# Concise Maths Solutions Circles Chapter-17 ICSE Class 10

## Selina Publishers Concise Maths Solutions Chapter-17 Circles

**Concise Maths Solutions Circles** Chapter-17 ICSE Class 10.** Solutions** of Exercise – 17 (A), Exercise – 17 (B), Exercise – 17 (C)for** Concise **Selina Maths of ICSE Board Class 10th. **Concise Maths Solutions Circles **Chapter-17 for ICSE Maths Class 10 is available here. All **Solutions **of **Concise** Selina **Maths** of **Circles **Chapter-17 has been solved according instruction given by council. This is the **Solutions **of **Circles **Chapter-17 for ICSE Class 10th. ICSE **Maths** text book of **Concise** is In series of famous ICSE writer in maths publications. **Concise** is most famous among students

**Concise Solutions Circles Chapter-17 Selina Maths ICSE Class 10**

The **Solutions** of **Concise** **Mathematics** **Circles **Chapter-17 for ICSE Class 10 have been solved. Experience teachers Solved Chapter-17 **Circles** to help students of class 10th ICSE board. Therefore the ICSE Class 10th **Maths** **Solutions** of **Concise** Selina Publishers helpful on various topics which are prescribed in most ICSE Maths textbook

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**How to Solve Concise Maths Selina Publications Chapter-17 Circles ICSE Maths Class 10**

Note:- Before viewing **Solutions** of Chapter-17 **Circles**** **of **Concise **Selina **Maths** read the Chapter Carefully then solve all example of your text book**. **The Chapter-17 **Circles**** **is main Chapter in ICSE board

**Concise Maths Solutions Circles Selina Maths ICSE Class 10 EXERCISE – 17(A)**

**Question 1.**

**In the given figure, O is the centre of the circle. ∠OAB and ∠OCB are 30° and 40° respectively. Find ∠AOC. Show your steps of working.**

**Answer 1 **

In circle with centre O, ∠BAO = 30°, ∠BCO = 40°.

Join BO.

OA = OB = OC (Radii of the circle)

∠OBA = ∠OAB = 30° and ∠OBC = ∠OCB = 40°

∠ABC = 30° + 40° = 70°

Now, AOC is at the centre and ∠ABC is on the remaining part of the circle.

∠AOC = 2 ∠ABC = 2 x 70° = 140°.

**Question 2.**

**In the given figure, ∠BAD = 65°, ∠ABD = 70°, ∠BDC = 45°**

**(i) Prove that AC is a diameter of the circle.**

**(ii) Find ∠ACB.**

**Answer 2**

(i) In ΔABD

65° + 70° + ∠ADB = 180°

∠ADB = 180° – 65° – 70° = 45°

∠ADC = 45° + 45° = 90°

AC is diameter [Angle in semi circle is 90°]
(ii) ∠ACB = ∠ADB = 45° [angle in same segment]

**Question 3.**

**Given O is the centre of the circle and ∠AOB = 70°. Calculate the value of:**

**(i) ∠OCA,**

**(ii) ∠OAC.**

**Answer 3**

O is the centre of the circle,

∠AOB = 70°

arc AB subtends ∠AOB at the centre

and ∠ OCA is at the remaining part of circle

∠AOB = 2 ∠OCA

or ∠OCA = ∠AOB = x 70° = 35°

In ΔOAC,

OC = OA (Radii of the same circle)

∠OAC = ∠OCA = 35°

**Question 4.**

**In each of the following figures, O is the centre of the circle. Find the values of a, b, and c.**

**Answer 4**

(i) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part

∠AOB = 2 ∠ACB

or ∠ACB = ∠AOB = x 130° = 65°

or b = 65°

But a + b = 180° (Opposite angles of a cyclic quad.)

a = 180° – b = 180° – 65° = 115°

a = 115°, b = 65°

(ii) Arc AB subtends ∠ AOB at the centre and ∠ACB at the remaining part.

Reflex ∠AOB = 2 ∠ACB

or ∠ACB = (reflex ∠AOB) = [360°- 112°]
= x 248° = 124°

Hence, c = 124°.

**Question 5. Concise Maths Solutions Circles**

**In each of the following figures, O is the centre of the circle. Find the value of a, b, c and d.**

**Answer 5**

(i) BOD is a diameter

∠BAD = 90° (Angle in a semi-circle)

∠ADB = 180° – (90° + 35°) = 180° – 125° = 55°

But ∠ACB = ∠ADB = 55° (Angles in the same segment)

a = 55°.

(ii) In ΔEBC.

Ext. 120° = 25° + ∠BCE

∠BCE = 120° – 25° = 95°

But ∠ADB = ∠ACB = 95° (Angles in the same segment)

b = 95°.

(iii) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part,

∠ AOB = 2 ∠ ACB = 2 x 50° = 100°

In ΔAOB,

OA = OB (Radii of the same circle)

∠OAB = ∠OBA

But ∠OAB + ∠OBA = 180° – 100° = 80°

c = ∠OAB = ~ x 80° = 40°.

(iv) In the given figure, O is the centre of the circle.

AOB is its diameter and ∠ABP = 45°

Q is any point and BQ, PQ are joined

In ΔABP,

∠APB = 90° (Angle in a semicircle)

∠PAB + ∠PBA = 90°

⇒ ∠PAB + 45° = 90°

⇒ ∠PAB = 90° – 45°

⇒ ∠PAB = 45°

Now ∠PAB = ∠PQB (Angle in the same segment)

BPQB = 45°

⇒ d = 45°

**Question 6. Concise Maths Solutions Circles**

**In the figure, AB is common chord of the two circles. If AC and AD are diameters; prove that D, B and C are in a straight line. O1 and 02 are the centres of two circles.**

**Answer 6**

Given- Two circles with centre O1 and O2 intersect each other at A and B.

AC and AD are the diameters of the circles.

To Prove- D, B, C are in the same straight line.

Construction- Join AB.

Proof- AO1C is diameter.

∠ABC = 90°. (Angle in a semi-circle)

Similarly ∠ABD = 90°,

Adding, we get:

∠ABC + ∠ABD = 90° + 90° = 180°

DBC is a straight line.

or D, B, C are in the same line.

**Question 7.**

**In the figure given beow, find :**

**(i) ∠BCD,**

**(ii) ∠ADC,**

**(iii) ∠ABC.**

Show steps of your workng.

**Answer 7**

ABCD is a cyclic quadrilateral

∠A + ∠C = 180°.

∠C = 180° – ∠A = 180° – 105° = 75°

or ∠BCD = 75°

DC || AB

∠ADC + ∠DAB = 180° (Angles on the same side of the transversal of || lines)

∠ADC = 180° – ∠DAB = 180° – 105° = 75°

But ∠ADC + ∠ABC = 180° (opposite angles of a cyclic quad.)

∠ABC = 180° – ∠ADC = 180° – 75° = 105°

**Question 8.**

**In the given figure, O is centre of the circle. If ∠ AOB = 140° and ∠ OAC = 50°; find :**

**(i) ∠ ACB,**

**(ii) ∠OBC,**

**(iii) ∠OAB,**

**(iv) ∠CBA**

**Answer 8**

O is the centre of circle ∠AOB = 140° and ∠OAC = 50°.

AB is joined

Reflex ∠AOB = 360° – 140° = 220°

But ∠ACB = Reflex ∠AOB = x 220° = 110°

In quad. OACB,

∠AOB + ∠OAC + ∠ACB + ∠OBC = 360°

⇒ 140° + 50° + 110° + ∠OBC = 360°

⇒ 300° + ∠OBC = 360°

⇒ ∠OBC = 360° – 300° = 60°

In ∆OAB,

∠AOB + ∠OAB + ∠OBA = 180°

But ∠OBA = ∠OAB (Angles opposite to equal sides)

140° + ∠OAB + ∠OAB = 180°

2 ∠OAB = 180° – 140° = 40°

∠OAB = = 20°

∠OAB = ∠OBA = 20°.

⇒ ∠OBC = ∠CBA + ∠ABO

and ⇒ 60° = ∠CBA + 20°

hence ⇒ ∠CBA = 40°

**Question 9. Concise Maths Solutions Circles**

**Calculate :**

**(i) ∠ CDB,**

**(ii) ∠ ABC,**

**(iii) ∠ ACB.**

**Answer 9**

∠CDB = ∠BAC (Angles is the same segment) = 49°

∠ABC = ∠ADC (Angles in the same segment) = 43°

∠ADB = ∠ADC + ∠BDC = 43° + 49° = 92°

and ∠ADB + ∠ACB = 180° (opposite angles of a cyclic quad.)

∠ACB = 180° – ∠ADB = 180° – 92° = 88°.

**Question 10.**

**In the figure given below, ABCD is a eyclic quadrilateral in which ∠BAD = 75°; ∠ABD = 58° and ∠ADC = 77°. Find:**

**(i) ∠BDC,**

**(ii) ∠BCD,**

**(iii) ∠BCA.**

**Answer 10**

In cyclic quad. ABCD,

∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°.

∠A + ∠C = 180° (opposite angles of a cyclic quad.)

∠C = 180° – ∠A = 180° – 75° = 105° or ∠BCD = 105°

In ΔABD,

∠BAD + ∠ABD + ∠ADB = 180°

⇒ 75° + 58° + ∠ ADB = 180°

⇒ 133° + ∠ADB = 180°

⇒ ∠ADB = 180° – 133° = 47°

∠BDC = 77° – ∠ADB = 77° – 47° = 30°

But ∠BCA = ∠BDA (Angles in the same) = 47°

**Question 11.**

**In the following figure, O is centre of the circle and ΔABC is equilateral. Find :**

**(i) ∠ADB**

**(ii) ∠AEB**

**Answer 11**

∠ACB and ∠ADB are in the same segment.

∠ADB = ∠ACB. = 60°. (Angle of an equilateral triangle)

AEBD is a cyclic quadrilateral

∠AEB + ∠ADB = 180°

⇒ ∠AEB + 60° = 180°

⇒ ∠AEB = 180° – 60° = 120°.

**Question 12.**

**Given- ∠CAB = 75° and ∠CBA = 50°. Find the value of ∠DAB + ∠ABD**

**Answer 12**

In ΔABC, ∠CBA = 50°, ∠CAB = 75°,

∠ACB = 180° – (∠CBA + ∠CAB) = 180° – (50° = 75°) = 180° – 125° = 55°

Bui ∠ADB = ∠ACB = 55° (Angles in the same segment)

Now in ΔABD,

∠DAB + ∠ABD + ∠ADB = 180°.

⇒ ∠DAB + ∠ABD + 55° = 180°

⇒ ∠DAB + ∠ABD = 180° – 55° = 125°.

**Question 13.**

**ABCD is a cyclic quadrilateral in a circle with centre O. If ∠ADC = 130°; find ∠BAC.**

**Answer 13**

ABCD is a cyclic quadrilateral and ∠ADC = 130°

O is centre of the circle, AOB is diameter.

∠ABC = 180° – 130° = 50°.

In ΔABC,

∠ACB = 90° (angle in semicircle)

∠BAC + ∠CBA = 90°.

and ∠BAC + 50° = 90°

so ∠BAC = 90° – 50° = 40°.

**Question 14.**

**In the figure given below, AOB is a diameter of the circle and ∠AOC = 110°. Find ∠BDC.**

**Answer 14**

∠AOC + ∠COB = 180° (Linear pair)

∠COB = 180° – ∠AOC = 180° – 110° = 70°

Arc BC subtends ∠COB at the centre and x at the remaining part of circle

∠COB = 2x

⇒ x = ∠COB = x 70° = 35°

**Question 15. Concise Maths Solutions Circles**

**In the following figure, O is centre of the circle, ∠AOB = 60° and ∠BDC = 100°. Find ∠OBC.**

**Answer 15**

** **

Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of circle,

∠AOB = 2 ∠ACB

or ∠ACB = ∠AOB = x 60° = 30°

Now in ΔDBC,

∠DBC + ∠ACB + ∠BDC = 180°

⇒ ∠DBC + 30° + 100° = 180°

and ⇒ ∠DBC = 180° – 130° = 50°

hence ∠OBC = 50°.

**Question 16.**

**ABCD is a cyclic quadrilateral in which ∠DAC = 27°; ∠DBA = 50° and ∠ADB = 33°. Calculate :**

**(i) ∠DBC,**

**(ii) ∠DCB,**

**(iii) ∠CAB.**

**Answer 16**

(i) ∠ CBD = ∠ DAC = 27° (Angles in the same segment)

(ii) In ΔADB,

∠ABD + ∠BAD + ∠BDA = 180°.

⇒ 50° + ∠BAD + 33° = 180°

so ⇒ ∠BAD + 83° = 180°

hence ⇒ ∠BAD = 180° – 83° = 97°

In cyclic quad. ABCD,

∠BAD + ∠DCB = 180°

⇒ 97° + ∠DCB = 180°

⇒ ∠DCB = 180°- 97° = 83°

(iii) ∠BAD = 97°

⇒ ∠BAC + ∠CAD = 97°

and ⇒ ∠BAC + 27° = 97°

hence ⇒ ∠BAC = 97° – 27° = 70°

∠CAB = 70°.

** PQ**

**In the figure given below, AB is diameter of the circle whose centre is O. Given that:**

**∠ECD = ∠EDC = 32°. Show that ∠COF = ∠CEF.**

**Answer **

Given- AB is the diameter of a circle with centre O

and ∠ECD = ∠EDC = 32°

To Prove- ∠COF = ∠CEF

Proof- Arc CF subtends ∠COF at the centre and ∠CDF at the remaining part of the circle.

∠COF = 2 ∠CDF = 2 x ∠EDC = 2 x 32° = 64° ….. (i)

In ΔCED,

Ext. ∠CEF = ∠CDF + ∠DCE = ∠EDC + ∠ECD = 32° + 32° = 64° ….(ii)

from (i) and (ii)

∠CDF = ∠CEF

**Question 17**

**In the figure given below, AB and CD arc straight lines through the centre O of a circle. If ∠AOC = 80° and ∠CDE = 40°, find the number of degrees in:**

**(i) ∠DCE,**

**(ii) ∠ABC**

**Answer 17**

In circle, COD is the diameter.

∠CED = 90° (Angle in a semi circle)

In right A CDE,

∠ DCE + ∠ EDC = 90°

⇒ ∠ DCE + 40° = 90°

∠ DCE = 90° – 40° = 50°

In ΔOBC.

Ext. ∠COA = ∠OBC + ∠OCB

⇒ 80° = ∠OBC + 50°

⇒ ∠OBC = 80° – 50° = 30°

or ∠ABC = 30°

**Question 18**

**In the given figure, AC is a diameter of a circle, whose centre is O. A circle is described on AO as diameter. AE, a chord of the larger circle, intersects the smaller circle at B. Prove that AB = BE.**

**Answer 18**

**
**Given- AC is a diameter of a circle with centre O.

AE is a chord which intersects the smaller circle with AO as diameter at B.

To Prove- AB = BE.

Construction- Join OB.

Proof- ∠ABO = 90° (Angle in a semi circle)

OB ⊥ AE

OB bisects chord AE

Hence, AB = BE.

**Question 19**

**In the following figure,**

**(i) if ∠ BAD = 96°, find ∠BCD and ∠BFE,**

**(ii) Prove that AD is parallel to FE.**

**Answer 19**

Given- In the figure, ∠BAD = 96°

To Prove-

(i) Find ∠BCD and ∠BFE

(ii) AD || EF

Proof- ABCD is a cyclic quadrilateral.

∠BAD + ∠BCD = 180°

⇒ 96° + ∠BCD = 180°

⇒ ∠BCD = 180° – 96° = 84°

Again BCEF is a cyclic quadrilateral,

Ext. ∠BCD = Int. opposite ∠BFE

∠BFE = 84°.

∠BAD + ∠BFE = 96° + 84° = 180°

But these are on same side of the transversal.

AD || FE.

**Question 20**

**Prove that**

**(i) the parallelogram, inscribed in a circle, is a rectangle.**

**(ii) the rhombus, inscribed in a circle, is a square.**

**Answer 20**

(i) ABCD is a parallelogram in a circle with centre O.

To Prove- ABCD is a rectangle.

Proof- ABCD is a cyclic parallelogram.

∠A + ∠C = 180°.

But ∠A = ∠C (opposite angles of a ||gm)

∠A = ∠C = 90°

Similarly we can prove that

∠B = ∠D = 90°

Each angle of a ||gm is right angle

Hence ABCD is a rectangle.

(ii) Given- ABCD is a cyclic rhombus.

To Prove- ABCD is a square.

Proof- ABCD is cyclic rhombus

∠A + ∠C = 180°

But ∠A = ∠C (opposite angles of rhombus)

∠A = ∠C = 90°

Similarly we can prove that ∠B = ∠D = 90°

Each angle of a rhombus is a right angle

ABCD is a square.

**Question 21.**

**In the following figure, AB = AC. Prove that DECB is an isosceles trapezium.**

**Answer 21**

Given- In the figure, AB = AC.

To Prove- DECB is an isosceles trape∠ium.

Proof- In ∆ABC,

AB = AC

∠B = ∠C

DECB is a cyclic quadrilateral.

∠B + ∠DEC = 180°

∠C + ∠DEC = 180°

But this is the sum of interior angles on one side of a transversal.

DE || BC ….(i)

But ∠ADE = ∠B

and ∠AED = ∠C (Corresponding angles)

∠ADE = ∠AED (∠B = ∠C)

AD = AE (Opposite to equafangles)

But AB = AC (Given)

AB – AD = AC – AE

⇒ DB = EC ….(ii)

From (i) and (ii)

DECB is an isosceles trape∠ium.

**Question 22**

**Two circles intersect at P and Q. Through P diameters PA and PB of the two circles are draw n. Show that the points A, Q and B are collinear.**

**Answer 22**

Given- Two circles with centres O and O’ intersect each other at P and Q.

From P, PA and PB are two diameters are drawn.

To Prove- A, Q and B are collinear.

Construction-

Join PQ, AQ and BQ.

Proof- In first circle.

∠PAQ = 90° (Angle in a semi circle) ….(i)

Similarly, in second circle, ∠PBQ = 90° ….(ii)

Adding (i) and (ii).

∠PAQ + ∠PBQ = 90° + 90° = 180°

But, there are adjacent angles

AQB is a straight line.

Hence A, Q and B are collinear.

Hence proved.

**PQ**

**ABCD is a quadrilateral inscribed in a circle, having ∠A = 60°; O is the centre of the circle. Show that:**

**∠OBD + ∠ODB = ∠CBD + ∠CDB.**

**Answer **

Given- ABCD is a cyclic quadrilateral in which ∠A = 60°

and O is the centre of the circle.

BD, OB and OD are joined.

To Prove- ∠OBD + ∠ODB = ∠CBD + ∠CDB

Proof- Arc BCD subtends ∠BOD at the centre and ∠BAD at remaining part of the circle.

∠BOD = 2 ∠BAD = 2 x 60° = 120°

In ∆BOD,

∠BOD = 120°

∠OBD + ∠ ODB = 180° – 120° = 60° …. (i)

ABCD is a cyclic quadrilateral

∠A + ∠C = 180°

⇒ 60° + ∠C = 180°

⇒ ∠C = 180° – 60° = 120°

In ∆BCD,

∠CBD + ∠CDB + ∠ C = 180°.

and ∠CBD + ∠CDB + 120° = 180°

∠CBD + ∠CDB = 180° – 120° = 60° ….(ii)

From (i) and (ii),

∠OBD + ∠ODB = ∠CBD + ∠CDB

**Question 23**

**The figure given below, shows a circle with centre O.**

**Given- ∠AOC = a and ∠ABC = b.**

**(i) Find the relationship between a and b. :**

**(ii) Find the measure of angle OAB, if OABC is a parallelogram.**

**Answer 23**

(i) ∠AOC = a, ∠ABC = b

Reflex ∠AOC = 360° – a

Now arc AC subtends reflex ∠AOC at the centre and ∠ABC at the remaining part of the circle

∠ABC = ref. ∠AOC

b = (360° – a)

⇒ 2b = 360° – a

⇒ a + 2b = 360° ….(i)

(ii) If OABC is a || gm,

then ∠AOC = ∠ABC

⇒ a = b

Substituting the value of a, in ….(i)

b + 2b = 360°

⇒ 3b = 360°

⇒ b = 120°

But ∠OAB + ∠ABC = 180° (Angles in a || gm)

⇒ ∠OAB + b = 180°

and ⇒ ∠OAB + 120° = 180°

hence ⇒ ∠OAB = 180° – 120° = 60°.

**Question 24**

**Two chords AB and CD intersect at P inside the circle. Prove that the sum of the angles subtended by the arcs AC and BD at the centre O is equal to twice the angle APC.**

**Answer 24**

Given- Two chords AB and CD intersect each other at P inside the circle, OA, OB, OC and OD are joined.

To Prove- ∠AOC + ∠BOD = 2 ∠APC.

Construction- Join AD

Proof- Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining pari of the circle

∠AOC = 2 ∠ADC ….(i)

Similarly, ∠BOD = 2 ∠BAD ….(ii)

Adding (i) and (ii),

∠AOC + ∠BOD = 2 ∠ADC

⇒ 2 ∠ BAD = 2 (∠ADC + ∠BAD) ….(iii)

But in ∆PAD,

Ext. ∠APC = ∠PAD + ∠ADC = ∠ADC + ∠BAD …(iv)

from (iii) and (iv)

∠AOC + ∠BOD = 2 ∠APC

**Question 25**

**In the given figure, RS is a diameter of the circle. NM is parallel to RS and ∠MRS = 29°.**

**Calculate:**

**(i) ∠RNM,**

**(ii) ∠NRM.**

**Answer 25**

** **

Join RN and MS. .

(i) RS is the diameter

∠RMS = 90° (Angle in semi circle)

and ∠RSM + ∠MRS = 90°

∠RSM = 90° – 29° = 61°

But ∠RSM + ∠RNM = 180° (Angles in a cyclic quad.)

61° + ∠RNM = 180°

⇒ ∠RNM = 180° – 61 = 119°

NM || RS

∠NMR = ∠MRS = 29° (Alt. angles)

In ∆RNM,

∠NRM + ∠RNM + ∠NMR = 180°

⇒ ∠NRM + 119° + 29° = 180°

so ⇒ ∠NRM + 148° = 180°

⇒ ∠NRM = 180° – 148° = 32°.

**Question 26**

**In the figure given alongside, AB // CD and O is the centre of the circle. If ∠ADC = 25°; find the angle AEB. Give reasons in support of your answer.**

**Answer 26**

AB || CD.

∠BAD = ∠ADC (Alternate angles) = 25° (∠ADC = 25° given)

Join AC and BD.

∠CAD = 90° (Angle in semi circle)

∠CAB = ∠CAD + ∠DAB = 90° + 25° = 115°

Now in cyclic quad. CABD.

∠CAB + ∠BDC = 180°

⇒ ∠CAB + ∠BDA + ∠ADC = 180°

and ⇒ 115° + ∠BDA + 25° = 180°

so ⇒ ∠BDA + 140° = 180°

⇒ ∠BDA = 180° – 140° = 40°

∠AEB and ∠BDA are in tire same segment of a circle

∠AEB = ∠BDA = 40° (proved)

Hence ∠AEB = 40°.

**Question 27.**

**Two circles intersect at P and Q. Through P, a straight line APB is drawn to meet the circles in A and B. Through Q, a straight fine is drawn to meet the circles at C and D. Prove that AC is || to BD.**

**Answer 27**

Given- Two circles intersect each other at P and Q. Through P, a line APB is drawn to meet the circles in A and B. Through Q, another straight line CQD is drawn meeting the circles in C and D.

AC, BD are joined.

To Prove- AC || BD.

Construction- Join PQ

Proof- APQC is a cyclic quadrilateral.

∠A + ∠PQC = 180° …… (i)

In cyclic quad. PBDQ,

Ext. ∠PQC = ∠ B …… (ii)

from (i),

∠A + ∠B = 180°.

But these are interior angles on the same side of a transversal.

AC || BD.

**Question 28**

**ABCD is a cyclic quadrilateral in which AB and DC on being produced, meet at P**

**Answer 28**

Given- The sides AB and DC of a cyclic quad. ABCD are produced to meet at P and PA = PD.

To Prove- AD || BC

Proof- In ∆PAD,

PA = PD (given)

∠A = ∠D (angles opposite to equal sides)

ADCB is a cyclic quad.

Ext. ∠PCB = ∠A = ∠D

But these are corresponding angles. ,

BC || AD or AD || BC.

**Question 29**

**AB is a diameter of the circle APBR as shown in the figure. APQ and RBQ are straight lines. Find:**

**(i) ∠PRB,**

**(ii) ∠ PBR,**

**(iii) ∠ BPR.**

**Answer 29**

∠PRB = ∠BAP (Angles in the same segment)

∠PRB = 35°

In ∆ABP,

∠APB = 90° (Angle in semi circle)

∠BPQ = 90°.

In ∆PQR,

∠R + ∠Q + ∠RPQ = 180°

⇒ 35° + 25° + ∠RPQ = 180°

and ⇒ ∠RPQ = 180° – 60° = 120°

⇒ ∠BPR = ∠RPQ – ∠BPQ = 120° – 90° = 30°

In ∆PBR,

∠PBR = 180° – (∠R + ∠BPR) = 180° – (35° + 30°) = 180° – 65° = 115°

**Question 30**

**In the given figure SP is bisector of ∠RPT and PQRS is a cyclic quadrilateral. Prove that SQ = SR.**

**Answer 30**

Given- SP is the bisector of ∠RPT and PQRS is a cyclic quadrilateral.

To Prove SQ = SR.

Proof- In cyclic quad. PQRS,

Ext. ∠SPT = ∠QRS

But ∠RPS = ∠SPT (PS is the bisector of ∠RPT)

∠QRS = ∠RPS ….(i)

But ∠RPS = ∠RQS (Angles in the same segment)

∠QRS = ∠RQS

Now in ∆QRS,

∠QRS = ∠RQS (proved)

SQ = SR (Sides opposite to equal angles)

**Question 31.**

**In the figure, O is the centre of the circle, ∠AOE = 150°, ∠DAO = 51°. Calculate the si∠es of the angles CEB and OCE.**

**Answer 31**

In the figure, ∠AOE = 150°, ∠DAO = 51°

Now in cyclic quad. ADEB,

Ext. ∠CEB = Int. Opp ∠DAO = 51°.

In ∆OEB,

Ext. ∠AOE = ∠OBE + ∠OEB

= ∠OBE + ∠OBE (OB = OE) = 2 ∠OBE

2 ∠OBE = 150°

⇒ ∠ OBE = 75°

∠EBC = 180° – 75° = 105°

Now in ∆EBC,

∠CEB + ∠OCE + ∠EBC = 180°

⇒ 51° + ∠OCE + 105° = 180°

and ⇒ ∠OCE + 156° = 180°

hence ⇒ ∠OCE = 180° – 156° = 24°.

**Question 32.**

**In the figure, given below, P and Q arc the centres of two circles intersecting at B and C. ACD is a straight line. Calculate the numerical value of x.**

**Answer 32**

In circle with centre P,

Arc AB subtends ∠APB at the centre and ∠ACB at the remaining part of the circle.

∠APB = 2 ∠ACB

⇒ ∠ACB = ∠APB = x 150° = 75°

But ∠ACB + ∠DCB = 180° (Linear pair)

∠DCB = 180° – ∠ACB = 180° – 75° = 105°

In circle with centre O,

Arc BD subtends ∠ BQD at the centre and ∠ DCB at the remaining part of the circle

∠BQD = 2 ∠DCB = 2 x 105° = 210°

But x + ∠BQD = 360° (Angles at a point)

⇒ x + 210° = 360°

⇒ x = 360° – 210° = 150°.

**Question 33. Concise Maths Solutions Circles **

**The figure shows two circles which intersect at A and B. The centre of the smaller circle is O and lies on the circumference of the larger circle. Given ∠APB = a°. ****Calculate, in terms of a°, the value of :**

**(i) obtuse ∠AOB,**

**(ii) ∠ACB,**

**(iii) ∠ADB.**

**Give reasons for your answers clearly.**

**So**

**Answer 33**

Arc AB in small circle subtends ∠AOB at the centre and ∠APB at the remaining part of the circle.

(i) ∠AOB = 2 ∠APB = 2a° (∠APB = a°)

∠APB = (2a)°

(ii) In larger circle, AOBC is a cyclic quad.

∠AOB + ∠ACB = 180°.

⇒ 2a° + ∠ACB = 180°

∠ACB = 180° – 2a° = (180° – 2a°)

(iii) But ∠ ACB and ∠ ADB are in the same segment

∠ADB = ∠ ACB = (180° – 2a°)

**Question 34.**

**In the given figure, O is the centre of the circle and ∠ABC = 55°. Calculate the values of x and y.**

**Answer 34**

In ∆OBC,

OB = OC (radii of the same circle)

∠OBC = ∠BCO or ∠ABC = ∠BCO

∠BCO = ∠ABC = 55°

Now in ∆OBC,

Ext. AOC = ∠OBC + ∠BCO = 55° + 55° = 110°

x = 110°

Now in cyclic quad, ABCD,

∠ADC + ∠ABC = 180°

⇒ y + 55° = 180°

⇒ y = 180° – 55° = 125°

**Question 35.**

**In the given figure, A is the centre of the circle, ABCD is a parallelogram and CDE is a straight line. Prove that: ∠BCD = 2 ∠ABE.**

**Answer 35**

Given- A is the centre of the circle and ABCD is a parallelogram.

CDE is a straight line.

To Prove- ∠BCD = 2 ∠ABE.

Proof- AB || DC (opposite sides of a || gm)

∠ABE = ∠BED (Alternate angles) ….(i)

ABCD is a || gm (given)

∠BAD = ∠BCD (opposite angles of a ||gm) ….(ii)

Now arc BD subtends ∠BAD at the centre and ∠BED at the remaining part of the circle

∠BAD = 2 ∠BED

from (i) and (ii)

∠BCD = 2 ∠ABE

**Question 36.**

**ABCD is a cyclic quadrilateral in which AB is parallel to DC and AB is a diameter of the circle. Given ∠BED = 65°; calculate :**

**(i) ∠DAB,**

**(ii) ∠BDC.**

**Answer 36**

∠DAB and ∠BED are in the same segment of the circle.

∠DAB = ∠BED = 65° (∠BED = 65° given)

DC || AB (Given)

∠BDC = ∠DBA (Alternate angles)

In ∆ADB,

AOB is the diameter

∠ADB = 90° (Angle in semi circle)

∠DAB + ∠DBA = 90°

⇒ 65° + ∠DBA = 90°

⇒ ∠DBA = 90° – 65° = 25°

But ∠DBA = ∠BDC (proved)

∠BDC = 25°

**Question 37. Concise Maths Solutions Circles **

**In the given figure, AB is a diameter of the circle. Chord ED is parallel to AB and ∠EAB = 63°. Calculate:**

**(i) ∠EBA,**

**(ii) ∠BCD.**

**Answer 37**

AOB is the diameter,

∠AEB = 90°

and ∠EAB + ∠EBA = 90°

⇒ 63° + ∠EBA = 90°

⇒ ∠EBA = 90° – 63° = 27°

ED || AB (given)

∠DEB = ∠EBA (Alternate angles) = 27°

In cyclic quad. EBCD,

∠DEB + ∠BCD = 180° (opposite angles of a cyclic quad.)

⇒ 27° + ∠BCD = 180°

⇒ ∠BCD = 180° – 27° = 153°.

**PQ**

**The sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E; the sides DA and CB arc produced to meet at F. If ∠BEG = 42° and ∠BAD = 98°; calculate :**

**(i) ∠AFB,**

**(ii) ∠ADC.**

**Answer **

In ∆AED,

∠ADE + ∠AED + ∠EAD = 180° (Angles of a triangle)

⇒ ∠ADE + 42° + 98° = 180°

⇒ ∠ADE + 140° = 180°

⇒ ∠ ADE = 180° – 140° = 40° or ∠ADC = 40°

In cyclic quad. ABCD.

∠BAD + ∠BCD = 180°

⇒ 98° + ∠BCD = 180°

⇒ ∠BCD = 180° – 98° = 82°

Now in ∆FCD,

∠DFC + ∠FDC + ∠FCD = 180°

⇒ ∠AFB + ∠ADC + ∠BCD = 180°

and ⇒ ∠AFB + 40° + 82° = 180°

so ⇒ ∠AFB + 122° = 180°

hence ⇒ ∠AFB = 180° – 122° = 58°

**Question 38**

**In the following figure, AB is a diameter of the circle with centre O. DO is parallel to CB and ∠DCB = 120°. Calculate :**

**(i) ∠DAB,**

**(ii) ∠DBA,**

**(iii) ∠DBC,**

**(iv) ∠ADC. Also, show that the ∆AOD is an equilateral triangle.**

**Answer 38**

(i) ABCD is a cyclic quadrilateral,

∠DCB + ∠DAB = 180°

⇒ 120° + ∠DAB = 180°

∠DAB =180° – 120° = 60°

##### (ii) AOB is a diameter.

∠ADB = 90° (Angle in a semi circle)

∠ DAB + ∠DBA = 90°

60° + ∠ DBA = 90°

∠DBA = 90° – 60° = 30°

##### (iii) In ∆OBD,

OD = OB (radii of the same circle)

∠ ODB = ∠ OBD

or ∠ABD = 30° (from ii)

But DO || CB (given)

∠ODB = ∠DBC (Alternate angles)

⇒ 30° = ∠DBC or ∠DBC = 30°

##### (iv)

∠ABD + ∠DBC = 30° + 30° = 60°

⇒ ∠ABC = 60°

Again in cyclic quad. ABCD,

∠ADC + ∠ABC = 180°

∠ADC + 60° = 180°

∠ ADC = 180° – 60° = 120°

In ∆AOD,

OA = OD (radii of the same circle)

∠ AOD = ∠ DAO or ∠ DAB = 60° (proved in (i))

and ∠ADO = 60° (Third angle)

∠ADO = ∠AOD = ∠DAO = 60°

∆AOD is an equilateral triangle.

**Question 39**

**In the given figure, I is the incentre of ∆ABC. BI when produced meets the circum circle of ∆ABC at D.**

**Given ∠BAC = 55° and ∠ACB = 65°; calculate:**

**(i) ∠DCA,**

**(ii) ∠DAC,**

**(iii) ∠DCI,**

**(iv) ∠AIC**

**Answer 39 Concise Maths Solutions Circles **

Join AD, DC, AI and Cl,

In ∆ABC,

∠BAC = 55°, ∠ACB = 65°

∠ABC = 180° – (∠BAC + ∠ACB) = 180°- (55° + 65°) = 180° – 120° = 60°

In cyclic quad. ABCD,

∠ABC + ∠ADC = 180°

⇒ 60° + ∠ADC = 180°

∠ADC = 180° – 60°= 120°

In ∆ADC,

∠ DAC + ∠ DCA + ∠ ADC = 180°

⇒ ∠ DAC+ ∠ DCA + 120° = 180°

⇒ ∠ DAC+ ∠ DCA = 180° – 120° = 60°

But ∠ DAC = ∠ DCA (I lies on the bisector of ∠ ABC)

∠ DAC = ∠ DCA = 30°

DI is perpendicular bisector of AC

∠ AIC = ∠ ADC= 120°

IC is the bisector of ∠ ACB

∠ ICA = = 32.5°

∠DCI = ∠DCA + ∠ACI = 30° + 32.5° = 62.5° = (62.5)° = 60° 30′.

**Question 40.**

**A triangle ABC is inscribed in a circle. The bisectors of angles BAC, ABC and ACB meet the circum circle of the triangle at points P, Q and R respectively. Prove that:**

**(i) ∠ ABC = 2 ∠ APQ,**

**(ii) ∠ ACB = 2 ∠ APR,**

**(iii) ∠ QPR = 90° – ∠BAC.**

**Answer 40**

Given- ∆ABC is inscribed in a circle. Bisectors of ∠BAC, ∠ABCand ∠ACB meet the circumcircle of the∆ABC at P, Q and R respectively.

To Prove-

(i) ∠ ABC = 2 ∠APQ.

(ii) ∠ ACB = 2 ∠ APR.

(iii) ∠ QPR = 90° – ∠BAC.

Construction-

Join PQ and PR.

Proof- ∠ABQ and ∠APQ are in the same segment of the circle.

∠ ABQ = ∠ APQ

But ∠ ABQ = – ∠ ABC (BQ is the angle bisector of ∠ ABC)

∠ ABC = ∠ APQ

Or ∠ ABC = 2 ∠APQ …,(i)

Similarly, ∠ APR and ∠ ACR are in the same segment of the circle.

∠ APR = ∠ ACR

But ∠ ACR = ∠ ACB (CR is the angles bisector of ∠ ACB)

∠ ACB = ∠ APR

Or ∠ ACB = 2 ∠ APR ….(ii)

Adding (i) and (ii)

∠ ABC + ∠ ACB = 2 ∠ APQ + 2∠ APR = 2 (∠ APQ + ∠ APR) = 2 ∠ PQR

Or 2 ∠ PQR = ∠ ABC + ∠ ACB

∠ PQR = (∠ ABC + ∠ ACB) ….(iii)

But ∠ ABC + ∠ ACB + ∠ BAC = 180° (Angles of a triangle)

∠ ABC + ∠ ACB = 180° – ∠ BAC ….(iv)

from (iii) and (iv) we get,

∠ PQR = (180° – ∠ BAC) = 90° – ∠ BAC

**Question 41.**

**Calculate the angles x, y and z if :**

**Solution:**

**Answer 41**

Ext. ∠ ADC = x + z ….(i)

and in ΔBPC,

Ext. ∠ ABC = y + x ….(ii)

(∠ BCP = ∠ DCQ = x vertically opposite angles)

Adding (i) and (ii),

x + z + y + x = ∠ ADC + ∠ ABC.

But ∠ ADC + ∠ ABC = 180° (opposite angles of a cyclic quad)

2x + y + z = 180°

⇒ 2 x 3k + 4k + 5k = 180°

and ⇒ 6k + 4k + 5k = 180°

so ⇒ 15k = 180°

⇒ k = 12°

x = 3k = 3 x 12° = 36°= x = 36°

y = 4k = 4 x 12° = 48°= y = 48°

z = 5k = 5 x 12° = 60° = z = 60°

**Question 42. Concise Maths Solutions Circles **

**In the given figure, AB = AC = CD and ∠ADC = 38°. Calculate :**

**(i) Angle ABC**

**(ii) Angle BEC [1995]**

**Answer 42**

In the figure, AB = AC = CD, ∠ ADC = 38°

BE is joined.

In ΔACD, AC = CD

∠CAD = ∠CDA = 38°

Ext. ∠ ACB = ∠ CAD + ∠ CDA = 38° + 38° = 76°

But in ΔABC,

AB = AC (given)

∠ ABC = ∠ ACB = 76°

and ∠ BAC =180° – (76° + 76°) = 180° – 152° = 28°

But ∠ BEC = ∠ BAC (Angles in the same segment)

∠BEC = 28°.

**Question 43.**

**In the given figure. AC is the diameter of circle, centre O. Chord BD is perpendicular to AC. Write down the angles p, q and r in terms of x. [1996]**

**Answer 43**

Arc subtends ∠ AOB at the centre and ∠ACB at the remaining part of the circle.

∠ AOB = 2 ∠ ACB

⇒ x = 2q ⇒ q =

But ∠ ADB and ∠ ACB are in the same segment

∠ ADB = ∠ ACB – q

Now in ΔAED,

p + q + 90° = 180° (sum of angles of a Δ)

⇒ p + q = 90°

and ⇒ p = 90° – q

⇒ p = 90° –

Arc BC subtends ∠ BOC at the centre and ∠ ADC at the remaining part of the circle

∠BOC = 2 ∠BDC = 2r.

r = ∠ BOC = (180° – x)

(∠ AOB + ∠ BOC = 180°)

r = 90° – x. = 90° –

**Question 44.**

**In the given figure, AC is the diameter of the circle with centre O. CD and BE are parallel. Angle ∠ AOB = 80° and ∠ ACE = 10°. Calculate:**

**(i) Angle BEC,**

**(ii) Angle BCD,**

**(iii) Angle CED. [1998]**

**Answer 44**

Arc AB subtends ∠ AOB at the centre and ∠ACB at the remaining part of the circle.

∠ ACB = ∠AOB = x 80° = 40°

But ∠ BOC + ∠ AOB = 180° (A linear pair)

∠ BOC + 80° = 180°

⇒ ∠ BOC = 180° – 80° = 100°

(i) Arc BC subtends ∠ BOC at the centre and ∠ BEC at the remaining part of the circle

∠ BEC = ∠ BOC = x 100° = 50°

(ii) EB || DC

∠ DCE = ∠ BEC (Alternate angles) = 50°

∠ BCD = ∠ BCA + ∠ ACE + ∠ ECD = 40° + 10° + 50° = 100°

(iii) In cyclic quad. CDE,

∠ BED + ∠ BCD = 180°

⇒ ∠ BEC + ∠ CED + ∠ BCD = 180°

and ⇒ 50° + ∠ CED + 100° = 180° (Proved in (i) and (ii))

so ⇒ ∠ CEb + 150° = 180°

∠ CED = 180° – 150° = 30°.

**Question 45.**

**In the given figure, AE is the diameter of the circle. Write down the numerical value of ∠ ABC + ∠ CDE. Give reasons for your answer. [1998]**

**Solutio**

**Answer 45**

** **

Join OA, OB, OC, OD.

In ΔOAB,

OA = OB (Radii of the same circle)

∠ 1 = ∠ 2

Similarly we can prove that

∠3 = ∠4,

∠5 = ∠6,

∠7 = ∠8

In A OAB,

∠1 + ∠2 + ∠a = 180° (Angles of a triangle)

Similarly ∠3 + ∠4 + ∠b = 180°

∠5 + ∠6 + ∠c = 180°

∠7 + ∠8 + ∠d = 180°

Adding we get

∠1 + ∠2 + ∠3 + ∠4 + ∠5 + ∠6 + ∠7 + ∠8 + ∠a + ∠b + ∠c + ∠d = 4 x 180° = 720°

⇒∠2 + ∠2 + ∠3 + ∠3 + ∠6 + ∠6+ ∠ 7 + ∠7 + ∠a + ∠b + ∠c + ∠d = 720°

and ⇒ 2 ∠2 + 2 ∠3 + 2 ∠6 + 2 ∠7 + ∠a + ∠ b + ∠ c + ∠ d = 720°

so ⇒ 2 [∠2 + ∠3] + 2 [∠6 + ∠7| + 180° = 720° ( ∠a + ∠b + ∠c + ∠d = 180°)

therefore ⇒ 2 ∠ ABC + 2 ∠ CDE = 720° – 180° = 540°

⇒ 2 (∠ ABC + ∠ CDE) = 540°

Hence ⇒ ∠ ABC + ∠ CDE = 270°

**Question 46. Concise Maths Solutions Circles **

**In the given figure, AOC is a diameter and AC is parallel to ED. If ∠ CBE = 64°, calculate ∠ DEC. [1991]**

**Answer 46**

Join AB.

AOC is the diameter

∠ ABC = 90° (Angle in a semi circle)

⇒ ∠ ABE + ∠ CBE = 90°

⇒ ∠ ABE + 64° = 90°

∠ ABE = 90° – 64° = 26° …(i)

AC || ED

∠ DEC = ∠ ACE (alternate angles)

But ∠ ACE = ∠ ABE (Angles in the same segment)

∠ DEC = ∠ ABE = 26° [from (i)]

**Question 47.**

**Use the given figure to find :**

**(i) ∠ BAD,**

**(ii) ∠ DQB. [1987]**

**Answer 47**

In ΔAPD,

∠ ADP + ∠ DPA + ∠ PAD = 180°

85° + 40° + ∠ PAD = 180°

∠ PAD = 180° – (85° + 40°) = 180° – 125° = 55° or ∠ BAD = 55°

In cyclic quad. ABCD,

∠ ADC + ∠ ABC = 180°

85° + ∠ ABC = 180°

∠ ABC = 180° – 85° = 95°

Now, in ΔAQB,

∠ QAB + ∠ ABC + ∠ BQA = 180°

⇒ 55° + 95° + ∠ BQA = 180°

and ⇒ 150° + ∠ BQA = 180°

hence⇒ ∠DQB = ∠ BQA = 180° – 150° = 30°

**Question 48.**

**In the given figure, AOB is a diameter and DC is parallel to AB. If ∠ CAB = x°; find (in terms of x) the values of:**

**(i) ∠ COB,**

**(ii) ∠DOC,**

**(iii) ∠DAC,**

**(iv) ∠ADC. [1991]**

**Answer 48**

Join CB.

In ΔAOC,

OA = OC (radii of the same circle)

∠ OCA = 4 OAC = x

Ext. ∠ COB = ∠ OAC + ∠ OCA = x + x = 2x

In ΔACB, ∠ ACB = 90° (Angle in semi circle)

∠ OBC = 90° – ∠ OAC = 90° – x

In cyclic quad. ABCD,

∠ ABC + ∠ ADC = 180°

⇒ ∠OBC + ∠ADC =180°

⇒ (90 – x) + 4 ADC = 180°

∠ADC = 180° – 90° + x = 90° + x

DC || AB

∠ DCO = ∠ COB = 2x (alternate angle)

And ∠ DCA = ∠ CAB = x (alternate angles)

In ΔADC,

∠ DAC + ∠ DCA + ∠ ADC = 180°

∠ DAC + x + 90 + x = 180°

2x + 90° + ∠ DAC = 180°

∠ DAC = 180° – 90° – 2x = 90° – 2x

In ΔOCD,

∠ DOC + ∠ OCD + ∠ CDO = 180°

and ∠ DOC + 2x + 2x = 180°

so ∠ DOC = 180° – 4x

Hence ∠ COB = 2x,

therefpre ∠ DOC = 180° – 4x

∠ DAC = 90° – 2x

and ∠ ADC = 90° + x

**Question 49.**

**In the given figure, AB is the diameter of a circle with centre O. ∠BCD = 130°. Find :**

**(i) ∠DAB**

**(ii) ∠DBA**

**Answer 49**

Join DB

(i) ∠DAB + ∠DCB = 180° [Opposite angles of a cyclic quadrilateral are supplementary]
∠DAB = 180° – 130° = 50°

(ii) In ΔADB,

∠ADB = 90° [Angle in a semi-circle is 90°]
So, ∠DBA = 180° – (∠DAB + ∠ADB) = 180° – (50° + 90°) = 40°

**Question 50.**

**In the given figure, PQ is a diameter of the circle whose centre is O. Given ∠ROS = 42°, calculate ∠RTS. [1992]**

**Answer 50**

In ΔOPR, OR = OP (radii of the same circle)

∠ OPR = ∠ ORP = x (Say)

∠POR = 180° – 2x

Similarly in ΔOQS,

OS = OQ

∠ OSQ = ∠ SQO = y (say)

∠ SOQ = 180° – 2y

POQ is a straight line,

∠ POR + ∠ ROS + ∠ SOQ = 180°

and ⇒ 180° – 2x + 42° + 180° – 2y = 180°

so ⇒ 222° – 2x – 2y = 0

hence ⇒ 2 (x + y) = 222°

x + y = 111° ….(i)

In. ΔPQT,

⇒ ∠P + ∠Q + ∠T = 180°

and ⇒ ∠ OPR + ∠ SQO + ∠ RTS = 180°

so ⇒ x + y + ∠RTS = 180°

hence ⇒ ∠ RTS = 180° – (x + y) = 180° – 1110 [From(i)] = 69°

**Question 51. Concise Maths Solutions Circles **

**In the given figure, PQ is a diameter. Chord SR is parallel to PQ. Given that ∠ PQR = 58°, Calculate:**

**(i) ∠RPQ,**

**(ii) ∠STP. [1989]**

**Answer 51**

Join PR,

In ΔPQR,

∠ PRQ = 90° (Angle in a semi-circle)

∠ RPQ + ∠ RQP = 90°

⇒ ∠RPQ + 58° = 90°

⇒ ∠RPQ = 90° – 58° = 32°

SR || PQ (given)

∠ SRP = ∠ RPQ (Alternate angles) = 32° [from(i)]
Now, in cyclic quad. PRST,

∠ STP + ∠ SRP = 180°

⇒ ∠STP + 32° = 180°

⇒ ∠STP = 180° – 32° = 148°

**Question 52.**

**AB is the diameter of the circle with centre O. OD is parallel to BC and ∠ AOD = 60°. Calculate the numerical values of: [1987]**

**(i) ∠ ABD**

**(ii) ∠ DBC**

**(iii) ∠ ADC**

**Answer 52**

Join BD,

Arc AD, subtends ∠ AOD at the centre and ∠ ABD at the remaining part of the circle

∠ ABD = ∠ AOD = x 60° = 30°

In ΔOBD,

OB = OD (Radii of the same circle)

∠ ODB = ∠ OBD = ∠ ABD = 30°

OD||BC (given)

∠ ODB = ∠ DBC (Alternate angles)

∠ DBC = ∠ ODB = 30° .

Again OD || BC

∠ AOD = ∠ OBC (Corresponding angles)

⇒ ∠ OBC = ∠ AOD = 60°

Now, in cyclic quad. ABCD,

∠ ADC+ ∠ ABC = 180°

⇒ ∠ ADC + 60° = 180°

⇒ ∠ ADC = 180° – 60° = 120°

**Question 53.**

**In the given figure, the centre O of the small circle lies on the circumference of the bigger circle. If ∠ APB = 75° and ∠ BCD = 40°. find:**

**(i) ∠AOB,**

**(ii) ∠ACB,**

**(iii) ∠ABD,**

**(iv) ∠ADB. [1984]**

**Answer 53**

** **

Join AB, AD.

(i) Arc AB of the smaller circle subtends ∠ AOB at the centre and ∠ APB at the remaining part of the circle.

∠ AOB = 2 ∠ APB = 2 x 75° = 150°

(ii) OACB is a cyclic quad.

∠AOB + ∠ACB = 180°

⇒ 150° + ∠ACB = 180°

⇒ ∠ACB = 180° – 150° = 30°

(in) Again, ABDC is a cyclic quad,

∠ ABD + ∠ ACD = 180°

⇒ ∠ABD + (30° + 40°) = 180° (∠ ACD = ∠ ACB + ∠ BCD)

and ⇒ ∠ ABD + 70° = 180°

so ⇒ ∠ ABD = 180° – 70° = 110°

(iv) ∠ ACB and ∠ADB are in the same segment

∠ ADB = ∠ ACB = 30°

**Question 54. Concise Maths Solutions Circles **

**In the given figure, ∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°. Find :**

**(i) ∠BCD**

**(ii) ∠ACB**

**Hence, show that AC is a diameter.**

**Answer 54**

In circle ABCD is a cyclic quadrilateral AC andBD are joined.

∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°

ABCD is a cyclic quadrilateral

65° + ∠BCD = 180°

⇒ ∠BCD = 180° – 65° = 115°

Arc AB subtends ∠ABD and ∠ACD in the same segment

∠ACD = ∠ABD = 70° (∠ABD = 70°)

∠ACB = ∠BCD – ∠ACD = 115° – 70° = 45°

But arc AB subtends ∠ADB and ∠ACD in the same segment

∠ADB = ∠ACB = 45°

∠ADC = ∠ADB + ∠BDC = 45° + 45° = 90°

Now in segment ADC,

∠ADC = 90°

Segment ADC is a semi-circle

so AC is the diameter of the circle.

Hence proved.

**Question 55.**

**In a cyclic quadrilateral ABCD, ∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5; find each angle of the quadrilateral.**

**Answer 55**

In a cyclic quadrilateral ABCD,

∠A : ∠C = 3 : 1 and ∠B : ∠D = 1 : 5

Let ∠A = 3x, ∠C = x,

But ∠A + ∠C = 180° (Opposite angles of a cyclic quadrilateral)

and ⇒ 3x + x = 180°

so ⇒ 4x = 180°

therefore ⇒ x = 45°

∠A = 3x = 3 x 45° = 135°

∠C = x = 45°

∠B = ∠D = 1 : 5

Similarly, Let ∠B = y and ∠D = 5y

But ∠B + ∠D = 180°

y + 5y = 180°

⇒ 6y = 180°

⇒ y = 30°

∠B = y = 30°

and ∠D = 5y = 5 x 30° = 150°

Hence ∠A = 135°, ∠B = 30°, ∠C = 45° and ∠D = 150°

Hence Proved.

**Question 56. Concise Maths Solutions Circles **

**The given figure shows a circle with centre O and ∠ABP = 42°. Calculate the measure of:**

**(i) ∠PQB**

**(ii) ∠QPB + ∠PBQ**

**Answer 56**

In the figure

∠ABP = 42°.

Join PO, QO

Arc PA subtends ∠POA at the centre and ∠PBA at the remaining part.

∠POA = 2 ∠PBA = 2 x 42° = 84°

But ∠AOP + ∠BOP = 180° (Linear pair)

and ⇒ ∠POA+ ∠POB = 180°

so ⇒ 84° + ∠POB = 180°

hence ⇒ POB = 180° – 84° = 96°

Similarly, arc BP subtends ∠BOP on the centre and ∠PQB at the remaining part of the circle

∠PQB = ∠POB = x 96° = 48°

But in ΔPBQ,

∠QPB + ∠PBQ + ∠PQB = 180° (Angles of a triangle)

∠QPB + ∠PBQ + 48° =180°

⇒ ∠QPB + ∠PBQ = 180°

and ⇒ ∠QPB + ∠PBQ = 180° – 48° = 132

Hence (i) PQB = 48° and

(ii) ∠QPB + ∠PBQ = 132°

**Question 57. Concise Maths Solutions Circles **

**In the given figure, M is the centre of the circle. Chords AB and CD are perpendicular to each other.**

**If ∠MAD = x and ∠BAC = y:**

**(i) express ∠AMD in terms of x.**

**(ii) express ∠ABD in terms of y.**

**(iii) prove that: x = y.**

**Sol**

**Answer 57**

In the figure, M is the centre of the circle chords AB and CD are perpendicular to each other at L.

∠MAD = x and ∠BAC = y

##### (i) In ΔAMD,

AM = DM (Radii of the same circle)

∠MDA = ∠MAD (Angles opposite to equal sides) = x

But, in ΔAMD,

∠MAD + ∠MDA + ∠AMD = 180° (Sum of angles of a triangle)

and⇒ x + x + ∠AMD = 180°

so ⇒ 2x + ∠AMD = 180°

hence ⇒ ∠AMD = 180°

∠AMD = 180° – 2x

##### (ii)

Arc AD subtends ∠AMD at the circle and ∠ABD at the remaining part of the circle

∠AMD = 2∠ABD

⇒ ∠ABD = ∠ABD = [180°- 2x] = 90° – x

AB ⊥s CD

∠ALC = 90°

In ΔALC,

∠LAC + ∠LCA = 90°

and⇒ ∠BAC + ∠DAC = 90°

so ⇒ y = ∠DAC = 90°

hence ⇒ ∠DAC = 90° – y

#### (iii)

But ∠DAC ∠ABD (Angles in the same segment)

∠ABD = 90° – y

But ∠ABD = 90° – x, Proved

90°- x = 90°- y

⇒ ∠x = yHence proved.

**PQ**

**In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that ∠APB = 60°.**

**Solution:**

**Answer **

Given : In circle with centre O,

ABCD is cyclic quadrilateral in which CD is equal radius of the circle and AB is diameter.

CD = AB

AD and BC are produced to meet at P.

To prove: ∠APB = 60°

Construction : Join DO, CO and PB

In ΔDOC,

DO = CO = DC (Radii of the circle)

ΔDOC is an equilateral triangle

∠DOC = 60° ….(i)

Now, arc DC subtends ∠DOC at the centre arc ∠DBC at the remaining part of the circle

∠DBC = ∠DOC = x 60° = 30° …(ii)

But ∠ADB = 90° (Angle in a semi-circle)

∠PDB = 90° (∠PDB x ∠ADB = 180°, Linear pair)

Now in ΔPDB,

∠PDB = 90° (Proved)

and ⇒ ∠DPB = ∠DBC = 90°

and ∠DPB + 30° = 90°

so ⇒ hence ∠DPB = 90° – 30° = 60°

⇒ APB = 60°

Hence proved.

**EXERCISE -17 (B) **Circles Concise Solutions of ICSE Maths Class 10 Selina Publishers

**P.Q.**

**In the given diagram, chord AB = chord BC.**

**(i) What is the relation between arcs AB and BC?**

**(ii) What is the relation between ∠AOB and ∠BOC?**

**(iii) If arc AD is greater than arc ABC, then what is the relation between chords AD and AC?**

**(iv) If ∠AOB = 50°, find the measure of angle BAC.**

**Answer PQ**

Join OA, OB, OC and OD,

(i) Arc AB = Arc BC (∵ Equal chords subtends equal arcs)

(ii) ∠AOB = ∠BOC (∵ Equal arcs subtends equal angles at the centre)

(iii) If arc AD > arc ABC, then chord AD > AC.

(iv) ∠AOB = 50°

But ∠ BOC = ∠AOB (ftom (ii) above)

∴ ∠BOC = 50°

Now, arc BC subtends ∠BOC at the centre and ∠BAC at the remaining part of the circle.

∴ ∠BOC = ∠BOC = x 50°= 25°

**P.Q.**

**In ∆ ABC, the perpendiculars from vertices A and B on their opposite sides meet (when produced) the circumcircle of the triangle at points D and E respectively.**

**Prove that: arc CD = arc CE.**

Circle Ex 17 B Ans 10

**Answer PQ**

**Given:** In ∆ ABC, perpendiculars from A and B are drawn on their opposite sides BC and AC at L and M respectively and meets the circumcircle of ∆ ABC at D and E respectively on producing.

**To Prove:** Arc CD = Arc CE

**Construction:** Join CE and CD

**Proof:** In ∆ APM and ∆ BPL,

∠AMP = ∠BLP (Each = 90°)

∠1 = ∠2 (Vertically opposite angles)

∴ ∆ APM ~ ∆ BPL (AA postulate)

And ∴ Third angle = Third angle

so ∴ ∠3 = ∠4

therefore∵ Arc which subtends equal angle at the circumference of the circle, are also equal.

hence ∴ Arc CD = Arc CE Q.E.D

**Question 1.**

**In a cyclic-trape∠ium, the non-parallel sides are equal and the diagonals are also equal. Prove it.**

**Answer 1**

**Given:** A cyclic-trape∠ium ABCD in which AB || DC and AC and BD are joined

**To Prove:**

(i) AD = BC

(ii) AC = BD

**Proof:**

∵AB || DC (given)

∴ ∠ABD = ∠BDC (Alternate angles)

∵ Chord AD subtends ∠ABD and chord BC subtends ∠BDC at the circumference of the circle

But ∠ABD = ∠BDC (Proved)

∴ Chord AD = Chord BC

⇒ AD = BC

Now in ∆ADC and ∆BDC

DC = DC (common)

AD = BC (proved)

and ∠CAD = ∠CBD (Angle in the same segment)

∴ ∆ADC ≅ ∆BDC (ASS axiom)

∴ AC = BD (c.p.c.t.)

**Question 2.Concise Maths Solutions Circles **

**In the following figure, AD is the diameter of the circle with centre Q. Chords AB, BC and CD are equal. If ∠ DEF = 110°, calculate:**

**(i) ∠AEF,**

**(ii) ∠FAB.**

**Answer 2**

Join AE, OB and OC

(i) ∵ AOD is the diameter

∴ ∠AED = 90° (AngIe in a semi-circle)

But ∠DEF = 110° (given)

∴ ∠AEF = ∠DEF – ∠AED =110° – 90° = 20°

(ii) ∵ Chord AB = Chord BC = Chord CD (given)

∴ ∠AOB = ∠BOC = ∠COD (Equal chords subtends equal angles at the centre)

But ∠AOB + ∠BOC + ∠COD = 180° (AOD is a straight line)

∴ ∠AOB – ∠BOC = ∠COD = 60°

In ∆ OAB, OA = OB (Radii of the same cirlce)

∴ ∠OAB = ∠OBA

But ∠OAB + ∠OBA = 180° – ∠AOB

= 180° – 60°= 120″

∴ ∠OAB = ∠OBA = 60°

In cyclic quad. ADEF,

∴ ∠DEF + ∠DAFJ= 180°

⇒ 110° + ∠DAF = 180°

∴ ∠DAF = 180° – 110° = 70°

Now, ∠FAB = ∠DAF + ∠OAB = 70° + 6Q° = 130°

**P.Q.**

**In the given figure, if arc AB = arc CD, then prove that the quardrilateral ABCD is an isosceles-trapezium (O is the centre of the circle).**

**Answer PQ **

**Given:** In the figure, O is the centre of a circle and arc AB = arc CD

**To Prove:** ABCD is an isosceles trapezium.

**Construction:** Join BD, AD and BC.

**Proof:** Since, equal arcs subtends equal angles at the circumference of a circle.

∴ ∠ADB = ∠DBC ( ∵ arc AB = arc dD)

But, these are alternate angles.

∴ AD || BC.

and ∴ ABCD is a trapezium.

so ∵ Arc AB = Arc CD (Given)

therefore∴ Chord AB = Chord CD

hence ∴ ABCD is an isosceles trapezium. Q.E.D.

** P.Q.**

**In the given figure, ABC is an isosceles triangle and O is the centre of its circumcirclc. Prove that AP bisects angle BPC.**

**Answer PQ**

**Given:** ∆ ABC is an isosceles triangle in * which AB = AC and O is the centre of the circumcircle.

**To Prove:** AP bisects ∠BPC

**Proof:** Chord AB subtends ∠APB and hord AC subtends ∠APC at the circumference of the circle.

But chord AB = chord AC.

∴ ∠APB ∠APC x

∴ AP is the bisector of ∠BPC Q.E.D.

**Question 3.**

**If two sides of a cyclic-quadrilateral are parallel; prove that:**

**(i) its other two sides are equal.**

**(ii) its diagonals are equal.**

**Answer 3**

**Given:** ABCD is a cyclic quadrilateral in whicnAB || DC. AC and BD are its diagonals.

**To Prove:**

(i) AD = BC, (ii) AC = BD.

**Proof:** AP || CD.

∴ ∠DCA = ∠CAB (Alternate angles)

Now, chord AD subtends ∠DCA and chord BC subtends ∠CAB at the circumference of the circle.

∵∠DCA = ∠CAB (Proved)

∴ Chord AD = chord BC or AD = BC.

Now, in A ACB and A ADB,

AB = AB (Common),

BC = AD (Proved)

∠ACB = ∠ADB (Angles in the same segment)

∆ ACB ≅ ∆ ADB (SAA postulate)

∴ AC = BT (C. P. C. T) Q.E.D.

**Question 4.**

**The given figure shows a circle with centre O. Also, PQ = QR = RS and ∠ PTS = 75°.**

**Calculate:**

**(i) ∠POS,**

**(ii) ∠QOR,**

**(iii) ∠ PQR**

**Answer 4**

Join OP, OQ, OR and OS.

∵ PQ = QR = RS.

∴ ∠POQ = ∠QOR = ∠ROS

(Equal chords subtends equal angles at the centre)

Arc PQRS subtends ∠POS at the centre and ∠PTS at the remaining part of the circle

∴ ∠POS = 2 ∠PTR = 2 x 75° = 150°

⇒ ∠POQ + ∠QOR + ∠ROS = 150°

**Question 5. Concise Maths Solutions Circles **

**In the given figure, AB is a side of a regular six-sided polygon and A****C is a side of a regular eight-sided polygon inscribed in the circle with centre O. Calculate the sizes of:**

**(i) ∠ AOB,**

**(ii) ∠ ACB,**

**(iii) ∠ ABC.**

**Answer 5**

Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of the circle.

**Question 6.**

**In a regular pentagon ABODE, inscribed in a circle; find ratio between angle EDA and angle ADC. [1990]**

**Answer 6**

Arc AE subtends ∠AOE at the centre and ∠ADE at the centre and ∠ADE at the remaining part of the circumference.

**Question 7. Concise Maths Solutions Circles **

**In the given figure, AB = BC = CD and ∠ ABC = 132°. Calculate :**

**(i) ∠AEB,**

**(ii) ∠AED,**

**(iii) ∠ COD. [1993]**

**Answer 7**

In the figure, O is the centre of circle AB = BC = CD and ∠ABC = 132°

Join BE and CE

(i) In cyclic quadrilateral ABCE ∠ABC + ∠AEC = 180°

(sum of opposite angles)

⇒ 132° + ∠AEC = 180°

⇒ ∠AEC = 180° – 132° = 48°

∵ AB = BC (given)

∴ ∠AEB = ∠BEC

(equal chords subtends equal angles)

**Question 8.**

**In the figure, O is centre of the circle and the length Of arc AB is twice the length of arc BC. if angle AOB = 108°, find :**

**(i) ∠ CAB,**

**(ii)∠ADB. [1996]**

**Answer 8**

(i) Join AD and DB.

∵ Arc AB = 2 arc BC. and ∠ AOB = 108° 1 1

∴ ∠ BOC = ∠ AOB = x 108° = 54°

Now, arc BC subtends ∠ BOC at the centre and ∠CAB at the remaining part of die circle.

∴ ∠CAB = ∠BOC = x 54° = 27°

(ii) Again arc AB subtends ∠ AOB at the centre and ∠ ACB at the remaining part of the circle

∴ ∠ACB = ∠AOB = x 108° = 54°

In cyclic quad. ADBC,

∠ ADB + ∠ ACB = 180°

⇒ ∠ ADB+ 54° =180°

∴ ∠ ADB = 180° – 54° = 126°

**Question 9.**

**The figure shows a circle with centre O, AB is the side of regular pentagon and AC is the side of regular hexagon.**

**Find the angles of triangle ABC.**

**Answer 9**

Join OA, OB and OC.

**Question 10. Concise Maths Solutions Circles **

**In the given figure, BD is a side ol’ a regular hexagon, DC is a side of a regular pentagon, and AD is a diameter. Calculate :**

**(i) ∠ ADC,**

**(ii) ∠ BDA,**

**(iii) ∠ ABC,**

**(iv) ∠AEC. [1984]**

**Answer 10**

Join BC, BO, CO, and EO.

**Concise Maths Solutions EXERCISE – 17(C) Circles for ICSE Class 10**

**Question 1**

**In the given circle with diameter AB, find the value of x. (2003)**

**Answer 1**

∠ABD = ∠ACD = 30° (Angle in the same segment)

Now in ∆ ADB,

∠BAD + ∠ADB + ∠DBA = 180° (Angles of a ∆)

But ∠ADB = 90° (Angle in a semi-circle)

∴ x + 90° + 30° = 180°

⇒ x + 120° = 180°

∴ x = 180°- 120° = 60°

**PQ**

**In the given figure, O is the centre of the circle with radius 5 cm, OP and OQ are perpendiculars to AB and CD respectively. AB = 8cm and CD = 6cm. Determine the length of PQ.**

**Answer PQ**

Radius of the circle whose centre is O = 5 cm

OP ⊥ AB and OQ ⊥ CD, AB = 8 cm and CD = 6 cm.

Join OA and OC, then OA = OC=5cm.

∴ OP ⊥ AB

∴ P is the mid-point of AB.

Similarly, Q is the mid-point of CD.

In right ∆OAP,

OA² = OP² + AP² (Pythagorous theoram)

⇒ (5)² =OP² +(4)² ( ∵ AP = AB = x 8 = 4cm)

⇒ 25 = OP² + 16

⇒ OP^{3} = 25 – 16 = 9 = (3)²

∵ OP = 3 cm

Similarly, in right ∆ OCQ,

OC^{2} = OQ^{2} + CQ^{2}

⇒ (5)^{2} =OQ^{2}+(3)^{2} (∵ CQ = CD = x 6 = 3cm)

and ⇒ 25 = OQ^{2} + 9

⇒ OQ^{3} = 25 – 9 = 16 = (4)^{2}

∴ OQ = 4 cm

Hence, PQ = OP + OQ = 3-4 = 7 cm.

**PQ**

**The given figure shows two circles with centres A and B; and radii 5cm and 3cm respectively, touching each other internally. If the perpendicular bisector of AB meets the bigger circle in P and Q, find the length of PQ.**

**Answer PQ**

Join AP and produce AB to meet the bigger circle at C.

AB = AC – BC = 5 cm – 3 cm = 2 cm.

But, M is the mid-point of AB

∴ AM = = 1cm.

Now in right ∆APM,

AP2 = MP2 + AM2 (Pythagorous theorem)

⇒ (5)^{2} = MP^{2} -1- (1 )^{2}

⇒ 25 = MP: + 1

⇒ MP: = 25 – 1 = 24

**Question 2**

**In the given figure, ABC is a triangle in which ∠BAC = 30°. Show that BC is equal to the radius of the circumcircle of the triangle ABC, whose centre is O.**

**Answer 2**

**Given:** In the figure ABC is a triangle in winch ∠A = 30°

**To Prove:** BC is the radius of circumcircle of ∆ABC whose O is the centre.

**Const:** Join OB and OC.

**Proof:** ∠BOC is at the centre and ∠BAC is at the remaining part of the circle

∴ ∠BOC = 2 ∠BAC = 2 x 30° = 60°

Now in ∆OBC,

OB = OC (Radii of the same circle)

∴ ∠OBC = ∠OCB

But ∠OBC + ∠OCB + ∠BOC – 180°

∠OBC + ∠OBC + 60°- 180°

⇒ 2 ∠OBC = 180°- 60° = 120°

⇒ ∠OBC = = 60°

∴ ∆OBC is an equilateral triangle.

∴ BC = OB = OC

But OB and OC are the radii of the circumcircle

∴ BC is also the radius of the circumcircle.

**Question 3.**

**Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base.**

**Answer 3**

**Given:** In ∆ABC, AB-AC and a circle with AB as diameter is drawn which intersects the side BC and D.

**To Prove:** D is the mid point of BC.

**Const:** Join AD.

**Proof:** ∠ 1 = 90° (Angle in a semi-circle)

But ∠ 1 + ∠ 2 – 180° (Linear pair)

∴ ∠ 2 = 90°

Now, in right ∆ ABD and ∆ ACD,

Hyp. AB – Hyp. AC (Given)

Side AD – AD (Common)

∴ ∆ABD = ∆ACD (RHS criterion of congruency)

∴ BD = DC (c.p.c.t.)

Hence D is he mid point of BC. Q.E.D.

**Question 4. Concise Maths Solutions Circles **

**In the given figure, chord ED is parallel to diameter AC of the circle. Given ∠CBE = 65°, calculate ∠DEC.**

**Answer 4**

Join OE,

Arc EC subtends ∠EOC at the centre and ∠EBC at the remaining part of the circle.

∴ ∠EOC = 2 ∠EBC = 2 x 65° = 130°

Now, in ∆OEC, OE = OC (radii of the same circle)

∴ ∠OEC = ∠OCE

But ∠ OEC + ∠ OCE + ∠ EOC = 180°

⇒ ∠ OCE + ∠ OCE + ∠ EOC = 180°

**PQ**

**Chords AB and CD of a circle intersect each other at point P, such that AP = CP. S.how that AB = CD.**

**Answer PQ**

**Given:** Two chords AB and CD intersect each other at P inside the circle with centre O and AP = CP.

**To Prove:** AB = CD.

**Proof:** ∵ Two chords AB and CD intersect each other inside the circle at P.

∴ AP x PB = CP x PD ⇒ =

But AP = CP ….(i) (given)

∴ PD = PB or PB = PD …,(ii)

Adding (i) and (ii),

AP + PB = CP + PD

⇒ AB = CD Q.E.D.

**Question 5**

**The quadrilateral formed by angle bisectors of a cyclic quadrilateral is also cyclic. Prove it.**

**Answer 5**

Given: ABCD is a cyclic quadrilateral and PRQS is a quadrilateral formed by the angle bisectors of angle ∠A, ∠B, ∠C and ∠D.

To Prove: PRQS is a cyclic quadrilateral.

Proof: In ∆APD,

∠1 +∠2 + ∠P= 180° ,…(i)

Similarly, in ∆BQC.

∠3 + ∠4 + ∠Q= 180° ….(ii)

Adding (i) and (ii), we get:

∠1 + ∠2 + ∠P + ∠3 + ∠4 ∠Q = 180° + 180° = 360°

⇒ ∠1 + ∠2 + ∠3 + ∠4 + ∠P + ∠Q = 360° ..(iii)

But ∠1+∠2+∠3+∠4 = (∠A+∠B+∠C+∠D)

= x 360°= 180°

∴ ∠P + ∠Q = 360° – 180° = 180° [From (iii)]
But these are the sum of opposite angles of quadrilateral PRQS

∴ Quad. PRQS is a cyclic quadrilateral. Q.E.D.

**Question 6**

**In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:**

**(I) ∠BDC**

**(ii) ∠BEC**

**(iii) ∠BAC (2014)**

**Answer 6**

∠DBC = 58°

BD is diameter

∴ ∠DCB=90° (Angle in semi circle)

(i) In ∆BDC

∠BDC + ∠DCB + ∠CBD = 180°

∠BDC = 180°- 90° – 58° = 32°

(ii) ∠BEC =180°-32°

(opp. angle of cyclic quadrilateral)

= 148°

(iii) ∠BAC = ∠BDC = 32°

(Angles in same segment)

**Question 7 Concise Maths Solutions Circles **

**D and E are points on equal sides AB and AC of an isosceles triangle ABC such that AD = AE. Prove that the points B, C, E and D are concyclic.**

**Answer 7**

** **

**Given:** In ∆ABC, AB = AC and D and E are points on AB and AC such that AD = AE, DE is joined.

**To Prove:** B,C,E,D. are concyclic.

**Proof:** In ∆ABC, AB = AC

∴ ∠B = ∠C (Angles opposite to equal sides)

Similarly in ∆ADE, AD = AE (given)

∴ ∠ADE = ∠AED

In ∆ABC,

∵ =

∴ DE || BC.

∴ ∠ADE = ∠B (Corresponding angles)

But ∠B = ∠C (Proved)

∴ Ext. ∠ADE = its interior opposite ∠C.

∴ BCED is a cyclic quadrilateral.

Hence B,C, E and D are concyclic.

**Question 8**

**In the given figure, ABCD is a cyclic quadrilateral. AF is drawn parallel to CB and DA is produced to point E. If ∠ADC = 92°, ∠FAE’= 20°; determine ∠BCD. Give reason in support of your answer.**

**Answer 8**

** **

In cyclic quad. ABCD,

AF || CB and DA is produced to E such that

∠ADC = 92° and ∠FAE – 20°

Now, we have to find the measure of ∠BCD In cyclic quad. ABCD,

∠B – ∠D = 180° ⇒ ∠B + 92° = 180″

⇒∠B = 180°-92° = 88°

∵ AF || CB.

∴∠FAB = ∠B = 88°

But ∠FAE – 20° (Given)

Ext. ∠BAE – ∠BAF + ∠FAE

= 88° + 20° = 108°

But Ext. ∠BAE – ∠BCD

∴ ∠BCD = 108°

**Question 9**

**If I is the incentre of triangle ABC and AI when produced meets the circumcircle of triangle ABC in point D. If ∠BAC = 66° and ∠ABC – 80°, calculate**

**(i) ∠DBC,**

**(ii) ∠IBC,**

**(iii) ∠BIC.**

**Answer 9**

** **

Join DB and DC, IB and IC.

∠BAC = 66°, ∠ABC = 80°. I is the incentre of the ∆ABC.

(i) ∵ ∠DBC and ∠DAC are in the same segment

∴ ∠DBC – ∠DAC.

But ∠DAC = ∠BAC = x 66° = 33°

∴ ∠DBC = 33°.

(ii) ∵ I is the incentre of ∆ABC.

∴ IB bisect ∠ABC

∴ ∠ IBC = ∠ABC = x 80° = 40°.

(iii) ∴∠BAC = 66° ∠ABC = 80°

∴ In ∆ABC,

∠ACB = 180° – (∠ABC + ∠CAB)

= 180°-(80°+ 66°)= 180°- 156° = 34°

∵ IC bisects the ∠C

∴ ∠ ICB = ∠C = x 34° = 17°.

Now in ∆IBC,

∠ IBC + ∠ ICB + ∠ BIC = 180°

⇒ 40° + 17° + ∠BIC = 180°

and ⇒ ∠ BIC = 180° – (40° + 17°) = 180° – 57°

hence = 123°

**Question 10. Concise Maths Solutions Circles **

**In the given figure, AB = AD = DC= PB and ∠DBC = x°. Determine in terms of x :**

**(i) ∠ABD**

**(ii) ∠APB**

**Hence or otherwise prove that AP is parallel to DB.**

**Answer 10**

**Given:** In figure, AB = AD = DC = PB.

∠DBC = x. Join AC and BD.

**To Find :** the measure of ∠ABD and ∠APB.

**Proof:** ∠DAC=∠DBC= x(angles in the same segment)

But ∠DCA = ∠DAC (∵ AD = DC)

= x

But ∠ABD = ∠DAC (Angles in the same segment)

In ∆ABP, ext. ∠ABC = ∠BAP + ∠APB

But ∠ BAP = ∠APB (∵ AB = BP)

2 x x = ∠APB + ∠APB = 2∠APB

∴ 2∠APB = 2x

⇒ ∠APB = x

∵ ∠APB = ∠DBC = x

But these are corresponding angles

∴ AP || DB. Q.E.D.

**Question 11. Concise Maths Solutions Circles**

**In the given figure, ABC, AEQ and CEP are straight lines. Show that ∠APE and ∠CQE are supplementary.**

**Answer 11**

**Given:** In the figure, ABC, AEQ and CEP are straight lines.

**To prove:** ∠APE + ∠CQE = 180°.

**Const:** Join EB.

**Proof:** In cyclic quad. ABEP,

∠APE+ ∠ABE= 180° ….(i)

Similarly in cyclic quad. BCQE

∠CQE +∠CBE = 180° ….(ii)

Adding (i) and (ii),

∠APE +∠ABE + ∠CQE +∠CBE = 180° + 180° = 360°

⇒ ∠APE + ∠CQE + ∠ABE + ∠CBE = 360°

But ∠ABE + ∠CBE = 180° (Linear pair)

∴ ∠APE + ∠CQE + 180° = 360°

⇒ ∠APE + ∠CQE = 360° – 180° = 180°

Hence ∠APE and ∠CQE are supplementary. Q.E.D.

**Question 12.**

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

**If ∠ADC = 32°, find angle BOC.**

**Answer 12**

Arc AC subtends ∠AOC at the centre and ∠ ADC at the remaining part of the circle

∴ ∠AOC = 2 ∠ADC

= 2 x 32° = 64°

∵ ∠AOC + ∠ BOC = 180° (Linear pair)

⇒ 64° + ∠ BOC = 180°

⇒ ∠ BOC=180° – 64° =116° .

**Question 13. Concise Maths Solutions Circles **

**In a cyclic-quadrilateral PQRS, angle PQR = 135°. Sides SP and RQ produced meet at point A : whereas sides PQ and SR produced meet at point B.**

**If ∠ A : ∠ B = 2 : 1 ; find angles A and B.**

**Answer 13**

PQRS is a cyclic-quadrilateral in which ∠ PQR =135°

Sides SP and RQ are produced to meet at A and sides PQ and SR are produced to meet at B.

**PQ**

**In the following figure, AB is the diameter of a circle with centre O and CD is the chord with length equal to radius OA. If AC produced and BD produced meet at point P ; show that ∠ APB = 60°.**

**Answer PQ**

**Given :** In the figure, AB is the diameter of the circle with centre O.

CD is the chord with length equal to the radius OA.

AC and BD are produced to meet at P.

**To prove :** ∠ APB = 60°

**Const :** Join OC and OD

**Proof :** ∵ CD = OC = OD (Given)

∴ ∆OCD is an equilateral triangle

∴ ∠ OCD = ∠ ODC = ∠ COD = 60°

In ∆ AOC, OA = OC (Radii of the same circle)

∴ ∠ A = ∠ 1

Similarly, in ∆ BOD,

OB = OD

∴∠2= ∠B

Now in cyclic quadrilateral ACDB,

∠ A CD + ∠B = 180°

∠60°+ ∠ 1 + ∠B = 180°

= ∠ 1 + ∠B = 180° – 60°

⇒∠ 1 + ∠B = 120°

But ∠ 1 = ∠ A

∴ ∠ A + ∠B = 120° …(i)

Now, in ∆ APB,

∠ P + ∠ A + ∠ B = 180° (Sum of angles of a triangle)

⇒ ∠P+120°=180° [From (i)]
⇒ ∠P = 180°- 120°= 60°

Hence ∠ P = 60° or ∠ APB = 60° Hence proved.

**Question 14 Concise Maths Solutions Circles**

**In the following figure,**

**ABCD is a cyclic quadrilateral in which AD is parallel to BC.**

**If the bisector of angle A meets BC at point E and the given circle at point F, prove that :**

**(i) EF = FC**

**(ii) BF = DF**

**Answer 14**

** **

**Given :** ABCD is a cyclic quadrilateral in which AD || BC.

Bisector of ∠ A meets BC at E and the given circle at F. DF and BF are joined.

**To prove :**

(i) EF = FC

(ii) BF = DF

**Proof :** ∵ ABCD is a cyclic -quadrilateral and AD || BC

∵ AF is the bisector of ∠ A

∴ ∠ BAF = ∠ DAF

∴ Arc BF = Arc DF (equal arcs subtends equal angles)

⇒ BF = DF(equal arcs have equal chords)

Hence proved

**Question 15. Concise Maths Solutions Circles **

**ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point E ; whereas sides BC and AD produced meet at point F. If ∠ DCF : ∠ F : ∠ E = 3 : 5 : 4, find the angles of the cyclic quadrilateral ABCD.**

**Answer 15**

**Given :** In a circle, ABCD is a cyclic quadrilateral AB and DC are produce to meet at E and BC and AD are produced to meet at F.

**Question 16.**

**The following figure shows a circle with PR as its diameter.**

**If PQ = 7 cm, QR = 3 cm, RS = 6 cm. find the perimeter of the cyclic quadrilateral PQRS. (1992)**

**Answer 16**

In the figure, PQRS is a cyclic quadrilateral in which PR is a diameter

PQ = 7 cm,

QR = 3 RS = 6cm

∴ 3 RS = 6cm

and RS = = 2cm

Now in ∆ PQR,

∠ Q = 90° (Angle in a semi-circle)

∴ PR^{2} = PQ^{2} + QR^{2} (Pythagoras theorem)

= (7)^{2} + (6)^{2} = 49 + 36 = 85

Again, in right ∆ PSQ, PR^{2} = PS^{2} + RS^{2}

⇒ 85 = PS^{2} + (2)^{2}

and ⇒ 85 = PS^{2} + 4

hence ⇒ PS^{2} = 85 – 4 = 81 = (9)^{2}

∴ PS = 9cm

Now, perimeter of quad. PQRS = PQ + QR + RS + SP = (7 + 9 + 2 + 6) cm = 24cm

**Question 17.**

**In the following figure, AB is the diameter of a circle with centre O.**

**If chord AC = chord AD, prove that :**

**(i)arc BC = arc DB**

**(ii) AB is bisector of ∠ CAD. Further, if the length of arc AC is twice the length of arc BC, find :**

**(a) ∠ BAC**

**(b) ∠ ABC**

**Answer 17**

**Given :** In a circle with centre O, AB is the diameter and AC and AD are two chords such that AC = AD.

**To prove :** (i) arc BC = arc DB

(ii) AB is the bisector of ∠CAD

(iii) If arc AC = 2 arc BC, then find

(a) ∠BAC (b) ∠ABC

**Construction:** Join BC and BD.

**Proof :** In right angled A ABC and A ABD

Side AC = AD (Given)

Hyp. AB = AB (Common)

∴ ∆ ABC ≅ ∆ ABD (R.H.S. axiom)

(i) ∴ BC = BD (C.P.C.T)

∴ Arc BC = Arc BD (equal chords have equal arcs)

(ii) ∠ BAC = ∠ BAD (C.P.C.T)

∴ AB is the bisector of ∠CAD.

(iii) If arc AC = 2 arc B

Then ∠ABC = 2 ∠BAC

But ∠ABC – 2 ∠BAC = 90°

∴ 2 ∠BAC + ∠BAC = 90°

⇒ 3 ∠BAC = 90° ⇒ ∠BAC = 30°

and ∠ABC = 2 ∠BAC = 2 * 30° = 60°

**Question 18. Concise Maths Solutions Circles **

**In cyclic-quadrilateral ABCD ; AD = BC,**

**∠ BAC = 30° and ∠ CBD = 70°, find:**

**(i) ∠ BCD**

**(ii) ∠ BCA**

**(iii) ∠ ABC**

**(iv) ∠ ADC**

**Answer 18**

ABCD is a cyclic-quadrilateral and AD = BC

∠ BAC = 30°, ∠ CBD = 70°

∠ DAC = ∠ CBD , (Angles in the same segment)

But ∠ CBD = 70°

∴ ∠ DAC = 70°

⇒ ∠ BAD = ∠ BAC + ∠ DAC = 30° + 70° = 100°

But ∠ BAD + ∠ BCD = 180°

(Sum of opposite angles of a cyclic quad.)

⇒100°+ ∠ BCD=180° ⇒ ∠ BCD=180° – 100° = 80°

∴ ∠ BCD = 80°

∵ AD = BC (Given)

∴ ∠ ACD = ∠ BDC

(Equal chords subtends equal angles)

But ∠ ACB = ∠ ADB

(Angles in the same segment)

∴ ∠ ACD + ∠ ACB = ∠ BDC + ∠ ADB

⇒ ∠ BCD = ∠ ADC = 80° (∵ ∠ BCD = 80°)

∴ ∠ ADC = 80°

But in ∆ BCD,

∠ CBD + ∠ BCD + ∠ BDC = 180° (Angles of a triangle)

⇒ 70° + 80° + ∠ BDC = ∠ 180°

⇒ 150°+ ∠ BDC = 180°

∴ ∠ BDC = 180° – 150° = 30°

⇒ ∠ ACD = 30° (∵ ∠ ACD = ∠ BDC)

∴ ∠ BCA = ∠ BCD – ∠ ACD = 80° – 30° = 50°

∠ ADC + ∠ABC = 180°

(Sum of opp. angles of a cyclic quadrilateral)

⇒ 80°+ABC = 180°

⇒ ∠ ABC=180°-80° = 100°

**Question 19.**

**In the given figure, if ∠ ACE = 43° and ∠CAF = 62°. Find the values of a, b and c.**

**Answer 19**

Now, ∠ ACE = 43° and ∠ CAF = 62° (given)

In ∆ AEC,

∠ ACE + ∠ CAE + ∠ AEC = 180°

∴ 43° + 62° + ∠ AEC = 180°

105° + ∠ AEC = 180°

⇒ ∠ AEC = 180°- 105° = 75°

Now, ∠ ABD + ∠AED=180°

(Opposite ∠ s of a cyclic quad, and ∠ AED = ∠ AEC)

⇒ 0 + 75°= 180° a = 180° – 75° = 105°

∠ EDF = ∠ BAE (Angles in the alternate segments)

∴ c = 62°

In ∆BAF, ∠a + 62° + ∠b = 180°

⇒ 105°+ 62°+ ∠b= 180°

so ⇒ 167° + ∠6 = 180°

therefore ⇒ ∠b= 180°-167°= 13°

Hence, a= 105°, 6=13° and c = 62°.

**Question 20.**

**In the given figure, AB is parallel to DC, ∠BCE = 80° and ∠BAC = 25° .**

**Find:**

**(i) ∠CAD**

**(ii)∠CBD**

**(iii) ∠ADC**

**Answer 20**

** **

In the given figure,

ABCD is a cyclic quad, in which AB || DC

∴ ABCD is an isosceles trapezium AD = BC

(i) Join BD

and Ext. ∠BCE = ∠BAD

{ Ext. angle of a cyclic quad, is equal to interior opposite angle}

∴ ∠BAD = 80° (∵ ∠BCE = 80°)

But ∠BAC = 25°

∴ ∠CAD = ∠BAD – ∠BAC = 80° – 25° = 55°

(ii) ∠CBD = ∠CAD (Angles in the same segment)

= 55°

(iii) ∠ADC = ∠BCD (Angles of the isosceles trapezium)

= 180°- ∠BCE =180°- 80° = 100°

**Question 21 Concise Maths Solutions Circles **

**ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AJD and BC produced meet at P, show that ∠APB = 60°.**

**Answer 21**

** **

**Given :** In a circle, ABCD is a cyclic quadrilateral in which AB is the diameter and chord CD is equal to the radius of the circle

**To prove:** ∠APB = 60°

**Construction :** Join OC and OD

**Proof:** ∵ chord CD = CO = DO (radii of the circle)

∴ ∆DOC is an equilateral triangle

∠DOC = ∠ODC = ∠OCD – 60°

Let ∠A = x and ∠B = y

∵ OA = OD = OC = OB (radii of the same circle)

so ∴ ∠ODA = ∠OAD = x and ∠OCB = ∠OBC =y

∴∠AOD = 180° – 2x and ∠BOC = 180° – 2y

But AOB is a straight line

∴ ∠AOD + ∠BOC + ∠COD = 180°

180°- 2x + 180° -2y + 60° = 180°

⇒ 2x + 2y = 240°

⇒ x + y = 120°

But ∠A + ∠B + ∠P = 180° (Angles of a triangle)

⇒ 120° + ∠P = 180°

⇒ ∠P = 180° – 120° = 60°

Hence ∠APB = 60°

**Question 22**

**In the figure, given alongside, CP bisects angle ACB. Show that DP bisects angle ADB.**

**Answer 22**

**Given :** In the figure,

CP is the bisector of ∠ACB

**To prove :** DP is the bisector of ∠ADB

**Proof:** ∵ CP is the bisector of

∴ ∠ACB ∠ACP = ∠BCP

But ∠ACP = ∠ADP {Angles in the same segment of the circle}

and ∠BCP = ∠BDP

But ∠ACP = ∠BCP

∴ ∠ADP = ∠BDP

∴ DP is the bisector of ∠ADB

**Question 23.**

**In the figure, given below, AD = BC, ∠BAC = 30° and ∠CBD = 70°. Find :**

**(i) ∠BCD**

**(ii) ∠BCA**

**(iii) ∠ABC**

**(iv) ∠ADB**

**Answer 23**

In the figure,

ABCD is a cyclic quadrilateral

AC and BD are its diagonals

∠BAC = 30° and ∠CBD = 70°

Now we have to find the measures of ∠BCD, ∠BCA, ∠ABC and ∠ADB

∠CAD = ∠CBD = 70°

(Angles in the same segment)

Similarly ∠BAC = ∠BDC = 30°

∴ ∠BAD = ∠BAC + ∠CAD = 30° + 70° = 100°

(i) Now ∠BCD + ∠BAD = 180° (opposite angles of cyclic quad.)

⇒ ∠BCD + 100°= 180°

⇒ ∠BCD = 180°-100° = 80°

(ii) ∵ AD = BC (given)

∴ ∠ABCD is an isosceles trapezium

and AB || DC

∴ ∠BAC = ∠DCA (alternate angles)

⇒ ∠DCA = 30°

∠ABD = ∠D AC = 30° (Angles in the same segment)

∴ ∠BCA = ∠BCD – ∠DAC = 80° – 30° = 50°

(iii) ∠ABC = ∠ABD + ∠CBD = 30° + 70° = 100°

(iv) ∠ADB = ∠BCA = 50° (Angles in the same segment)

**P.Q.**

**In the given below figure AB and CD are parallel chords and O is the centre.**

**If the radius of the circle is 15 cm, find the distance MN between the two chords of length 24 cm and 18 cm respectively.**

**Answer PQ**

**Given :** AB = 24 cm, CD = 18 cm

⇒AM = 12 cm, CN = 9 cm

Also, OA = OC = 15 cm

Let MO = y cm, and ON = x cm

In right angled ∆AMO

(OA)^{2} = (AM)^{2} + (OM)^{2}

(15)^{2} = (12)^{2} + (y^{2}

⇒ (15)^{2}-(12)^{2}

⇒ y^{2} = 225-144

⇒ y^{2} = 81 = 9 cm

In right angled ∆CON

(OC)^{2} = (ON)^{2} + (CN)^{2}

⇒(15 )^{2}= x2 + (9)^{2}

⇒ x^{2} = 225-81

⇒x^{2}= 144

⇒ x = 12 cm

Now, MN = MO + ON =y + x = 9 cm + 12 cm = 21 cm

**Question 24 Concise Maths Solutions Circles **

**In the figure given below, AD is a diameter. O is the centre of the circle. AD is parallel to BC and ∠CBD = 32°. Find :**

**(i) ∠OBD**

**(ii) ∠AOB**

**(iii) ∠BED (2016)**

**Answer 24**

(i) AD is parallel to BC, that is, OD is parallel to BC and BD is transversal.

∴ ∠ODB = ∠CBD = 32° (Alternate angles)

In ∆OBD,

OD = OB (Radii of the same circle)

⇒ ∠ODB = ∠OBD = 32°

(ii) AD is parallel to BC, that is, AO is parallel to BC and OB is transversal.

∴∠AOB = ∠OBC (Alternate angles)

∠OBC = ∠OBD + ∠DBC

⇒ ∠OBC = 32° + 32°

⇒ ∠OBC = 64°

∴ ∠AOB = 64°

(iii) In AOAB,

OA = OB(Radii of the same circle)

∴ ∠OAB = ∠OBA = x (say)

∠OAB + ∠OBA + ∠AOB = 180°

⇒ x + x + 64° = 180°

and ⇒ 2x = 180° – 64°

so ⇒ 2x= 116°

hence ⇒ x = 58°

∴ ∠OAB = 58°

That is ∠DAB = 58°

∴ ∠DAB = ∠BED = 58°

(Angles inscribed in the same arc are equal)

**PQ.**

**In the given figure PQRS is a cyclic quadrilateral PQ and SR produced meet at T.**

**(i) Prove ∆TPS ~ ∆TRQ**

**(ii) Find SP if TP = 18 cm, RQ = 4 cm and TR = 6 cm.**

**(iii) Find area of quadrilateral PQRS if area of ∆PTS = 27 cm2.(2016)**

**Answer PQ**

(i) Since PQRS is a cyclic quadrilateral ∠RSP + ∠RQP = 180°

(Since sum of the opposite angles of a cyclic quadrilateral is 180°)

⇒ ∠RQP = 180° – ∠RSP …(i)

∠RQT + ∠RQP = 180°

(Since angles from a linear pair)

⇒ ∠RQP = 180° – ∠RQT …(ii)

From (i) and (ii),

180° – ∠RSP = 180° – ∠RQT

⇒ ∠RSP = ∠RQT …(iii)

In ∆TPS and ∆TRQ,

∠PTS = ∠RTQ (common angle)

∠RSP = ∠RQT [From (iii)]
∴ ATPS ~ ATRQ (AA similarity criterion)

(ii) Since ∆TPS ~ ∆TRQ implies that corresponding sides are proportional that

#### Question 25

In the figure given, O is the centre of the circle. ∠DAE = 70°. Find giving suitable reasons, the measure of

i.∠BCD

ii ∠BOD

iii. ∠OBD

#### Answer 25

∠DAE and ∠DAB are linear pair

So,

∠DAE + ∠DAB = 180°

∴∠DAB = 110°

Also,

∠BCD + ∠DAB = 180°……Opp. Angles of cyclic quadrilateral BADC

∴∠BCD = 70°

∠BCD = ∠BOD…angles subtended by an arc on the center and on the circle

∴∠BOD = 140°

In ΔBOD,

OB = OD……radii of same circle

So,

∠OBD =∠ODB……isosceles triangle theorem

∠OBD + ∠ODB + ∠BOD = 180°……sum of angles of triangle

2∠OBD = 40°

∠OBD = 20°

**End Of Concise Maths Solutions Circles** Chapter-17

Return to :- Concise Selina Maths Solutions for ICSE Class-10

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