CIRCLE (CLASS 10) EXERCISES

                    Exercise 12                     

1. If O is the Centre of the following circles, find the value of x.

 a)

Solution:

Given: ∡BAC = 30°, ∡DBC = 70°& ∡BCD = x°

 →∡CAD = ∡CBD = 70° [inscribed Angles subtended by same arc CD]

→ ∡BCD + ∡BAD = 180° [Since the sum of opposite angles of a cyclic quadrilateral are supplementary]

OR, x°+30°+70° = 180°

OR, x° = 180° - 30°- 70°

OR, x = 80°

b)

Solution:

Given: ∡PBA = 35° & ∡PQB = x°

→ ∡PAB = ∡PQB = x° [inscribed Angles subtended by same arc PB]

→ ∡APB = 90° [Inscribed angle in a semicircle is 90°]

→ ∡APB + ∡PAB + ∡PBA = 180° [sum of angles of a triangle]

OR, 90° + x +35° = 180°

OR, x = 180°-90° -35°

OR, x°= 55°

c)

Solution:

Given: ∡ABO = 25°, ∡BOC = x° & ∡ACO = 35°

Construction: Let's produce AO up to D

→∡OAB = ∡OBA = 25° [Since, OA=OB]

→ ∡OAC = ∡OCA = 35° [Since, OA=OC]

→ ∡BOD = ∡0AB + ∡OBA [ An exterior angle of a triangle is equal to the sum of its opposite interior angles]

OR, ∡BOD = 25° +25° = 50°

→∡OAC + ∡OCA = ∡COD [An exterior angle of a triangle is equal to the sum of its opposite interior angles.]

OR, ∡COD = 35° +35° = 70°

⇒ x = ∡BOD + ∡COD [Whole parts axiom]

OR, x°= 50° +70°

OR, x° = 120°

d)

Solution:

Given: ∡AOB = 120°, ∡AOC = 90° & ∡BAC = x°

 ⇒ ∡OAB = ∡OBA {Since, OB=OA}

OR, ∡OAB + ∡OBA + ∡BOA = 180° [Sum of angles of a triangle]

OR, ∡0AB + ∡0AB + 120° = 180°

OR, 2∡OAB = 180° - 120°

OR, ∡OAB = (60°)/2

OR, ∡OAB = 30° → ∡OAC = ∡OCA [Since, OA=OC]

OR, ∡0AC + ∡0CA + ∡COA = 180° [Sum of angles of a triangle]

OR, ∡0AC+∡0AC +90° = 180°

OR, 2∡OAC = 180° - 90°

OR, ∡OAC = (90°)/2

OR, ∡OAC = 45°

⇒ x = ∡0AB + ∡0AC [Whole-part axiom]

OR, x° = 30° + 45°

OR, x°= 75°

e)

Solution:

Given: ∡CAB = 50° & ∡OBD = x°

⇒ ∡CAB = ∡CDB = 50° [inscribed Angles subtended by same arc CB]

⇒ ∡ODB = ∡CDB = 50° [Same angle]

⇒ ∡OBD = ∡ODB [Since, OB=OD]

OR, x = 50°

f)

Solution:

Given: ∡COA = 130° & ∡ABC = x°

⇒ ∡COA + reflex ∡COA = 360° [Complete angle]

OR, 130° + reflex ∡COA = 360°

OR, reflex ∡COA = 360° - 130°

OR, reflex ∡COA = 230°

⇒ ∡ABC = (reflex ∡COA)/2 [an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle]

OR, x° = 230°/2

OR, x° = 115°

 

2. In the following figure, find the value of x.

a)

Solution:

Given, ∡ADC = 90°, ∡CAD = 50° & ∡DCB = x°

⇒ ∡ACB = 90° [Inscribed angle in a semicircle is 90°]

OR, ∡DAC + ∡ADC + ∡DCA = 180° [sum of angles of a triangle]

OR, 50°+90°+∡DCA = 180°

OR, ∡DCA = 180°- 90° - 50°

OR, ∡DCA = 40°

⇒x = ∡ACB - ∡DCA

OR, x° = 90°- 40°

OR, x = 50°

b)

Solution:

Given: ∡AOC = 130°, ∡CDB = x°

∡AOC + ∡BOC = 180°

OR, 130°+∡BOC = 180°

OR, ∡BOC = 180° - 130°

OR, ∡BOC = 50° ⇒ ∡CDB = 1/2 ∡BOC [an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc on the circle.]

OR, x° = 1/2 x 50°

OR, xº = 25°

c)

Solution:

Given: ∡CAB = 40°, ∡DCB = x°& ∡DBC = 80°

⇒ ∡CDB = ∡CAB = 40° [inscribed Angles subtended by same arc CB]

→ ∡DCB + ∡CBD + ∡BDC = 180° [Sum of angles of a triangle]

OR, x°+80°+ 40° = 180°

OR, x° =180°-80° -40°

OR, x° = 60°

d)

Solution:

Given: ∡OAC = 50°, ∡ADB = x°

→ ∡BCA = ∡BDA = x° [inscribed Angles subtended by same arc AB]

→ ∡OCA = ∡0AC [Since, OA=OC]

OR, x° = 50°

3.

a)In the adjoining figure, O is the centre of circle, ∠OBA = 25° and ∠OCA = 30° find the value of obtuse ∠BOC.

Solution:

Given: ∡ABO = 25°& ∡ACO = 30°

Construction: Let's produce OA upto D.

⇒ ∡OAB = ∡OBA = 25° [Since, OA=OB]

→ ∡OAC = ∡OCA = 30° [Since, OA=OC]

⇒ ∡BOD = ∡OAB + ∡OBA [An exterior angle of a triangle is equal to the sum of its opposite interior angles.]

OR, ∡BOD = 25°+25° = 50°

→ ∡OAC + ∡OCA = ∡COD [An exterior angle of a triangle is equal to the sum of its opposite interior angles.]

OR, ∡COD = 30°+30° = 60°

⇒ x° = ∡BOD + ∡COD [Whole parts axiom]

OR, x°= 50°+ 60°

OR, x° = 110°

b) In the adjoining figure, O is the centre of the circle. If ∠PQR= 40° and ∠PSQ = x°, find the value of x.

Solution:

Given: ∡PQR = 40° & ∡PSQ = x°

→ ∡PRQ = ∡PSQ = x° [inscribed Angles subtended by same arc PQ]

⇒ ∡QPR = 90° [Inscribed angle in a semicircle is 90°]

⇒ ∡QPR + ∡PRQ + ∡PQR = 180° [Sum of angles of a triangle]

OR, 90° + x° + 40° = 180°

OR, x° = 180°- 90°- 40°

OR, x° = 50°

c) In the given figure, O is the centre of the circle. FAEC is a cyclic quadrilateral. If ∠CED= 68°, then (i) Find the value of ∠AFC. (ii) Find the reflex ∠AOC. 

Solution:

Given: ∡CED = 68°, ∡AFC =? & reflex ∡AOC =?

i. ∡AEC + ∡CED = 180° [Being Straight angle]

OR, ∡AEC+ 68° = 180°

OR, ∡AEC = 180° - 68°

OR, ∡AEC = 112°

→∡AFC + ∡AEC = 180° [Since the opposite angles of a cyclic quadrilateral are supplementary]

OR, ∡AFC +112° = 180°

OR, ∡AFC = 180°- 112°

OR, ∡AFC = 68°

ii. Reflex ∡AOC = 2∡AEC [an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc AFC on the circle]

= 2 × 112° = 224°

Also, ∡AOC = 2∡AFC [an angle θ inscribed in a circle is half of the central angle 2θ that subtends the same arc AEC on the circle]

= 2 x 68° = 136°

d) In the given figure, AOB is a diameter of the circle. If ∠ADC= 100°, then find the value of ∠BAC.

Solution:

Given: ∡ADC = 100°, ∡CAB =? ∡ACB = 90° [Inscribed angle in a semicircle is 90°]

→ ∡ADC + ∡ABC = 180° [Since the opposite angles of a cyclic quadrilateral are supplementary]

OR, 100°+∡ABC = 180°

OR, ∡ABC = 180°-100°

OR, ∡ABC = 80°

→ ∡ABC + ∡ACB + ∡BAC = 180° [sum of angles of a triangle]

OR, 80°+90° + ∡BAC = 180°

OR, ∡BAC = 180°-170°

OR, ∡BAC = 10°

4.

a) In the adjoining figure, BC = DE and AB = FE, Prove that ∠ACB = ∠FDE.

Solution:

Given: BC = DE, arc AB = arc FE,

To prove: ∡ACB = ∡FDE,

Construction: joined A with B and F with E,

Proof:

s.n	Statements	s.n	Reasons
1.	In ΔABC and ΔFED	1	…
a)	BC = DE (Side.)	a)	Given
b)	∡CBA = ∡DEF (Angle)	b)	Arc EFA=Arc BAF and being angle subtended on those arcs.
c)	AB = FE (side)	c)	Arc AB=Arc FE
2.	ΔACB ≅ ΔFDE	2.	From S.A.S axiom
3.	∡ACB = ∡FDE	3.	Being corresponding angles of congruent triangle

Hence, Proved.

b) In the given figure, if ∠APC = ∠BQD, then prove that AB//CD.

Solution:

1. Given: ∡APC = ∡BQD

2. To prove: AB || CD

3. Proof:

s.n	Statements	s.n	Reasons
1.	∡APC = ½ arc AC	1.	The relationship between an arc and the circumference angle subtended by that arc
2.	∡BQD = ½ arc BD	2.	The relationship between an arc and the circumference angle subtended by that arc
3.	½ arc AC = ½ arc BD Or, arc AC=arc BD	3.	∡APC = ∡BQD, from Statements (1) and (2)
4.	AB || CD	4.	From Statement (3)

Hence, Proved

c) In the adjoining figure, O is the centre of the circle. If arc PQ = arc PB, then prove that AQ// OP.

Solution:

1. Given: PQ = PB

2. To prove: AQ || OP

3. Proof:

s.n	Statements	s.n.	Reasons
1.	Arc PB = ∡POB	1.	The relationship between an arc and the central angle subtended by that arc
2.	Arc QPB = 2∡QAB	2.	The relationship between an arc and the central angle subtended by that arc
3.	2 arc PB = arc QPB	3.	Since, PQ = PB
4.	2∡POB = 2∡QAB→ ∡POB = ∡QAB	4.	From statements (1), (2) and (3)
5.	AQ || OP	5.	From Statement (4), corresponding angles being equal.

Hence, Proved.

a.   d. In the given figure, chords AC and BD are intersected at a point P. If PB = PC then prove that:

(i)   Chord AB = chord DC.

(ii) Chord AC = chord BD.

(iii) Arc ABC = arc BCD

Solution:

1.     Given: Chords AC and BD intersect each other at point P. Also PB=PC

2.     To prove: i) AB=CD, ii) AC=BD iii) arc ABC= arc BCD

Proof:

s.n	statements	s.n.	Reasons
1.	In triangle ABP and DCP	1.	...
a)	∡BAP = ∡CDP (Angle.)	a)	Being angles inscribed on arc BC
b)	∡ABP = ∡DCP (Angle.)	b)	Being angles inscribed on arc BC
c)	BP=PC (Side)	c)	Given
2.	ΔABP ≅ ΔDCP	2.	From A.A.S axiom
3.	AB=CD, AP=PD	3.	Being corresponding sides of congruent triangles
4.	AP+ PC =BP+PD → AC=BD	4.	From statement 1. c) and 3
5.	Arc AB=Arc DC	5.	Because, AB=CD
6.	Arc AB+ Arc BC=Arc DC +Arc BC→arc ABC=arc BCD	6.	Adding arc BC on both sides of Statement (5)

5. In the figure, O is the centre of the circle. If the chords DE and FG are intersected at a point H, prove that: ∠DOF + ∠EOG = 2∠EHG.

Solution:

1. Given: Chords DE and FG are intersected at point H. 

2. To prove: ∡DOF + ∡EOG = 2 ∡EHG 

3. Proof:

s.n	Statements	s.n.	Reasons
1.	∡EOG= 2∡EDG	1.	The relationship between central angle and inscribed angle made on same arc EG.
2.	∡DOF = 2∡FGD	2.	The relationship between central angle and inscribed angle made on same arc FD.
3.	∡FGD + ∡EDG = ∡EHG	3.	In ΔHDG, Exterior angle of a triangle is equal to sum of two opposite interior angles.
4.	½ ∡DOF+ ½ ∡EOG = ∡EHG	4.	From statements (1), (2) and (3)
5.	∡DOF + ∡EOG = 2∡EHG	5.	From statement (4)

Hence, Proved

6.  In the figure, chords MN and RS of the circle intersect externally at the point X. Prove that: ∠MXR = ½ (arc MR – arc NS).

Solution:

1. Given: Two chords MN and RS of a circle intersects at point X outside of circle.

2. To prove: ∡MXR = ½ (arc MR – arc NS)

3. Construction: Joined M and S

4. Proof:

s.n	Statements	s.n.	Reasons
1.	∡NMS = ½ arc NS	1.	From the relation between inscribed angle and its opposite arc of the circle.
2.	∡RSM = ½ arc MR	2.	From the relation between inscribed angle and its opposite arc of the circle.
3.	∡MXS + ∡XMS = ∡RSM	3.	In ΔXSM, Exterior angle of a triangle is equal to sum of two opposite interior angles.
4.	∡MXR + ½ arc NS = ½ arc MR	4.	From statement (1), (2) and (3)
5.	∡MXR = ½ (arc MR – arc NS)	5.	From statement (4)

Hence, Proved.

7. PQRS is a cyclic quadrilateral. If the bisectors of the ∠QPS and ∠QRS meet the circle at A and B respectively, prove that AB is a diameter of the circle.

Solution:

1. Given: PQRS is a cyclic quadrilateral. ∡QPA=∡APS, ∡QRB = ∡BRS

2. To prove: AB is diameter of circle.

3. Construction: Joined B and P

4. Proof:

s.n.	Statements	s.n	Reasons
1.	∡QPS + ∡QRS = 180 ̊	1.	Opposite angles of cyclic quadrilateral
2.	2∡QPA + 2∡BRQ = 180 ̊ Or, ∡QPA + ∡BRQ = 90 ̊	2.	Since, ∡QPS = 2∡QPA, ∡QRS= 2∡BRQ
3.	∡BRQ = ∡BPQ	3.	Angles inscribed on same arc BQ
4.	∡QPA + ∡BPQ = 90 ̊	4.	From statement (2) and (3)
5.	∡BPA = 90 ̊	5.	From statement (4)
6.	AB is diameter of circle	6.	Since, Angle at circumference of circle inscribed at arc AB is 90 ̊

Hence, Proved.

8. In the given figure, NPS, MAN and RMS are straight lines. Prove that PQRS is a cyclic quadrilateral.

Solution:

1. Given: NPS, MAN and RMS are straight lines.

2. To prove: PQRS is cyclic quadrilateral

3. Construction: A and Q are Joined.

4. Proof:

s.n.	Statements	s.n	Reasons
1.	∡NPQ = ∡NAQ	1.	Being angles at circumference of circles standing on same arc.
2.	∡QRM = ∡NAQ	2.	The exterior angle of a cyclic quadrilateral is equal to interior opposite angle.
3.	∡SPQ = 180 ̊ - ∡NPQ	3.	Because, ∡SPQ + ∡NPQ gives straight angle of 180 ̊
4.	∡SPQ = 180 ̊ - ∡QRM	4.	From statements 1, 2, 3
5.	∡SPQ + ∡QRS = 180 ̊	5.	From statement (4) and ∡QRM = ∡QRS
6.	PQRS is a cyclic quadrilateral.	6.	From statement 5, as the sum of opposite angles is 180 ̊

Hence, Proved.

9. In the given figure, PQRS is a parallelogram. Prove that UTRS is a cyclic quadrilateral.

Solution:

1. Given: PQRS is parallelogram

2. To prove: UTRS is a cyclic quadrilateral

3. Proof:

s.n.	Statements	s.n	Reasons
1.	∡PUT=∡PQT	1.	Being angles at circumference of circles standing on same arc PT.
2.	∡SRQ = 180 ̊- ∡PQR	2.	∡SRQ and ∡PQR are Co-interior angles
3.	∡SRQ = 180 ̊ - ∡PUT	3.	From statement (1), (2)
4.	∡SRQ + ∡SUT = 180 ̊	4.	From statement (3)
5.	UTRS is cyclic quadrilateral	5.	From statement (4), the sum of opposite angles is 180 ̊

Hence, Proved.