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

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### 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.

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.

(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.

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 = $\frac { 1 }{ 2 }$ ∠AOB = $\frac { 1 }{ 2 }$ 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.

(i) Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part
∠AOB = 2 ∠ACB
or ∠ACB = $\frac { 1 }{ 2 }$ ∠AOB = $\frac { 1 }{ 2 }$ 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 = $\frac { 1 }{ 2 }$ (reflex ∠AOB) = $\frac { 1 }{ 2 }$ [360°- 112°] = $\frac { 1 }{ 2 }$ 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.

(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.

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°,
∠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,
(iii) ∠ABC.

∠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

O is the centre of circle ∠AOB = 140° and ∠OAC = 50°.
AB is joined
Reflex ∠AOB = 360° – 140° = 220°
But ∠ACB = $\frac { 1 }{ 2 }$ Reflex ∠AOB = $\frac { 1 }{ 2 }$ x 220° = 110°
∠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 = $\frac { 40 }{ 2 }$ = 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.

∠CDB = ∠BAC (Angles is the same segment) = 49°
∠ABC = ∠ADC (Angles in the same segment) = 43°
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.

∠A + ∠C = 180° (opposite angles of a cyclic quad.)
∠C = 180° – ∠A = 180° – 75° = 105° or ∠BCD = 105°
In ΔABD,
⇒ 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 :
(ii) ∠AEB

∠ACB and ∠ADB are in the same segment.
∠ADB = ∠ACB. = 60°. (Angle of an equilateral triangle)
⇒ ∠AEB + 60° = 180°
⇒ ∠AEB = 180° – 60° = 120°.

#### Question 12.

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

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.

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.

∠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 = $\frac { 1 }{ 2 }$ ∠COB = $\frac { 1 }{ 2 }$ 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.

Arc AB subtends ∠AOB at the centre and ∠ACB at the remaining part of circle,
∠AOB = 2 ∠ACB
or ∠ACB = $\frac { 1 }{ 2 }$ ∠AOB = $\frac { 1 }{ 2 }$ 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.

(i) ∠ CBD = ∠ DAC = 27° (Angles in the same segment)
∠ABD + ∠BAD + ∠BDA = 180°.
⇒ 50° + ∠BAD + 33° = 180°
so ⇒ ∠BAD + 83° = 180°
hence ⇒ ∠BAD = 180° – 83° = 97°
⇒ 97° + ∠DCB = 180°
⇒ ∠DCB = 180°- 97° = 83°
⇒ ∠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.

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

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.

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.

Given- In the figure, ∠BAD = 96°
To Prove-
(i) Find ∠BCD and ∠BFE
Proof- ABCD is a cyclic quadrilateral.
⇒ 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.

#### Question 20

Prove that
(i) the parallelogram, inscribed in a circle, is a rectangle.
(ii) the rhombus, inscribed in a circle, is a square.

(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.

Given- In the figure, AB = AC.
To Prove- DECB is an isosceles trape∠ium.
Proof- In ∆ABC,
AB = AC
∠B = ∠C
∠B + ∠DEC = 180°
∠C + ∠DEC = 180°
But this is the sum of interior angles on one side of a transversal.
DE || BC ….(i)
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.

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)
∠PAQ + ∠PBQ = 90° + 90° = 180°
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.

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)
∠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.

(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 = $\frac { 1 }{ 2 }$ ref. ∠AOC
b = $\frac { 1 }{ 2 }$ (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.

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.
Proof- Arc AC subtends ∠AOC at the centre and ∠ADC at the remaining pari of the circle
Similarly, ∠BOD = 2 ∠BAD ….(ii)

∠AOC + ∠BOD = 2 ∠ADC
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.

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.

AB || CD.
Join AC and BD.
∠CAD = 90° (Angle in semi circle)
∠CAB = ∠CAD + ∠DAB = 90° + 25° = 115°
∠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.

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)
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

Given- The sides AB and DC of a cyclic quad. ABCD are produced to meet at P and PA = PD.
PA = PD (given)
∠A = ∠D (angles opposite to equal sides)
Ext. ∠PCB = ∠A = ∠D
But these are corresponding angles. ,

#### 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.

∠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.

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

To Prove SQ = SR.
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.

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

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 = $\frac { 1 }{ 2 }$ ∠APB = $\frac { 1 }{ 2 }$ 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,

So

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.

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°

⇒ 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.

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
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.

∠DAB and ∠BED are in the same segment of the circle.
∠DAB = ∠BED = 65° (∠BED = 65° given)
DC || AB (Given)
∠BDC = ∠DBA (Alternate angles)
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.

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°
∠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,

In ∆AED,
⇒ ∠ADE + 42° + 98° = 180°
⇒ ∠ADE + 140° = 180°
⇒ ∠ ADE = 180° – 140° = 40° or ∠ADC = 40°
⇒ 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.

(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°
∠ 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°
⇒ 60° + ∠ADC = 180°
∠ADC = 180° – 60°= 120°
∠ 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 = $\frac { 65 }{ 2 }$ = 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° – $\frac { 1 }{ 2 }$ ∠BAC.

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° – $\frac { 1 }{ 2 }$ ∠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)
$\frac { 1 }{ 2 }$ ∠ ABC = ∠ APQ
Or ∠ ABC = 2 ∠APQ …,(i)
Similarly, ∠ APR and ∠ ACR are in the same segment of the circle.
∠ APR = ∠ ACR
But ∠ ACR = $\frac { 1 }{ 2 }$ ∠ ACB (CR is the angles bisector of ∠ ACB)
$\frac { 1 }{ 2 }$ ∠ ACB = ∠ APR
Or ∠ ACB = 2 ∠ APR ….(ii)
∠ ABC + ∠ ACB = 2 ∠ APQ + 2∠ APR = 2 (∠ APQ + ∠ APR) = 2 ∠ PQR
Or 2 ∠ PQR = ∠ ABC + ∠ ACB
∠ PQR = $\frac { 1 }{ 2 }$ (∠ ABC + ∠ ACB) ….(iii)
But ∠ ABC + ∠ ACB + ∠ BAC = 180° (Angles of a triangle)
∠ ABC + ∠ ACB = 180° – ∠ BAC ….(iv)
from (iii) and (iv) we get,
∠ PQR = $\frac { 1 }{ 2 }$ (180° – ∠ BAC) = 90° – $\frac { 1 }{ 2 }$ ∠ BAC

#### Question 41.

Calculate the angles x, y and z if :

Solution:

Ext. ∠ ADC = x + z ….(i)
and in ΔBPC,
Ext. ∠ ABC = y + x ….(ii)
(∠ BCP = ∠ DCQ = x vertically opposite angles)

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]

In the figure, AB = AC = CD, ∠ ADC = 38°
BE is joined.
In ΔACD, AC = CD
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]

Arc subtends ∠ AOB at the centre and ∠ACB at the remaining part of the circle.
∠ AOB = 2 ∠ ACB
⇒ x = 2q ⇒ q = $\frac { x }{ 2 }$
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° – $\frac { x }{ 2 }$
Arc BC subtends ∠ BOC at the centre and ∠ ADC at the remaining part of the circle
∠BOC = 2 ∠BDC = 2r.
r = $\frac { 1 }{ 2 }$ ∠ BOC = $\frac { 1 }{ 2 }$ (180° – x)
(∠ AOB + ∠ BOC = 180°)
r = 90° – $\frac { 1 }{ 2 }$ x. = 90° – $\frac { x }{ 2 }$

#### 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]

Arc AB subtends ∠ AOB at the centre and ∠ACB at the remaining part of the circle.
∠ ACB = $\frac { 1 }{ 2 }$ ∠AOB = $\frac { 1 }{ 2 }$ 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 = $\frac { 1 }{ 2 }$ ∠ BOC = $\frac { 1 }{ 2 }$ x 100° = 50°

(ii) EB || DC
∠ DCE = ∠ BEC (Alternate angles) = 50°
∠ BCD = ∠ BCA + ∠ ACE + ∠ ECD = 40° + 10° + 50° = 100°

∠ 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

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°
∠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]

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 :
(ii) ∠ DQB. [1987]

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

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
∠ ABC + ∠ 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)
∠ 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

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

∠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]

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]

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

Join BD,
Arc AD, subtends ∠ AOD at the centre and ∠ ABD at the remaining part of the circle
∠ ABD = $\frac { 1 }{ 2 }$ ∠ AOD = $\frac { 1 }{ 2 }$ 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°
∠ 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,

(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.

In circle ABCD is a cyclic quadrilateral AC andBD are joined.
∠BAD = 65°, ∠ABD = 70° and ∠BDC = 45°
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
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.

∠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

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 = $\frac { 1 }{ 2 }$ ∠POB = $\frac { 1 }{ 2 }$ 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

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 = $\frac { 1 }{ 2 }$ ∠ABD = $\frac { 1 }{ 2 }$ [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:

Given : In circle with centre O,
ABCD is cyclic quadrilateral in which CD is equal radius of the circle and AB is diameter.
CD = $\frac { 1 }{ 2 }$ 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 = $\frac { 1 }{ 2 }$ ∠DOC = $\frac { 1 }{ 2 }$ 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.

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 = $\frac { 1 }{ 2 }$∠BOC = $\frac { 1 }{ 2 }$ 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

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.

Given: A cyclic-trape∠ium ABCD in which AB || DC and AC and BD are joined
To Prove:
(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
DC = DC (common)
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.

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°
∴ ∠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).

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.
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.

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.

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)
Now, in A ACB and A ADB,
AB = AB (Common),
∠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

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 AC 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.

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]

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]

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,

∵ Arc AB = 2 arc BC. and ∠ AOB = 108° 1 1
∴ ∠ BOC = $\frac { 1 }{ 2 }$ ∠ AOB = $\frac { 1 }{ 2 }$ x 108° = 54°

Now, arc BC subtends ∠ BOC at the centre and ∠CAB at the remaining part of die circle.
∴ ∠CAB = $\frac { 1 }{ 2 }$ ∠BOC = $\frac { 1 }{ 2 }$ x 54° = 27°
(ii) Again arc AB subtends ∠ AOB at the centre and ∠ ACB at the remaining part of the circle
∴ ∠ACB = $\frac { 1 }{ 2 }$ ∠AOB = $\frac { 1 }{ 2 }$ x 108° = 54°
∠ ADB + ∠ ACB = 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.

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 :
(ii) ∠ BDA,
(iii) ∠ ABC,
(iv) ∠AEC. [1984]

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)

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

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 = $\frac { 1 }{ 2 }$ x 8 = 4cm)
⇒ 25 = OP² + 16
⇒ OP3 = 25 – 16 = 9 = (3)²
∵ OP = 3 cm
Similarly, in right ∆ OCQ,
OC2 = OQ2 + CQ2
⇒ (5)2 =OQ2+(3)2 (∵ CQ = CD = $\frac { 1 }{ 2 }$ x 6 = 3cm)
and ⇒ 25 = OQ2 + 9
⇒ OQ3 = 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.

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 = $\frac { 2 }{ 2 }$ = 1cm.
Now in right ∆APM,
AP2 = MP2 + AM2 (Pythagorous theorem)
⇒ (5)2 = MP2 -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.

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 = $\frac { { 120 }^{ circ } }{ 2 }$ = 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.

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.
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)
∴ ∆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.

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.

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 ⇒ $\frac { AP}{ CP }$ = $\frac { AD }{ PB }$
But AP = CP ….(i) (given)
∴ PD = PB or PB = PD …,(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.

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 = $\frac { 1 }{ 2 }$ (∠A+∠B+∠C+∠D)
$\frac { 1 }{ 2 }$ x 360°= 180°
∴ ∠P + ∠Q = 360° – 180° = 180° [From (iii)] But these are the sum of opposite angles of quadrilateral PRQS

#### Question 6

In the figure, ∠DBC = 58°. BD is a diameter of the circle. Calculate:
(I) ∠BDC
(ii) ∠BEC
(iii) ∠BAC (2014)

∠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°
= 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.

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)
In ∆ABC,
∵ $\frac { AP }{ AB }$ = $\frac { AE }{ AC }$
∴ 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.

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.

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 = $\frac { 1 }{ 2 }$∠BAC = $\frac { 1 }{ 2 }$ x 66° = 33°
∴ ∠DBC = 33°.
(ii) ∵ I is the incentre of ∆ABC.
∴ IB bisect ∠ABC
∴ ∠ IBC = $\frac { 1 }{ 2 }$ ∠ABC = $\frac { 1 }{ 2 }$ 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 = $\frac { 1 }{ 2 }$ ∠C = $\frac { 1 }{ 2 }$ 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.

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.

Given: In the figure, ABC, AEQ and CEP are straight lines.
To prove: ∠APE + ∠CQE = 180°.
Const: Join EB.
∠APE+ ∠ABE= 180° ….(i)

∠CQE +∠CBE = 180° ….(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.

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

= 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.

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°.

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
∠ 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

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.

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)

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 = $\frac { 6 }{ 3 }$ = 2cm
Now in ∆ PQR,
∠ Q = 90° (Angle in a semi-circle)
∴ PR2 = PQ2 + QR2 (Pythagoras theorem)
= (7)2 + (6)2 = 49 + 36 = 85
Again, in right ∆ PSQ, PR2 = PS2 + RS2
⇒ 85 = PS2 + (2)2
and ⇒ 85 = PS2 + 4
hence ⇒ PS2 = 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

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
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

∠ BAC = 30° and ∠ CBD = 70°, find:
(i) ∠ BCD
(ii) ∠ BCA
(iii) ∠ ABC

∠ 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°
∴ ∠ ACD = ∠ BDC
(Equal chords subtends equal angles)
But ∠ ACB = ∠ ADB
(Angles in the same segment)
∴ ∠ ACD + ∠ ACB = ∠ BDC + ∠ ADB
⇒ ∠ BCD = ∠ ADC = 80° (∵ ∠ BCD = 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.

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:
(ii)∠CBD

In the given figure,
ABCD is a cyclic quad, in which AB || DC
∴ ABCD is an isosceles trapezium AD = BC
(i) Join BD
{ 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°.

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.

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
∴ 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

In the figure,
AC and BD are its diagonals
∠BAC = 30° and ∠CBD = 70°
Now we have to find the measures of ∠BCD, ∠BCA, ∠ABC and ∠ADB
(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.

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 + (y2
⇒ (15)2-(12)2
⇒ y2 = 225-144
⇒ y2 = 81 = 9 cm
In right angled ∆CON
(OC)2 = (ON)2 + (CN)2
⇒(15 )2= x2 + (9)2
⇒ x2 = 225-81
⇒x2= 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)

(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)

(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

∠DAE and ∠DAB are linear pair

So,

∠DAE + ∠DAB = 180°

∴∠DAB = 110°

Also,

∴∠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°

Thanks