# Chapter – 7 : Theorems Related To Angles In A Circle

 Book Name : Ganit Prakash Subject : Mathematics Class : 10 (Madhyamik) Publisher : Prof Nabanita Chatterjee Chapter Name : Compound Interest And Uniform Rate Of Increase Or Decrease (7th Chapter)

## LET US DO – 7.1

Question 1

In the adjoining figure, ∠AMB, formed by the circular arc \widehat{APB}, is ____________ angle and ∠ANB, formed by circular arc ______________ is front angle of the circle.

Solution:

In the adjoining figure, ∠AMB formed by the circular arc \widehat{APB} is cyclic angle and ∠ANB formed by circular arc AQB is front angle of the circle.

Question 2

∠SLT in the adjoining figure at the point L by the chord ______________ is front angle. Again, since the point L lies on the circle, so angle formed by the chord ST is a front angle in the segment. Again, ∠SLT formed by the circular arc ______________ is a front angle of the circle. The four angles in the segment of a circle with centre O in the adjoining figure are ______________, ______________, ______________ and ______________.

Solution:

∠SLT in the adjoining figure at the point L by the chord ST is front angle.

Again, since the point L lies on the circle, so angle SLT formed by the chord ST is a front angle in the segment.

Again, ∠SLT formed by the circular arc SNT is a front angle of the circle. The four angles in the segment of a circle with centre O in the adjoining figure are

We see in no. (i), (ii) circles the angle at the centre formed by the arc AQB is ∠AOB and an angle in the segment is ∠APB. But in no. (iii) circle, the

angle at the centre formed by the arc ASB is ∠AOB and the angle in the segment formed by the arc ASQ is ∠APQ.

## LET US DO – 7.2

Question 1

I drew a circle and I drew two angles in the segment in that circle which (a) are formed with same arc; (b) are not formed with same arc.

Solution:

Question 2

I drew angles at the centre and the angle in the segment which (a) are formed with same arc; (b) are not formed with the same arc.

Solution:

(a) In fig (i), angle at the centre O} and angle in the segment are formed with same arc.

(b) In fig (ii) angle at the centre & angle in the segment are not formed with same arc.

Question 3

Let us look at the picture and let us give answer (0 is the centre of the circle) :

Solution:

(a) : (i) ∠AOB, is the angle at the centre on the arc APB.

(ii) ∠APB in the segment is formed with arc AQB

(iii) In fig (iii) ∠ADB is cyclic formed with arc AQB

(iv) In fig (iv) ∠ACB is cyclic formed with arc APB

(v) In fig (v) ∠ADB is not an angle in the segment formed with the arc AQB.

In fig. (i) two angles ∠ACB & ∠ADB are in the same arc.

In fig. (ii) ∠ACB & ∠ADB are two angles that lie in the segment ADCB.

Application 1.

I drew a circle with centre P and by drawing an angle ∠A P B at the centre and an angle ∠AQB at the point on the circle by the arc ACB, we shall prove that, ∠APB = 2 ∠AQB [Let me do it myself].

Solution:

Let on the arc ACB of the circle with centre P, ∠APB is the angle the centre & ∠AQB is the angle on the circumference. To prove, ∠APB = 2 ∠AQB.

Join Q, P and produce to R.

Proof: In ∆APQ, AP = PQ (Radii of same circle)

\therefore ∠PAQ = ∠AQP

Again, QP is produced to R.

\therefore External ∠APQ = ∠PAQ + ∠AQP = ∠2AQP

similarly, From ∠BQP, ∠BPR = 2 ∠BQP

\therefore ∠APB = ∠APR + ∠BPR = 2(∠AQP + ∠BQP})

= ∠AQB Proved

Application 3

The circle with centre O passes through three points A, B and C; if ∠ABO = 35° and ∠ACO = 45°, let us write by calculating the value of ∠BOC.

In ∆OAB, OA = OB(radii of same circle) OAB = \therefore ∠OBA = 35°

In ∆OAC, OA = OC (radii of same circle) OAC = \therefore ∠OCA = 45° [Let me do it myself]

Solution:

∠BAC = 35° + 45° = 80°

Application 5

Let us write by calculating the value of x and y From the two figures given below. [Let me do it myself]

Solution:

In fig. (i) ∠OAC = ∠OCA = \frac{180° - 140°}{2} = 20°

\therefore ∠OBA = ∠OAB} = \frac{180 - 100}{2} = 40°

\therefore x = ∠BAC} = 20° + 40° = 60°

In fig. (ii) ∠AOB = 360° – 120° = 240°

y° = ∠APB = \frac{1}{2} × 240° = 120°

Application 6

Solution:

BX = BY (Radius of equal circles)

Application 7

ABC is an isosceles triangle with sides AB = AC; we draw ∆DBC in such a way that ∆DBC and ∆ABC lie on the same side of BC and ∠BAC = 2 ∠BDC. Let us prove that the circle drawn with centre A and with radius AB is passing through the point D, i.e., the point D lies on the circle. [Let me do it myself]

Solution:

Proof: Let the circle is not passing through D. If cuts BD at D

Join C, D.

Now, ∠BAC is the angle at the centre of the circle with centre A and ∠BDC is the angle on the circumference on the same arc.

\therefore ∠BAC = 2 ∠BDC

But ∠BAC = 2 ∠BDC (given)

\therefore ∠BDC = ∠BDC

It is not possible until the point D. Coincide with D as an external angle of a triangle cannot be equal with its internal opposite angle.

\therefore The point D is on the circle.

## LET US WORK OUT – 7.1

Question 1

O is the circumcentre of the isoscles triangle ABC, whose AB = AC, the point and B, C are on opposite sides of the centre O. If ∠BOC = 100°, let us write b calculating the values of ∠ABC and ∠ABO.

Solution:

‘ O ‘ is the circumcentre of the issosceles triangle ABC, where AB = AC.

Join OB & OC.

Proof: \therefore ∠BOC is the angle at the centre & ∠B AC is the angle at the circumference on the same arc BC.

\therefore ∠BAC = \frac{1}{2} ∠BOC = \frac{1}{2} × 100° = 50°

Again, as AB = AC, \therefore ∠ABC = ∠ACB

In ∆ABC,

∠BAC + ∠ABC + ∠AC B = 180° or, 50° + ∠ABC + ∠ABC = 180°

\therefore 50° + 2 ∠ABC = 180°

\therefore 2 ∠ABC = 180° – 50° = 130°

\therefore ∠ABC = \frac{130°}{2} = 65°

Again, in ∆OBC, OB = OC (Radii of same circle)

\therefore ∠OBC = ∠OCB

In ∆BOC, ∠BOC + ∠OBC + ∠OCB = 180°

or, 100° + 2 ∠{OBC} = 180°

\therefore 2 ∠OBC = 180° – 100° = 80°

\therefore ∠OBC = \frac{1}{2} × 80° = 40°

Now, ∠ABO = ∠ABC – ∠O B C = 65° – 40° = 25°

\therefore ∠ABC = 65° & ∠ABO = 25°

Question 2

In the adjoining figure, if O is the centre of circumcircle of ∆ABC and ∠AOC 110°; let us write by calculating the value of ∠ABC

Solution:

In the fig. ∠AOC = 110°.

\therefore Reflex ∠AOC = 360° – 110° = 250°

\therefore ∠ABC = \frac{1}{2} of the angle at the centre

= \frac{1}{2} × reflex AOC

= \frac{1}{2} × 250° = 125°

Question 3

ABCD is A Cyclic quadrilateral of A Circle with centre O; DC is extended to the point P. If ∠BCP = 108°, let us write by calculating the value of ∠BOD.

Solution:

ABCD is A Cyclic quadrilateral with centre of the circle O.

Side DC is produced to P. such that ∠BCP = 108°

Find ∠BOD.

Ans. ∠BCD + ∠BCP = 180°

∠BCD = 180° – ∠BCP = 180° – 108° = 72°

∠BOD = 2 × ∠BCD

= 2 × 72° = 144°

& Reflex ∠BOD = 360° – 144° = 216°.

Question 4

O is the centre of the circle; ∠AOD = 40° and ∠ACB = 35°; let us write by calculating the value of ∠BCO and ∠BOD, and answer with reason.

Solution:

O is the centre of the circle ∠AOD = 40° & ∠ACB = 35°.

Find ∠BCO & ∠BOD.

Ans: Produce CD, which cuts the circle at E.

∠AOE is the angle at the centre and ∠ACE is the angle at the circumference on the same arc AE.

\therefore ∠AC E = \frac{1}{2} A O E = \frac{1}{2} × 40° = 20°

Now, ∠BCO = ∠BCA + ∠ACO = ∠ACB + ∠ACE = 35° + 20° = 55°.

Again, ∠AOB is the angle at the centre & ∠ACB is the angle at the circumference.

\therefore ∠AOB = 2 × ∠ACB = 2 × 35° = 70°

∠BOD = ∠AOB + ∠AOD = 70° + 40° = 110°

\therefore ∠BCO = 55° & ∠BOD = 110°

Question 5

O is the centre of circle in the picture beside, if ∠APB = 80°, let us find the sum of the measures of ∠AOB and ∠COD and answer with reason.

Solution:

‘O’ is the centre of the circle & ∠ADB = 80°.

To find the sum of ∠AOB + ∠COD.

Join B, C.

As ∠AOB is the angle at the centre and ∠DBC is the angle at the circumference on the same arc AB.

\therefore ∠AOB = 2 × ∠ACB

Again, ∠COD is the angle at the centre and ∠DBC is the angle at the circumference on the same arc CD.

∠COD = 2 × ∠CBD

∠AOB + ∠COD = 2 ×(∠ACB + ∠CBD) ———- (i)

Now, in ∆BCP, the side CP is produced to A.

\therefore External ∠BPA = ∠PCB + ∠PBC

∠ACB = ∠ACB + ∠DBC

∠ACB + ∠DBC = ∠APB = 80° ———- (ii)

Now, From (i) & (ii), ∠AOB + ∠COB = 2 ×(∠ACB + ∠DBC) = 2 × 80° = 160°.

Question 6

Like the adjoining figure, we draw two circles with centres C and D which inter. sect each other at the points A and B. We draw a straight line through the point A which intersects the circle with centre C at the point P and the circle with centre D at the point Q. Let us prove that :

(i) ∠PBQ = ∠CAD; (ii) ∠BPC = ∠BQD.

Solution:

Two circles with centres C & D intersect each other at A & B.

A straight line passing through A, cuts the circles at P & Q respectively.

To prove : \therefore (i) ∠PBQ = ∠CAD & (ii) ∠BPC = ∠BQD.

Join A, B; A, C; P, C; A, D; D, Q; B, P; B, Q; B, C; B, D.

In the circle with centre C,

∠ACP is the angle at the centre &

∠ABP is the angle at the circumference on the same arc AP.

∠ABP = \frac{1}{2} ∠AC P and ∠CAP = ∠CPA [as, CA = CP (same radius)]

Now, in ∠CAP; ∠CAP + ∠CPA + ∠ACP = 180°

or, 2 ∠CAP + ∠ACP = 180°

or, 2 ∠CAP = 180° – ∠ACP

\therefore ∠CAP = 90° - \frac{1}{2} ∠ACP = 90° - ∠ABP

\therefore ∠CAP = 90° - ∠ABP ---------- (i)

Again, in the circle with centre D,

∠ADQ is the angle at the centre and ∠ABQ is the angle at the circumference on the same arc AQ.

\therefore ∠DAQ = ∠DQA

\therefore ∠DAQ + ∠DQA = 2 ∠DAQ

or, ∠ADQ + 2 ∠DAQ = 180°

or, 2 ∠DAQ = 180° - ∠ADQ

\therefore ∠DAQ = 90° - \frac{1}{2} ∠ADQ = 90° - ABQ

\therefore ∠DAQ = 90° - ∠ABQ ---------- (ii)

∠CAP + ∠DAQ = 90° - ∠ABP + 90° - ∠ABQ

= 180° - (∠ABP + ∠ABQ) = 180° - ∠PBQ.

or, ∠PBQ = 180° - (∠CAP + ∠DAQ) ∠PBQ = ∠CAD Proved.

Again, In ∆ABC, CA = CB (Radii of same circle)

∠CAB = ∠CB A

& In ∠DAB, DA = DB (Radii of same circle)

∠DAB = DBA

∠CAB + ∠DAB = ∠CBA + ∠DBA

∠CBD + ∠PBD = ∠PAD + ∠PBQ

or, ∠CBP = ∠DBQ [as BC = PC (Radii of same circle)] and BD = DQ

∠BPC = ∠CBP and ∠DBQ = ∠BQD

∠BPC = ∠CBP = ∠DBQ = ∠BQD

∠BPC = ∠BQD Proved.

Question 7

If the circumcentre of triangle ABC is 0; let us prove that ∠OBC + ∠BAC = 90°.

Solution:

O is the circumcentre of the triangle ∠ABC.

To prove, ∠OBC + ∠BAC = 90°

Ans. BOC = 2 BAC

\therefore ∠B AC = \frac{1}{2} ∠BOC

In ∆OBC, ∠OBC = ∠OCB

∠OBC + ∠OCB + ∠BOC = 180°

\therefore 2 ∠OBC + ∠BOC = 180°

or, 2 ∠QBC = 180° - ∠BOC

or, ∠OBC = 90° = \frac{1}{2} ∠BOC

= 90 - ∠BAC

\therefore ∠OBC + ∠BAC = 90° Proved.

Question 8

Each of two equal circles passes through the centre of the other and the two circles intersect each other at the points A and B. If a straight line through the point A intersects the two circles at points C and D, let us prove that ∆BCD is an equilateral triangle.

Solution:

Let P & Q are the centres of two equal circles. They cut each other at A and B.

Join, A, P; A, Q; B, P; B, Q; A, B; P, Q.

\therefore AP = BP = ∆AB = AQ = PQ

∆APQ & BQ are equilateral triangles.

\therefore ∠PAQ = ∠APQ = ∠AQP = ∠PBQ = ∠BPQ = ∠BQP = 60°.

Now, ∠APB = ∠APQ + ∠BPQ = 60° + 60° = 120°

∠AQB = ∠AQP + ∠PQB = 60° + 60° = 120°

\therefore ∠APB = ∠AQB = 120°

In the circle with centre P, APB is the angle at the centre & ADB is the angle at the circumference on the same arc AQB.

\therefore ∠ADB = \frac{1}{2} ∠{APB} = \frac{1}{2} × 120° = 60°

\therefore ∠CDB = 60°

Again, in the circle with centre Q, ∠APB & ∠ACB are the angles at the circumference on the same arc ∠ACB = ∠APB = 120°

\therefore ∠BCD = 180° - ∠AC B = 180° - 120° = 60°

\therefore In ∆BCD, ∠CDB = ∠BCD = 60°

Remaining ∠DBC = 180° - (∠CDB} + ∠BCD)

= 180° - (60° + 60°) = 60°

\therefore All the angles of ∠BCD are equal.

\therefore ∆BCD is an equilateral triangle.

Question 9

S is the centre of the circumcircle of ∆ABC and if AD ⊥ BC, let us prove that ∠BAD = ∠SAC.

Solution:

S is the circumcentre of ∆ABC & AD ⊥ BC.

Join, S, A & S, C.

Proof: SA = SC (Radii of same circle)

Again, in ∆SAC, AS = SC

∠SAC = ∠SCA

Again, in ∆SAC,

∠ASC + ∠SAC + ∠SCE = 180°

or, ∠ASC + 2 ∠SAC = 180°[\because ∠SAC = ∠SCA]

\therefore 2 ∠SAC = 180° – ∠ ASC

or ∠SAC = 90° – \frac{1}{2} ∠ASC ———- (i)

Again, in the circle with centre S, ∠ASC is the angle at the centre & ∠ABC is the ang at the circumference on the same arc ∠AKC.

\therefore ∠ABC = \frac{1}{2} ∠ASC ———- (ii)

∠SAC = 90° – ∠ABC ———- (iii)

or, ∠BAD = 90° – ∠ADB = 90° – ∠ABC ———- (iv)

\therefore From (iii) & (iv), ∠SAC = ∠BAD Proved.

Question 10

Two chords AB and CD of a circle with centre O intersect each other at the point P, let us prove that ∠AOD + ∠BOC = 2 ∠BPC.

If ∠AOD and ∠BOC are supplementary to each other, let us prove that the th chords are perpendicular to each other.

Solution:

Two chords AB & CD of circle with centre O, intersect each other at P.

To prove,

∠AOD + ∠BOC = 2 BPC

If ∠AOD & BOC are supplementary to each other, then prove that the chords are perpendicular to each other.

Join, O, A; O, B; O, C; O, D; & B, D.

In the circle with centre O, AOB is the angle at the centre & ABD is the angle at the circumference, in the same arc AKD.

\therefore ∠ABD = \frac{1}{2} ∠A OD \\

or, ∠AOD = 2 × ∠ABD

Again, in the circle with centre O, ∠BOC is the angle at the centre and ∠BDC is the angle at the circumference on the same arc ∠BLC.

∠BOC = 2 × ∠BDC

∠AOD + ∠BOC = 2(∠ABD + ∠BDC) = (∠PBD + ∠BDP) ———- (i)

Now in the triangle PBD, Ext. ∠BPC

∠BPC = ∠PBD + ∠BDP

From (i), ∠AOD + ∠BOC = 2 ∠BPC (Proved 1st part)

Again, if ∠AOD + ∠BOC = 2 right angles

2 × ∠BPC = ∠AOD + ∠BOC = 2 right angles

∠BPC = 90°

\therefore BP ⊥ PC, i.e., Two chords AB & CD are perpendicular to each other. Proved – 2nd part.

Question 11

If two chords AB and CD of a circle with centre O, when produced, intersect each other at the point P, let us prove that ∠AOC – ∠BOD = 2 ∠BPC.

Solution:

Two chords AB and CD of the circle with centre O, when produced, cut each other at P, outside the circle.

To prove, ∠AOC – ∠BOD = 2 × ∠BDC

Join O, A; O, B; O, C; O, D; & A, D.

Proof: In the circle with centre O, ∠AOC is the angle at the centre & ∠ADC is the angle at the circumference on the same arc. AKC, \therefore ∠AOC = 2 ∠ADC

Again, in that circle ∠BAD is the angle at the centre & ∠BAD is the angle at the circumference on the same arc BD, \therefore ∠BOD = 2 ∠BAD

∠AOC – ∠BOD = 2 ∠APC[ From (i)]

Question 12

We drew a circle with the point A of quadrilateral ABCD as centre which passes through the points B, C and D. Let us prove that ∠CBD + ∠CDB = \frac{1}{2}∠BAD.

Solution:

The circle drawn with centre A of the quadrilateral ABCD, passing through B, C, D.

To prove, ∠CBD + ∠CDB = \frac{1}{2} ∠BAD

Proof: 2 ∠BCD = 360 – ∠BAD

\therefore ∠BCD = 180° – \frac{1}{2} ∠BAD ———- (i)

From ∆BCD, ∠BDC + ∠CBD = 180° – ∠BCD

or, ∠CBD + ∠CDB = 180° – (180° – \frac{1}{2} ∠BAD) = \frac{1}{2} ∠BAD Proved.

Question 13

O is the circumcentre of ∆ABC and OD is perpendicular on the side BC; let us prove that ∠BOD = ∠BAC

Solution:

O is the circumcentre & OD is perpendicular on BC.

To prove, ∠BOD = BAC.

Join O, A; O, B; O, C; O, D & A, D.

Proof: From ∆AOC,

∠AOC + ∠OAC + ∠OCA = 180°

\therefore 2 ∠OAC = 180° – ∠AOC = 90° – ∠ABC

∠BAD = 90° – ∠ABC = ∠OAC

\therefore ∠BAC = ∠OAC + ∠O A B = ∠BOD Proved.

## Very short answer type questions (V.S.A.) :

A. M.C.Q. :

Question 1

In the adjoining figure, if ‘ O ‘ is the centre of circle and PQ is a diameter then the value of x is

1. 140
2. 40
3. 80
4. 20

Solution:

∠ROQ = 180° – 140° = 40°

\therefore x = ∠RSQ = \frac{1}{2} × 40° = 20°

Ans. (d)

Question 2

In the adjoining figure, if O is the centre of circle, then the value of x is

1. 70
2. 60
3. 40
4. 200

Solution:

∠BOR = 360° – (140° + 80°) = 360° – 220° = 140°

\therefore ∠QPR = \frac{1}{2} ∠QOR = \frac{1}{2} × 140° = 70°

Ans. (a)

Question 3

In the adjoining figure, if O is centre of circle and BC is the diameter then the value of x is

1. 60
2. 50
3. 100
4. 80

Solution:

∠AOC = 1800° – 80° = 100°

\therefore ∠ADC = x° = \frac{100°}{2}= 50°

Ans. (b)

Question 4

If O is the circumcentre of ∆ABC and ∠OAB = 50°, then the value of ∠ACB is

1. 50°
2. 100°
3. 40°
4. 80°

Solution:

If ∠OAB = 50°

∠AOB = 180° – (50° + 50°) = 180° – 100° = 80°

\therefore ∠ACB = \frac{1}{2} AOB = \frac{1}{2}× 80° = 40°

Ans. (c)

Question 5

In the adjoining figure, if 0

1. 20°
2. 40°
3. 60°
4. 80°

Solution:

∠POQ = 180° – 20° = 160°

∠QOR = 180° – 80° = 100°

∠POR = 160° – 100° = 60°

Ans. (c)

B. Let us write whether the following statements are true or false :

Question 1

In the adjoining figure, if O is the centre of the circle, then ∠AOB = 2 ∠ACD.

Solution:

FALSE

Question 2

The point O lies within the triangular region ABC in such a way that OA = OB and ∠AOB = 2 ∠ACB. If we draw a circle with centre O and length of radius OA, then the point C lies on the circle

Solution:

TRUE

C. Let us fill in the blanks :

Question 1

The angle at the centre is same arc the angle on the circle, subtended by the

Solution:

Half

Question 2

The lengths of two chords AB and CD of a circle with centre O are equal. If APB and AQC are angles on the circle, then the values of the two angles are

Solution:

Equal.

Question 3

If O is the circumcentre of an equilateral triangle, then the value of the angle formed by an side of the triangle is

Solution:

120°.

## Short answer type questions (S.A.) :

Question 1

In the adjoining figure, O is centre of the circle, if ∠OAB = 30°, ∠ABC = 120°, ∠BCO = y° and ∠COA = x°, let us find x and y.

Solution:

∠AOC = x° = 360° – 2 × 120° = 120°

∠BCO = y° = 360° – 2(120° + 30° + 120°) = 360°

Question 2

O is the circumcentre of the triangle ABC and D is the mid point of the side BC. If ∠BAC = 40°, let us find the value of ∠BOD.

Solution:

∠BOC = 2 BAC = 2 × 40° = 80°

∠BOD = (90° – \frac{180 - 80}{2} = 90° – 50° = 40°

Question 3

Three points A, B and C lie on the circle with centre O in such a way that AOCB is a parallelogram, let us calculate the value of the side ∠AOC.

Solution:

∠AOC = 360° – 2 ∠ABC

= \frac{1}{3}× 360° = 120°

Question 4

O is the circumcentre of the isosceles triangle ABC and ∠ABC = 120°; if the length of the radius of the circle is 5cm, let us find the value of the side AB.

Solution:

∠ACB = \frac{180° - 120°}{2} = 30°

\therefore ∠AOB = 60°

As OA = OB

\therefore ∠OAB = ∠OBA = 60°

\therefore OA = OB = AB = 5cm

Question 5

Two circles with centres A and B intersect each other at the points C and D. The centre B of the other circle lies on the circle with center A. If ∠CQD = 70°, let us find the value of ∠COD.

Solution:

∠CAD = (360° – 4 × 70) = 80°

Example:

∠CPD = \frac{1}{2} ∠CAD = \frac{1}{2} × 80° = 40°

Each semi circle angle is one right angle.

All the semicircle angles are equal.

∠APB = ∠AQB = ∠ARB = ∠ASB.

## Let us do – 7.3

Question 1

Let us prove that all angles of a circle formed by same arc are equal.

Solution:

Let in the circle with centre O, ∠ACB ∠ADB are the two angles at the circumference on the same arc ∠APB.

Join O, A O, B.

As on the arc ∠APB, AOB is the angle at the centre and ∠ACB & ∠ADB are the angles on the circumference.

\therefore ∠AOB = 2 ∠ACB ∠AOB = 2 ∠ADB

Question 2

Let us prove that if all angles of a circle subtended by circular arcs be equal then the lengths of the arcs are equal.

Solution:

Let O is the centre of the circle the angles ∠ACB, ∠ADB, ∠AEB are the angles on the circumference are equal, i.e., ∠ACB = ∠ADB = ∠AEB.

To prove that the angles are on the arc of equal length.

Proof: If the angle on the circumference is the half of the angle at the centre then will be on the same arc.

\therefore AOB = 2 ∠ACB = 2 ∠ADB = 2 ∠AEB

\therefore The angles on the same arc ∠APB.

Application 8.

Let us see the figure of circle below and let us find the value of x. 0 is the centre of circle.

Solution:

(i) O is the centre of the circle.

∠ABC and ∠ADC are the front angles on the circle formed by the minor arc ∠CPA.

and in ∆ABC, ∠ABC + ∠ACB + ∠BAC = 180°

\therefore 40° + 100° + x = 180°

or, x° = 180° – 140° = 40°

\therefore x = 40.

Application 9

In the adjoining figure, ∠BDC = 50°, ∠APB = 65° . ∠CBD = 28°; Let us determine the values of ∠ADB, ∠ABD, ∠BAC, ∠ACB, ∠CAD and ∠ACD.

Solution:

∠BAC = ∠BDC = 50°

In ∆BPC , Exterior ∠APB = ∠PBC + ∠PCB

65° = 28° + ACB ; ACB = 37°

In ∆ABP – ∠ABP + ∠BPA + ∠PAB = 180°

∠ABP = ∠ACB = 37°

∠ACD = ∠ABD = 65°

## LET US WORK OUT – 7.2

Question 1

In the adjoining figure, ∠DBA = 40°, ∠BAC = 60° and ∠CAD = 20°; let us find the values of ∠DCA and BCA. Also let us see by calculating what the sum of ∠BAD and ∠DCB will be.

Solution:

In the figure, ∠DBA = 40° ; ∠BAC = 60° & ∠CAD = 20°.

∠DCA & ∠DBA are the angles on the circumference on the same arc CD.

\therefore ∠DCA = ∠DBA = 40°

Now, ∠BAD = ∠BAC + ∠CAD = 60° + 20° = 80°

The sum of the three angles of a triangle = 180°

\therefore In ∆ ABD, ∠BAD + ∠BDA + ∠ABD = 180°

\therefore ∠BDA = 180° – (BAD + ABD)

= 180 – (80° + 40°)

= 180° – 120° = 60°

\therefore On the arc A B, the angle on the circumference ∠BCA = ∠BDA 60°

\therefore ∠BCD = ∠BCA + ∠ACD = 60° + 40° = 100°

∠BCD + ∠BAD = 100° + 80° = 180°

\therefore ∠DCA = 40°, ∠BCA = 60° & Sum of the ∠BAD + ∠BCD = 180°

Question 2

In the adjoining figure, AOB is the diameter of the circle and O is the centre of the circle. The radius OC is perpendicular on AB. If P is any point on minor arc CB, Let us write by calculating the values of ∠BAC and ∠APC.

Solution:

As, AB ⊥ OC

∠AOC = ∠BOC = 90°

∆AOC is an isosceles triangle.

\therefore AO = OC

∠OAC = ∠OCA

∠OAC + ∠OCA = 90° [\because ∠AOC = 90°]

Or, ∠OAC + ∠OAC = 90°

or, 2 ∠OAC = 90°

or, ∠OAC = ∠BAC = 45°

Similarly, ∠ABC = 45°

Again, ∠ABC & ∠APC are the angles at the circumference.

\therefore  ∠APC = ∠ABC = 45°.

Question 3

O is the orthocentre of the triangle ABC and the perpendicular AD drawn on BC when extended, intersects the circumcircle of ∆ABC at the point G; let us prove that OD = DG.

Solution:

To prove, OD = DG.

Join B, G & C, G.

Proof: In the circumcircle of the ABC,

∠ACB is the angle on the circumference on the arc AB.

\therefore  ∠ACB = ∠AGB

or, ∠ACB = ∠OGB

∠ECD = ∠OGB = ∠BGO ———- (i)

Again, BE ⊥ AC & AD⊥ BC.

∠OEC = ∠ODC = 90°

∠OEC + ∠ODC = 90° + 90° = 180°

∠OEC + ∠ECD + ∠ODC + ∠DOE = 4 right angles

∠ECD + ∠DOE = 4right angles – (∠OEC + ∠ODC) =

4 right angles – 2 right angles = 2 right angles

∠ECD + ∠EOD = ∠EOD + ∠BOD

\therefore ∠ECD = ∠BOD = ∠BOG ———- (ii)

From (i) & (ii), BGO = BOG     BG = BO

Now in two right angled triangles,

Hypotenuse BG = Hypotenuse BO, BD is common.

\therefore  ∆BDG ≅ ∆BDO  \therefore  OD = DG Proved.

Question 4

I is the centre of the incircle of ΔABC; Al produced intersects the circumcircle that triangle at the point P, let us prove that PB = PC = PI

Solution:

To prove PB = PC = PI

Join B, I; C, I; & P, B; P, C.

\therefore The bisectors of the angles of a triangle meet at a point called In-centre.

\therefore ∠BAI = ∠CAI = \frac{A}{2}, ∠ABI = ∠CBI = \frac{B}{2}, ∠ACI = ∠BCI = \frac{C}{2}

arc the angles of the circumcircle of ΔABC, on the arc PC.

\therefore ∠PBC = ∠PAC

Now, ∠IBP = ∠PBC + ∠CBI = ∠PAC + ∠CBI = ∠CAI + ∠CBI = \frac{A}{2} + \frac{B}{2} ———- (i)

External angle of the ∆ABI, ∠BIP = ∠BAI + ∠ABI = \frac{A}{2} + \frac{B}{2} ———- (ii)

From (i) & (ii), ∠IBP = ∠BIP \therefore ln ∆ PBI, ∠IBP = ∠BIP ———- (iii)

\therefore BP = PI ———- (iv)

Similarly, in ∠PCI, ∠ICP = ∠CIP

\therefore From (iii) & (iv), PB = PC = PI Proved.

Question 5

Timir drew two circles which intersect each other at the points P and Q. Through the point P two straight lines are drawn so that they intersect one of the circles at the points A and B, and the other circle at the points C and D respectively; let us prove that ∠AQC = ∠BQD.

Solution:

To prove, ∠AQC = ∠BQD.

Join A, Q; B, Q; C, Q & D, Q

Proof : In the circle with centre L, ∠PCQ & ∠PDQ are the angles on the same circumference.

\therefore ∠PCQ = ∠PDC———- (i)

Again, in the circle with centre K, ∠PAQ = ∠PBQ ———- (ii)

As these are angles on the same arc.

∠PAQ + ∠PCQ = ∠PBQ + ∠PDC ———- (iii)

In ∆ACQ sum of the three angles = 180°

\therefore ∠CAQ + ∠ACQ + ∠AQC = 180°

or, ∠PAQ + ∠PCQ + ∠AQC = 180° ———- (iv)

Similarly in ∆BDC, the sum of three angles = 180°

\therefore ∠DBQ + ∠BDQ + ∠BQD = 180°

or, ∠PBQ + ∠PDQ + ∠BQD = 180°

From (iv) & (v),

∠PAQ + ∠PCQ + ∠AQC = ∠PBQ + ∠PDQ + ∠BQD

Subtracting (iii) From (vi),

∠AQC = ∠BQD Proved.

Question 6

Two chords AB and CD of a circle are perpendicular to each other. If a perpendicular drawn to AD From the point of intersection of those two chords AB and CD is produced to meet BC at the point E, let us prove that the point E is the mid point of BC.

Solution:

In the circle AB & CD are perpendicular to each other & intersect at O. The perpendicular From O on chord AD cuts the chord BC at E.

To prove, E is the mid point of BC.

Proof: In ∆AFD, ∠AFO = 90° as OF ⊥ AD

In ∆AFO, ∠FAO + ∠AOF = 90°

i.e., ∠DAQ + ∠ADO = 90° ———- (i)

Again, in ∆AOD, ∠AOD = 90° as AO ⊥ DO

\therefore ∠DAQ + ∠ADO = 90° ———- (ii)

From (i) & (ii), ∠AOF = ∠ADO

or, ∠ADO = ∠EOB as ∠AOF ∠EOB (vertically opposite)

\therefore ∠EOB = ∠ABC

(Angles on the same circumference)

\therefore ∠EOB = ∠OBC = ∠OBE

In ∆OEB, ∠EOB = ∠OBE      \therefore BE = OE ———- (iii)

∠AOF + ∠DOF = 90° (as .AOD = 90°)

and ∠AOF + ∠DAF = 90°    \therefore ∠DOF = ∠DAO

or, ∠COE = ∠DAB    [ ∠COE = vertically ∠DOF]

\therefore ∠COE = ∠DCB[ ∠COE & ∠DCB are angles on the same circumference]

i.e., ∠COE = ∠OCEIn∆OCE, ∠COE = ∠OCE

\therefore CE = OF ———- (iv)

From (iii) & (iv), BE = CE

\therefore E is the midpoint of BC.

Question 7

If in a cyclic quadrilateral ABCD, AB = DC, let us prove that AC = BD.

Solution:

ABCD is a cyclic quadrilateral, AB = DC. To prove AC = BD.

Proof: Join A, C & B, D.

Let AC & BD intersect at E.

∠CAB & ∠DAB are the angles on the same circumference.

\therefore ∠CAB = ∠CBD

i.e., ∠EAB = ∠CDE

In ∠AEB & ∠DEC,

∠EAB = ∠CDE

∠AEB = ∠DEC (vertically opposite)

& AB = DC (given)

\therefore ∆AEB ≅ ∆DCB. (AAS)

\therefore AE = DE & B E = CE

\therefore AC = A E + C E = DE + BE = BD

\therefore AC = BD.

Question 8

OA is the radius of circle with centre at O ; AQ is its chord and C is any point on the circle. A circle passes through the points O ; A ; C intersects the chord AQ at the point P; let us prove that CP = PQ.

Solution:

Join O, A & O, Q.

Proof: OQ & OA are two radii of the circle with centre O.

In the isosceles ∆OQA,

∠OQA = ∠OAQ

Again, ∆OCA is an isosceles triangle as OQ = OC (Radii of same circle)

\therefore ∠OQC = ∠OCQ

or, ∠OQA + ∠AQC = ∠PCQ + ∠PCO

or, ∠OQA + ∠PQC = ∠PCQ + ∠PCO

\therefore ∠PCO & ∠PAO are the angles on the same circumference OP.

∠PCO = ∠PAO = ∠OAQ (as ∠OQA = ∠OAQ)

\therefore ∠PCO = ∠OQA

\therefore ∠OAQ + ∠PQC = ∠PCQ + ∠OQA or, ∠PQC = ∠PCQ

\therefore CP = PQ (As ∆PCQ is an isosceles triangle).

Question 9

The triangle ABC is inscribed in a circle, the bisectors AX, BY and CZ of the angles ∠BAC, ∠ABC and ∠ACB intersect at the points X, Y, Z on the circle respectively, let us prove that AX is perpendicular to YZ.

Solution:

Join X, Y ; Y, Z ; Z, X

Let AX cuts YZ at P.

\therefore ∠AXY + Y = ∠AXY + ∠BYX + ∠BYZ

= ∠ABY + ∠BAX + ∠BCZ

= \frac{∠A}{2} + \frac{∠2}{B} + \frac{∠2}{C}

= \frac{∠A + ∠B + ∠C}{2}

= 1 rt. angle

\therefore In PYX, ∠YXP + ∠PYX = 1 rt. angle \therefore ∠P = 1 rt. angle

\therefore AX ⊥ YZ.

Question 10

The triangle ABC is inscribed in a circle. The bisectors of the angles ∠BAC, ∠ABC and ∠ACB intersect at the points X, Y and Z on the circle respectively, let us prove that in ∆XYZ, ∠YXZ = 90^{\circ}-\frac{∠B A C}{2}.

Solution:

∆ABC is in a circle. Bisectors of ∠BAC, ∠ABC, & ∠ACB meet the circle at X, Y, Z respectively.

To prove in ∆XYZ, ∠YXZ = 90^{\circ}-\frac{∠B A C}{2}

Proof: On the arc AY, ∠AXY = ∠ABY = \frac{1}{2} ∠B

Similarly, on the arc AZ, ∠AXZ = ∠ACZ = \frac{1}{2} ∠C.

\therefore Total ∠X = \frac{1}{2}∠B + \frac{1}{2} ∠C

In ∆ABC, \frac{1}{2}∠A + \frac{1}{2} ∠B + \frac{1}{2} ∠C = 90°

or, \frac{1}{2} ∠B + \frac{1}{2} ∠C = 90^{\circ}-\frac{1}{2} ∠A

or, ∠X = 90^{\circ}-\frac{1}{2} ∠A

\therefore ∠YXZ = 90^{\circ}-\frac{1}{2} ∠BAC.

Question 11

A perpendicualar drawn on BC from the point A of ∆ABC intersects the side BC at the point D and a perpendicular drawn on the side CA intersects the side CA at the point E; let us prove that four points A, B, D, E are concyclic.

Solution:

Join D, E.

Proof: AD ⊥ BC & AE ⊥ CA

∠AEB = ∠BEC = 90°

\therefore External ∠EDC = internal opposite ∠BAE

\therefore ∠BDE + ∠EDC = 2 rt. angles

i.e., ∠BDE + ∠BAE = 2 rt. angles

\therefore Opposite angles of the quadrilateral are supplementary.

\therefore Opposite angles or the cyclic quadrilateral are supplementary.

\therefore A, B, D, E arc concyclic.

## 12. Very short answer type questions (V.S.A.) :

A. M.C.Q. :

Question 1

In the adjoining figure, O is the centre of the circle, if ∠ACB = 30°, ∠ABC = 60°, ∠DAB = 35° and ∠DBC = x°, the value of x is

1. 35
2. 65
3. 70
4. 55

Solution:

∠BAC = 90°

\therefore ∠BDC = 90° – 35° = 55°, – – – – – – – – – – – – – – – – – – – (d)

Question 2

In the adjoining figure, O is the centre of the circle, if ∠BAD = 65°, ∠BDC = 45°, then the value of ∠CBD is

1. 65°
2. 40°
3. 45°
4. 20°

Solution:

\therefore ∠CAD = ∠CBD = 20° – – – – – – – – – – – – – – – – – – – (d)

Question 3

In the adjoining figure, the O is the centre of circle, if ∠AEB = 110° and ∠CBE = 30°, the value of ∠ADB is

1. 70°
2. 60°
3. 80°
4. 90°

Solution:

∠ACB = ∠ADB = 180° – (70° + 30°) = 80° – – – – – – – – – – – – – – – – – – – (c)

Question 4

In the adjoining figure, O is the centre of the circle, if ∠BCD = 28°, ∠AEC = 38°, then the value of ∠AXB is.

1. 56°
2. 86°
3. 38°
4. 28°

Solution:

∠CBE = 180° – (38° + 28°) = 114°

∠ABX = 180° – 114° = 66°

∠AXB = 180° – (66° + 28°) = 86° – – – – – – – – – – – – – – – – – – – (b)

Question 5

In the adjoining figure, O is the centre of the circle and AB is the diameter. If AB || CD, ∠ABC = 25°, the value of ∠CED is

1. 80°
2. 25°
3. 50°
4. 40°

Solution:

∠ABC = ∠BCD = 25° (alternate angle)

∠CED = 90° – 2 × 25° = 40° – – – – – – – – – – – – – – – – – – – (d)

B. Write true or false :

Question 1

In the adjoining figure AD and BE are the perpendiculars on sides BC and CA of the triangle ABC. A, B, D, E are concyclic.

Solution: TRUE

In ∆ABC, AD & BE are the perpendiculars on BC and AC, respectively points A, B, C, D, E are concyclic.

Question 2

In ABC, AB = AC, BE and CF are the bisectors of the angles ABC and ACB and they intersect AC and AB at the points E and F respectively. Four points B, C, E, F, are concyclic.

Solution:

In ∆ABC, AB = AC; BE & CF are the are the bisectors of the angles ∠ABC & ∠ACB, respectively which cut at E & F.

Points B, C, E, F are not concyclic.

C. Fill in the blanks :

Question 1

All angles in the same segment are ________________.

Solution:

Equal.

Question 2

If the line segment joining two points on the same side, then the four points are __________________.

Solution:

Concyclic.

Question 3

If two angles on the circle formed by two – arcs are equal then the lengths of arcs __________________________.

Solution:

Equal.

Question 1

In the adjoining figure, O is centre of the circle, AC is the diameter and chord DE is parallel to the diameter AC. If ∠CBD = 60°; let us find the value of ∠CDE.

Solution:

∠CDE = 90° – ∠CBD = 90° – 60° = 30°

Question 2

In the adjoining figure, QS is the bisector of an angle ∠PQR, if ∠SQR = 35° and ∠PRQ = 32°; let us find the value of ∠QSR.

Solution:

∠SRP = ∠SQR = ∠PSQ = 35°

\therefore ∠QSR = ∠QPR = 180° – (35° + 35° + 32°) = 78°

Question 3

In the adjoining figure, O is centre of the circle and AB is the diameter. If AB and CD are mutually perpendicular to each other and ∠ADC = 50°; let us find the value of ∠CAD.

Solution:

∠BAD = 90° – 50° = 40°

\therefore ∠CAD = 2 × 40° = 80°.

Question 4

In the adjoining figure, O is centre of the circle and AB = AC; if ∠ABC = 32°, let us find the value of ∠BDC.

Solution:

∠ABC = ∠ACB = 32° \therefore ∠BAC = 180° – (32° + 32°) = 116°

∠BDC = 180° – 116° = 64°

Question 5

In the adjoining figure, BX and CY are the bisectors of the angles ∠ABC and ∠ACB respectively. If AB = AC and BY = 4cm, let us find the length of AX.

Solution:

AX = BY = 4cm.

Application 17.

Let us prove that the angle in the segment of a circle which is less than a semicircle is an obtuse angle. [Let me do it myself]

Solution:

In the figure, O is the centre of the circle.

\therefore The segment ACB is less than a semicircle.

To prove ∠ACB is an obtuse angle.

\therefore ADB is a major arc.

The reflex ∠AOB, the angle at the centre standing on that arc is greater than 2 right angles.

\therefore ∠ACB is an angle at the circle also standing on the same arc.

\therefore ∠ACB is greater than 1 right angle.

\therefore ∠ACB is an obtuse angle. Proved.

Application 19.

Let us prove with reason that the circle drawn with hypotenuse of a right angled triangle as diameter passes through the right angular vertex. [Let me do it myself]

Solution:

To prove that the circle drawn with BC as diameter passes through the point A.

Proof : Let us suppose the circle does not pass through A. Then, let the circle intersects BA at the point D. angle)

\therefore ∠BDC is 1 right angle (\because angle in a semicircle is a right

\therefore ∠BAC = 1 right angle (by hypothesis)

\therefore ∠BAC = BDC.

Which is impossible, if the points D & A do not coincide, since the exterior angle ∠BDC of ∆ADC> the interior opposite angle ∠BAC.

\therefore The circle passes through the point A.

## LET US WORK OUT – 7.3

Question 1

In ∆ABC, ∠B is a right angle. A circle drawn taking AC as diameter intersects AB at the point P; let us write the correct information from the followings :

Solution:

We know angle in a semicircle angle = 90°

But ∠ABC = 90°

It is possible if point B & point D are the same point.

Question 2

Let us prove that the circle drawn triangle as diameter bisects the drawn with any one of the equal side of an isosceles triangle as diameter bisects the unequalside.

Solution:

In ∆ABC, AB = AC.

A circle with centre O and with A B as diameter is drawn.

The circle cuts BC at D.

To prove D bisects BC.

Proof : Join A, B.

As AB is the diameter, \therefore ∠ADB is one right angle.

Now, in two right angled ∆ABD & ∆ACD,

Hypotenuse AB = Hypotenuse AC (given) And AD is common.

\therefore ∆ABD ≅ ∆ACD       \therefore BD = CD

\therefore D bisects the unequal side of isosceles triangle ABC.

Question 3

Sahana drew two circles which intersect each other at the points P and Q. If the diameters of the two circles are PA and PB respectively, then let us prove that A, Q, B are collinear.

Solution:

Two circles cut each other at P & Q. If P A & PB are the diameters of the two circles, prove that the points A, Q, B are collinear.

Proof : Join Q, A ; Q, B & P, Q.

As, AP is the diameter & AQP is a semi circle angle,

\therefore ∠AQP = 1 rt. angle

Similarty, BP is the diameter.

\therefore ∠BQP = 1 rt. angle

\therefore ∠AQP+∠BQP = 2 r. angles

Now, AQ & BQ meet at Q & the sum of the adjacent angles = 2 rt. angles

\therefore AQ and BQ are on the same straight line.

\therefore A, Q & B are collinear.

Question 4

Rajat drew a line segment PQ whose mid point is R and two circles are drawn with PR and PQ as diameters. I drew a straight line through the point P which intersects the first circle at the point S and the second circle at the point T. Let us prove with reason that PS = ST.

Solution:

PQ is a straight line whose midpoint is R. Now two circles are drawn with PR & PQ as diameter.

A straight line passing through P is drawn which cuts the 1st circle at S & cuts the 2nd cricle at T.

To prove PS = ST.

Join R, S & Q, T.

As PR is the diameter of the 1st circle,

\therefore Semicircle angle PSR = 90°

\therefore SR ⊥ PT

Similarly, ∠PTQ = 90°

\therefore QT ⊥ PT

\therefore SR & QT are both perpendiculars on PT.

\therefore SR || QT

In ∆PQT, the midpoint PQ is R

and SR || QT.

\therefore S is the mid point of PT.

\therefore PS = ST.

Question 5

Three points P, Q, R lie on a circle. The two perpendiculars PQ and PR at the prove that RQ = point P intersect the circle at the points S and T respectively. Let us prove that RQ = ST.

Solution:

Let P, Q & R are the three points on a circle.

Perpendiculars drawn from P on the chords PQ & PR are PS & PT, which cut the circle at S & T respectively.

To prove RQ = S T.

Proof : Join R, Q ; S, T ; S, Q ; & T, R.

Let the straight lines RT & S Q intersect at O.

∠SPQ = 1 rt. angle [as PS ⊥ PQ (given)]

\therefore SQ is a diameter.

Again, ∠RPT = 1 rt. angle [as PT ⊥ PR (given)]

\therefore RT is a diameter.

\therefore O is the centre as diameters SQ & RT intersect each other.

\therefore OR = OQ = OT

\therefore In triangles ORQ & OST,

OR = OS (Radius of same circle)

OQ = OT (Radius of same circle)

and ∠ROQ = ∠SPT (vertically opposite)

∆ORQ ≅  ∆OST

\therefore RQ = ST. Proved.

Question 6

ABC is an acute angled triangle. AP is the diameter of the circumcircle of the triangle ABC; BE and CF are perpendiculars on AC and AB respectively and they intersect each other at the point Q. Let us prove that BPQC is a parallelogram.

Solution:

ABC is an acute angled triangle. AP is the diameter of the circumcircle of ∆ABC.

BE & CF are the perpendiculars on the sides AC & AB, respectively.

They meet at Q.

To prove BPCQ is a parallelogram.

Proof : CF      AB (given)

and ∠ABP is a semicircle angle.

\therefore BP ⊥ AB

\therefore CF || BP or CQ || BP

Again, BE ⊥ AC (given)

and ∠ACP is an angle in a semi-circle.

\therefore CP ⊥ AC

\therefore CP || BE or CP || BQ.

\therefore BPCQ is a parallelogram. Proved.

Question 7

The internal and external bisectors of the vertical angle of a triangle intersect th circumcircle of the triangle at the points P and Q. Let us prove that PQ is a diameter of the circle.

Solution:

ABC is a triangle, internal & external bisectors of ∠A of ∆ABC are AP & AC respectively, which cut the circumcircle of ∆ABC at P & Q respectively. To prove PQ is the diameter of the circle.

Proof : As AP & AQ are the internal & external bisectors of ∠A.

∠PAQ = 90°

PAQ is a semi circle angle.

PQ is the diameter. Proved.

Question 8

AB and  D are two diameters of circle. Let us prove that ABCD is a rectangular figure.

Solution:

Let AB & CD are two diameters of the circle with centre O Join AC, BD, AD & BC.

To prove ACBD is rectangle.

Proof : ∠ADB = ∠ACB = 1 rt. angle

& ∠CAD = ∠DBC = 1 rt. angle

∠A = ∠C = ∠B = ∠D = 1 it. angle

\therefore ACBD is a rectangle.

Question 9

Let us prove that if the circles are drawn having sides of a rhombus as diameter then the circles pass through a fixed point.

Solution:

If circles are drawn with diameters of the sides of a rhombus, they will pass a fixed point.

Let ABCD is a rhombus. The circle with AB as diameter cuts BC or produced BC at D. Join A, D.

\therefore ∠ADB = 1 rt. angle (as semicircle angle)

\therefore ∠AOC = 1 rt. angle

Again the circle as AC diameter will pass through the point D.

\therefore Two circles with AB & AC as diameter intersect each other at D.

\therefore Circles with diameters of the sides of a rhombus will pass through a fixed point. Proved.

## 10. Very short answer (V.S.A.) :

A. M.C.Q :

Question 1

PQ is a diameter of a circle with center O, and PR = RQ; the value of ∠RPQ is

1. 30°
2. 90°
3. 60°
4. 45°

Solution:

As PR = RQ,   \therefore P R Q = 90° (semi circle angle)

\therefore RPQ = RQP = \frac{180°-90°}{2} = \frac{90°}{2}

Ans. 45°

Question 2

QR is a chord of a circle and POR is a diameter of a circle. OD is perpendicular on QR. If OD = 4cm, the length of PQ is

1. 4 cm
2. 2 cm
3. 8 cm
4. none of these

Solution:

\therefore PQ = 2 × OD = 2 × 4 = 8 cm.

Ans. 8 cm.

Question 3

AOB is a diameter of a circle. The two chords AC and BD when extended meet at the point E. If ∠COD = 40°, the value of ∠CED is

1. 40°
2. 80°
3. 20°
4. 70°

Ans. 20°

Question 4

AOB is diameter of a circle. If AC = 3 cm, BC = 4 cm, then the length of AB is

1. 3 cm
2. 4 cm.
3. 5 cm
4. 8 cm.

Solution:

AB2 = \sqrt{A C^{2}+B C^{2}} = \sqrt{3^{2}+4^{2}} = \sqrt{25} = 5

Question 5

In the adjoining figure, O is centre of circle & AB is a diameter, if ∠BCE = 20° ∠CAE = 25°, the value of ∠AEC is

1. 50°
2. 90°
3. 45°
4. 20°

Solution:

ACE = 90° + 20° = 110° ; AEC = 180° – (110° + 25°) = 45°

Ans. 45°

B. Let us write true or false :

Question 1

The angle in the segment of a circle which is greater than a semi-circle is an obtuse angle.

Solution:

FALSE

Question 2

O is the mid point of the side AB of the triangle ABC, and OA = OB = OC; if we draw a circle with side AB as diameter, then the circle passes through the point C.

Solution:

TRUE

C. Let us fill in the blanks.

Question 1

Semicircular angle is a ___________________.

Solution:

Right angle.

Question 2

The angle in the segment of a circle which is less than a semicircle is an ________________ angle.

Solution:

Obtuse angle.

Question 3

The circle drawn with the hypotenuse of a right angled triangle as diamete passes through the ___________________.

Solution:

Vertices.

Question 1

In isosceles triangle ABC, AB = AC; a circle drawn taking AB as diameter meets the side BC at the point D. If BD = 4 cm, let us find the value of CD.

Solution:

ABC is an isosceles triangle, where AB = AC. A circle is drawn with AB as diam eter, the circle will’ cut BC at D. If BD = 4 cm.

Find CD.

BD = CD = 4 cm.

Question 2

Two chords AB and AC of a circle are mutually perpendieular to each other. AB = 4 cm, AC = 3, let us find the length of the radius of the circle.

Solution:

AB & AC the two chords of a circle are perpendicular to each other.

AB = 42 cm, AC = 3 cm, the radius of the circle = ?

Diameter (BC) = \sqrt{3^{2}+4^{2}} = \sqrt{25} = 5, radius = \frac{5}{2} = 2.5 cm.

Question 3

Two chords PQ and PR of a circle are mutually perpendicular to each other. the length of the radius of the circle is r cm, let us find the length of the chord QR.

Solution:

PQ & PR are two chords of a circle perpendicular to each other; if radius = r cm find the length of the chord QR.

QR – Diameter – 2 r cm[\therefore QPR = 1 rt. angle ]

Question 4

AOB is a diameter of a circle. The point C lies on the circle. If ∠OBC = 60°, Let us find the value of ∠OCA.

Solution:

AOB is a diameter of a circle. C is any point on the circle; if ∠OBC = 60°, find ∠OC

\therefore OB = OC \therefore ∠OBC = ∠OCB = 60°

\therefore ∠OCA = 90° – 60° = 30°

Question 5

In the picture beside, O is centre of the circle and AB is the diameter. The length of chord CD is equal to the length of the radius of the circle. AC and BD produce meet at the point P, let us find the value of ∠APB.

Solution:

OA = OC = OB  = OD = CD

\therefore ΔOCD is an equilateral triangle.

\therefore ∠COD = 60°

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