∠OCA = ∠OAC = 40°

The triangle thus formed, ΔAOD is an isosceles triangle with OA = OD as they are radii of the same circle.
Thus, ∠OAD = ∠ODA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°
So, ∠AOD + ∠OAD + ∠ODA = 180°

#### Question 7

In the given figure, AB is a side of a regular hexagon and AC is a side of a regular eight sided polygon. Find:

(i) ∠AOB

(ii) ∠AOC

(iii) ∠BOC

(iv) ∠OBC

#### Question 8

In the given figure, O is the center of the circle and the length of arc AB is twice the length of arc BC. If ∠AOB = 100°,
find: (i) ∠BOC (ii) ∠OAC We know that when two arcs are in ratio 2: 1 then the subtended by them is also in ratio 2: 1
As given arc AB is twice the length of arc BC.
Therefore, arc AB: arc BC = 2: 1
Hence, ∠AOB: ∠BOC = 2: 1

Now given that ∠AOB = 100°.
So, ∠BOC =

Now, ∠AOC = ∠AOB + ∠BOC = 100° + 50° = 150°.
The triangle thus formed, ∠AOC is an isosceles triangle with OA = OC as they are radii of the same circle.
Thus,
∠OAC = ∠OCA as they are opposite angles of equal sides of an isosceles triangle.
The sum of all the angles of a triangle is 180°.
So, ∠COA + ∠OAC + ∠OCA = 180°
2∠OAC + 150° = 180° as, ∠OAC = ∠OCA
2∠OAC = 180° – 150°
2∠OAC = 30°
∠OAC = 15°
as ∠OCA = ∠OAC So,
∠OCA = ∠OAC = 15°.

### Exercise-17 DCircle Theorem Class-9th Concise Selina ICSE Mathematics Solutions

#### Question 1

The radius of a circle is 13 cm and the length of one of its chords is 24 cm. Find the distance of the chord from the centre.

#### Question 2

Prove that equal chords of congruent circles subtend equal angles at their centre.

#### Question 3

Draw two circles of different radii. How many points these circles can have in common? What is the maximum number of common points?

So, the circle can have 0, 1 or 2 points in common.

The maximum number of common points is 2.

#### Question 4

Suppose you are given a circle. Describe a method by which you can find the centre of this circle.

#### Question 5

Given two equal chords AB and CD of a circle with centre O, intersecting each other at point P. Prove that:

(i) AP = CP

(ii) BP = DP

#### Question 6

In a circle of radius 10 cm, AB and CD are two parallel chords of lengths 16 cm and 12 cm respectively. Calculate the distance between the chords, if they are on:

(i) the same side of the centre.

(ii) the opposite sides of the centre.

#### Question 7

In the given figure, O is the centre of the circle with radius 20 cm and OD is perpendicular to AB.

If AB = 32 cm, find the length of CD

Question 8

In the given figure, AB and CD are two equal chords of a circle, with centre O.

If P is the mid-point of chord AB, Q is the mid-point of chord CD and ∠POQ = 150°, find ∠APQ.

It is given in the question that point.

P is the mid-point of the chord AB and Point Q is the mid-point of the CD.
⇒ ∠APO = 90°      …( as the straight line drawn from the center of a  circle to bisect a chord, which is not a diameter, is at the right angle to the chord. )

As chords, AB and CD are equal therefore they are equidistant from the center i.e; PO = OQ      …( ∵ Equal chords of a circle are equidistant from the center)

Now, the ΔPOQ is an isosceles triangle with OP = OQ as its two equal sides of an isosceles triangle.

The sum of all the angles of a triangle is 180°.
⇒ ∠POQ + ∠OPQ + ∠PQO = 180°
⇒ ∠OPQ + ∠POQ + 150° = 180° …( Given: ∠POQ = 150° )
⇒ 2∠OPQ = 180° – 150°  …( As, ∠OPQ = ∠PQO )
⇒ 2∠OPQ = 30°
⇒ ∠OPQ = 15°

As ∠APO = 90°
⇒ ∠APQ + ∠OPQ = 90°
⇒ ∠APQ = 90° – 15°     ….( As, ∠OPQ = 15° )
⇒ ∠APQ = 75°.

#### Question 9

In the given figure, AOC is the diameter of the circle, with centre O.

If arc AXB is half of arc BYC, find ∠BOC.

Given :
1. AOC is the diameter.
2. Arc AXB =  Arc BYC

From Arc AXB =  Arc BYC We can see that
Arc AXB : Arc BYC = 1 : 2
⇒ ∠BOA : ∠BOC = 1 : 2

Since AOC is the diameter of the circle hence,
∠AOC = 180°
Now,
Assume that ∠BOA = x° and ∠BOC = 2x°
∠AOC = ∠BOA + ∠BOC = 180°
⇒ x + 2x = 180°
⇒ 3x = 180°
⇒ x = 60°
Hence, ∠BOA = 60° and ∠BOC = 120°.

#### Question 10

The circumference of a circle, with center O, is divided into three arcs APB, BQC, and CRA such that:
……………………..

Find ∠BOC.

then
Arc APB = 2k, Arc BQC = 3k, Arc CRA = 4k
or
Arc APB : Arc BQC : Arc CRA = 2 : 3 : 4
⇒ ∠AOB : ∠BOC : ∠AOC = 2 : 3 : 4
and therefore,
and ∠AOB = 2k°, ∠BOC = 3k°, and ∠AOC = 4k°
Now,
Angle in a circle is 360°
So, 2k + 3k + 4k = 360°
⇒ 9k = 360°
⇒ k = 40°
Hence,
∠BOC = 3 x 40° = 120°.

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