RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise

RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise is part of RBSE Solutions for Class 10 Maths. Here we have given Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Miscellaneous Exercise.

Rajasthan Board RBSE Class 10 Maths Chapter 12 Circle Miscellaneous Exercise

Multiple choice Questions (1 to 20)
RBSE Solutions For Class 10 Maths Chapter 12 Miscellaneous Question 1.
The length of chord distant 6 cm (RBSESolutions.com) from center of a circle of radius 10 cm is :
(a) 16 cm
(b) 8 cm
(c) 4 cm
(d) 5 cm
Solution :
In right angled ∆OMA
RBSE Solutions For Class 10 Maths Chapter 12 Miscellaneous
OA2 = OM2 + AM2
(10)2 = (6)2 + AM2
AM2 = 102 – 62
= 100 – 36
AM2 = M
AM = \(\sqrt { 64 }\) = 8 cm
chord AB = 2 AM = 2 × 8 = 16 cm
Thus, (a) ¡s correct.

RBSE Solutions For Class 10 Maths Chapter 12 Question 2.
A chord of 24 cm length is drawn in a (RBSESolutions.com) circle of radius 13 cm. The distance of chord from center of circle is –
(a) 12 cm
(b) 5 cm
(c) 6.5 cm
(d) 12 cm
RBSE Solutions For Class 10 Maths Chapter 12
Solution :
AM = BM = \(\frac { 1 }{ 2 }\) × AM = \(\frac { 1 }{ 2 }\) × 24 = 12 cm
In right ∆OAM
OA2 = AM2 + OM2
(13)2 = (12)2 + (OM)2
OM2 = (13)2 – (12)2 = 169 – 144
OM2 = 25
OM = \(\sqrt { 25 }\) = 5 cm
Thus, distance of chord from center of circle = 5 cm.
Thus, (b) is correct.

RBSE Solutions For Class 12 Maths Chapter 10 Miscellaneous Question 3.
The degree measure of major arc is
(a) Less than 180°
(b) More than 180°
(c) 360°
(d) 270°
Solution :
(a) is correct.

Chapter 12 Class 10 Maths RBSE Question 4.
The degree measure of major arc is
(a) Less than 180°
(b) More than 180°
(c) 360°
(d) 90°
Solution :
(b) is correct.

RBSE Solutions For Class 10 Maths Chapter 12 In Hindi Question 5.
Chords equidistant (RBSESolutions.com) to each other from center of circle are –
(a) Double
(b) Triple
(c) Half
(d) Equal
Solution :
(d) is correct.

RBSE Class 10 Maths Chapter 12 Question 6.
The degree measure of arc of a circle is 180°, than arc is –
(a) Major
(b) Minor arc
(e) Circle
(d) Semicircle
Solution :
(d) is correct.

RBSE Solutions For Class 10 Maths Question 7.
Number of circles passes through three collinear points
(a) One
(b) Two
(c) Zero
(d) Infinite
Solution :
(a) is correct.

RBSE Solutions For Class 10th Maths Question 8.
If in a circle, \(\widehat { AB }\) = \(\widehat { BA }\), then arc is –
(a) Major arc
(b) Minor arc
(c) Semi circle
(d) Circle
Solution :
(c) is correct.

RBSE Solution Class 10th Maths Question 9.
If diameter of a circle (RBSESolutions.com) bisects two chords, then chords will be –
(a) Parallel
(b) Perpendicular
(e) Intersecting
(d) None of there
Solution :
(a) is correct.

RBSE Solution Of Class 10th Maths Question 10.
If two arcs in congruent circles arc equal, then their corresponding chords will be –
(a) Parallel
(b) Equal
(e) Perpendicular
(d) Intersecting
Solution :
(b) is correct.

Class 10th Maths RBSE Solution Question 11.
AD is diameter of a circle and AB is its chord. If AD = 34 cm, AB = 30 cm, then distance of AB from center of the circle is –
(a) 17 cm
(b) 15 cm
(c) 4 cm
(d) 8 cm
Solution :
Diameter (AD) = 34 cm
Radius (OA) = 17 cm
From center O, OM is (RBSESolutions.com) perpendicular to chord AB
RBSE Solutions For Class 12 Maths Chapter 10 Miscellaneous
OM ⊥ AM
AM = \(\frac { 1 }{ 2 }\) AB
AM = \(\frac { 1 }{ 2 }\) × 30
AM = 15 cm
In right angled triangle AOM
By Pythagoras theorem.
AO2 = OM2 + AM2
OM2 = AO2 – AM2
= 172 – 152
= (17 + 15) (17 – 15)
= 32 × 2
= 64
OM = \(\sqrt { 64 }\)
OM = 8 cm
Thus, (d) is correct.

RBSE Solutions Class 10th Maths Question 12.
In figure, If OA = 5 cm, AB = 8 cm and OD is (RBSESolutions.com) perpendicular to chord AB then CD equals –
Chapter 12 Class 10 Maths RBSE
(a) 2 cm
(b) 3 cm
(c) 4 cm
(d) 5 cm
Solution :
radius (OA) = 5 cm
OD ⊥ AB
∴ OC ⊥ AB
Perpendicular drawn from (RBSESolutions.com) center of circle bisects the chord.
AC = BC = AB
AC = \(\frac { 1 }{ 2 }\) × 8
AC = 4 cm
In right angled triangle OCA
OA2 = OC2 + AC2
OC2 = OA2 – AC2
= (5)2 – (4)2
= 25 – 16 = 9
OC = \(\sqrt { 9 }\)
OC = 3 cm.
∵ OD = OA = 5 = radius of circle
CD = OD – OC
= 5 – 3
CD = 2 cm
Thus, (a) is correct

RBSE Solutions Class 10 Maths Question 13.
If AB = 12 cm, BC = 16 cm and AB is (RBSESolutions.com) perpendicular to line segment BC then radius of a circle passing through A, B, and C is – .
(a) 6 cm
(b) 8 cm
(c) 10 cm
(d) 12 cm
Solution:
We know that angle in a semicircle is of 90°. Now, according to figure.
RBSE Solutions For Class 10 Maths Chapter 12 In Hindi
∴ ∠B = 90°
∴ In ΔABC
AC2 = AB2 + BC2
= (12)2 + (16)2
= 144 + 256 = 400
⇒ AC = \(\sqrt { 400 }\) = 20
∴ Diameter = 20 cm
or Radius = 10 cm
Hence, (RBSESolutions.com) option (c) is correct.

RBSE Solution Class 10 Maths Question 14.
In figure, if ∠ABC = 20°, then ∠AOC equals
RBSE Class 10 Maths Chapter 12
(a) 20°
(b) 40°
(c) 60°
(c) 10°
Solution : ∠ABC = 20°
We know that angle subtended by an arc (RBSESolutions.com) at the center of circle in double the angle subtended at remaining part.
∠AOC = 2∠ABC
⇒ ∠AOC = 2 × 20°
⇒ ∠AOC = 40°
Thus, (b) is correct.

RBSE Solutions For Class 10 Maths Chapter Question 15.
In figure, If AOB is diameter of circle and AC = BC Then ∠CAB, then
RBSE Solutions For Class 10 Maths
(a) 30°
(b) 60°
(c) 90°
(d) 45°
Solution :
∠ACB is angle is semicircle
∴ ∠ACB = 90°
AC = BC (Given)
∴ ∠CAB = ∠CBA (angles opposite to equal sides)
In ∆ACB
∠CAB + ∠ABC + ∠BCA = 1800
= ∠CAB + ∠CAB + 90° = 180°
2∠CAB = 180 – 90°
∠CAB = 45°
Thus, (d) is correct.

RBSE Solutions Of Class 10th Maths Question 16.
In figure, If ∠OAB = 40°, then ∠ACB equals
RBSE Solutions For Class 10th Maths
(a) 50°
(b) 40°
(c) 60°
(d) 70°
Solution :
∵ OA = OB (radius of circle)
∴ ∠OAB = ∠OBA (angles (RBSESolutions.com) opposite to equal sites)
⇒ ∠OBA = 40°
In right angled ∆OAB
∠OAB + ∠OBA + ∠AOB = 180°
⇒ 40° + 40° + ∠AOB = 180°
⇒ ∠AOB = 180° – 80°
⇒ ∠AOB = 100°
We know that
∠ACB = \(\frac { 1 }{ 2 }\)∠AOB
= \(\frac { 1 }{ 2 }\) x 100° = 50°
Thus, (a) is correct.

RBSE Solutions For Class 10 Hindi Chapter 12 Question 17.
In figure, ∠DAB = 60°, ∠ABD = 50°, then ∠ACB equal
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 9
(a) 60°
(b) 50°
(c) 70°
(d) 80°
Solution :
∠DAB = 60° and ∠ADB = 50°
In ΔADB
∠DAB + ∠ABD + ∠ADB = 180°
⇒ 60° + 50° + ∠ADB = 180°
⇒ ∠ADB = 180° – (60° – 50°)
⇒ ∠ADB = 70°
∠ACB and ∠ADB are angles in same segment.
So will be equal
⇒ ∠ACB = ∠ADB
= 70°
Thus, (c) is correct.

RBSE Solutions Maths Class 10 Question 18.
Side AB of a quadrilateral is a (RBSESolutions.com) diameter of its circumcircle and ∠ADC = 140°, then ∠BAC equals
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 10
(a) 80°
(b) 50°
(c) 40°
(d) 30°
Solution :
ABCD is a cyclic quadrilateral
∴ ∠D + ∠B = 1800
∠B = 180° – ∠D
∠B = 180° – 140°
∠B = 40°
∠ACB is angle in semicircle
∴ ∠ACB = 90°
Now in ∆ABC
∠ABC + ∠ACB + ∠BAC = 180°
⇒ 40° + 90° + ∠BAC = 180°
⇒ ∠BAC= 180° – (40° + 90°)
⇒ ∠BAC= 180° – 130°
⇒ ∠BAC = 50°
Thus, (b) is correct.

Class 10 Maths RBSE Solution Question 19.
In figure BC is diameter of (RBSESolutions.com) circle and ∠BAO = 60° then ∠ADC equal
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 11
(a) 30°
(b) 45°
(e) 60°
(d) 120°
Solution :
Since, ∠BAO = 60°
∴ ∠OBA = 60° ( OA = OB)
Then ∠ADC = 60° (angle of the same segment)
Thus, ∠ADC = 60°
Thus, option (c) is correct.

RBSE Solutions Of Class 10 Maths Question 20.
In fig. ∠AOB = 90° and ∠ABC = 30° then, ∠CAO equal to :
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 12
(a) 30°
(b) 45°
(c) 90°
(d) 60°
Solution :
∠AOB = 90° (given)
∵ OA = OB (Radius of circle)
∴ ∠OAB = ∠OBA = x (Let)
In ΔOAB
∠OAB + ∠OBA + ∠AOB = 180°
⇒ x + x + 90° = 180°
⇒ 2x = 180° – 90°
⇒ x = \(\frac { { 90 }^{ \circ } }{ 2 } \) = 45°
∴ ∠OAB = 45°
∠OBA = 45°
We know that angles (RBSESolutions.com) subtended by arc at centre of circle double the angle subtended at remaining part of circle.
∠AOB = 2∠ACB
∠ACB = \(\frac { 1 }{ 2 }\) ∠AOB = \(\frac { 1 }{ 2 }\) × 90° = 45°
Now, on ΔABC
∠ACB + ∠BAC + ∠CBA = 180°
∠ACB + [∠BAO + ∠CAO] + ∠CBA = 180°
⇒ 45° + (45° + ∠CAO) + 30° = 180°
⇒ ∠CAO = 180° – (30° + 45° + 45°)
⇒ ∠CAO = 180° – 120°
⇒ ∠CAO = 60°
Thus, (d) is correct.

Short/Long Answer Type Questions
Maths RBSE Solutions Class 10 Question 1.
If two equal chords of circle (RBSESolutions.com) intersects then Prove that segments of two chords are equal separately.
Solution :
Given :
chord AB = chord CD is circle with center O. AB and CD intersects each other at point E.
 RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 13
To Prove:
AE = CE
and BE = DE
Construction : Draw OL ⊥ AB and OM ⊥ CD and OE and join
Proof : OL ⊥ AB
Thus, AL = LB = \(\frac { 1 }{ 2 }\)AB
Similarly OM ⊥ CD
Thus, CM = MD = \(\frac { 1 }{ 2 }\) CD
But AB = CD
Thus AL = CM
and LB = MD ……(i)
and OL = OM (Equal chords arc equidistant from centre.)
Thus, in right angled ∆OLE and ∆OME
Hypotenuse OE = hypotenuse OE (common)
and OL = OM
∆OLE = ∆OME
Thus LE = ME [∵ AL = CM]
Adding,
LE + AL = ME + CM
AE = CE
Similarly BE = DE

Class 10 Maths RBSE Solutions Question 2.
If P, Q, and R arc mid (RBSESolutions.com) points of sides BC, CA and AB of a triangle ABC, respectively and AD ⊥ BC, then Prove that P, Q, R, D are cyclic.
Solution :
Given : In ∆ABC, P, Q, Rare mid points of sides BC, CA and AB respectively and AD ⊥ BC
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 14
To Prove : Points P Q, R and arc cyclic.
Construction : Join RD, RP PQ and RQ.
Proof : In ∆ABC
points R and Q arc mid points of sides AB and CA
∴ RQ || BC [By mid point theorem]
Similarly PQ || AB and PR || CA
In right BPQR
BP || RQ and PQ || BR (RQ || BC and PQ || AB)
∴ Quadrilateral BPQR is a (RBSESolutions.com) cyclic quadrilateral.
Similarly
Quadrilateral ARPQ will be a cyclic quadrilateral.
∴ ∠A = ∠RPQ [opposite angles of parallelogram]
PR || AC and PC is transversal
∠BPR = ∠C (corresponding angles)
∠DPQ = ∠DPR + ∠RPQ
= ∠A + ∠C ……(i)
RQ || BC and BR is transveral
∴ ∠ARO = ∠B (corresponding angles) …(ii)
In ΔABD
R is mid point of AB
OR || BD
∴ O, AD is mid (RBSESolutions.com) point of AD
OA = OD
In ΔAOR and ΔDOR
OA = OD
∠AOR = ∠DOR = 90°
OR = OR (common)
∴ ΔAOR = ΔDOR By (SAS Congruence)
⇒ ∠ARO = ∠DRO
∠DRO = ∠B(From equation (ii))
From quadrilateral PRQD
∠DRO + ∠DPQ = ∠B + (∠A + ∠C)
= ∠B + ∠A + ∠C
= 180°
⇒ ∠DRO + ∠DPQ = 180°
Thus, quadrilateral PRQD is a cyclic quadrilateral.
Points P, Q R and D are a cyclic.

RBSE Solution Of Maths Class 10 Question 3.
ABCD is a parallelogram. (RBSESolutions.com) A circles passes through A and B and cuts AD at P and BC at Q. Prove that P, Q, C and D arc cyclic.
Solution :
Given :
ABCD is a parallelogram and circle passes through A and B, cuts AD at P a BC at Q.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 15
To prove : P, Q C and D are cyclic
Construction : join PQ
Proof : ABCD is a parallelogram
∴ ∠ABC = ∠ADC …..(i) (opposite angles of parallelogram)
All the four vertices (RBSESolutions.com) of quadrilateral ABQP is circular
∴ ABQR is cyclic quadrilateral
∴ ∠ABQ + ∠APQ = 180° …..(ii)
BC || AD, PQ is transversal angles
Then, ∠APQ = ∠PQC (alternate angles) ……(iii)
from equation (i), (ii) and (iii)
∠ADC + ∠PQC = 180°
⇒ ∠PDC + ∠PQC = 180°
Similarly
∠QCD + ∠QPD = 180°
Thus, points P, Q, C and D will be cyclic.

Question 4.
Prove that if angle bisector of a triangle and (RBSESolutions.com) perpendicular bisector of its opposite sides intersects, then intersects at circumcircle of that triangle.
Solution :
Given
XY is perpendicular bisector of base BC of ∆ABC. ABDC is circumcircle of ∆ABC perpendicular bisector XY cuts circumcircle at D. XY cuts BC at M.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 16
To Prove : Bisector of ∠A will pass through D.
Construction : Join DB and DC
Proof : XY is perpendicular (RBSESolutions.com) bisector of BC and cuts circumcircle
∴ Point D, lies on circumcircle and XY
In right angled ABDM and ∆CDM
BM = CM (XY, BC is perpendicular bisector of)
MD = MD (common side)
∠BMD = ∠CMD (∵ XY ⊥ BC)
By SAS congruence criterion
∆BDM = ∆CDM
∴ BD = CD
∵ Point D lies on circumcircle.
∴ In circumcircle chord BD chord CD
∴ arc BD arc CD (Corresponding (RBSESolutions.com) sides of congruent that angles)
In a circle, equal chords cuts equal arcs.
∴ Angle subtended by arc BD at point A = Angles subtended by arc CD at point A
∴ ∠BAD = ∠CAD
∴ AD, ∠A is bisector of ∠A
Thus, bisector of ∠A, AD will pass through D.

Question 5.
If AB and CD two chords of circle AYDZBWCX Intersects at (RBSESolutions.com) right angles (see fig.) then prove that arc CXA + arc DZB = arc AYD + arc BWC = a semicircle.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 17
Solution :
Given : AB and CD two chords intersect each other at point M at right angle.
∴ ∠AMB = ∠BMD = ∠BMC
= ∠AMD = 90°
To Prove:
arc CXA + arc DZB = a semicircle
arc AID + arc BWC = a semicircle
Construction : Join AC, AD, BD and BC
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 18
Proof : We know that (RBSESolutions.com) angles in same segment by same arc are same.
∴ ∠ABC and ∠ADC are angles in same segment by same arc CXA.
∴ ∠ABC = ∠ADC
∠1 = ∠6
Similarly by arc DZB angle in same segment ∠BCD = ∠BAD
∠2 = ∠5 …(ii)
Adding equations (i) and (ii)
∠1 + ∠2 = ∠5 + ∠6
⇒ ∠1 +∠2 = ∠5 + ∠6 = 90°
[Since ∆AMD and ∆BMC arc right angles ∴ ∠5 + ∠6 = 90° = ∠1 + ∠2]
⇒ arc CXA + arc DZB = a semicircle
Similarly
By arc AYD angles is (RBSESolutions.com) same segment arc
∠ACD = ∠ABD
∠3 = ∠8 …..(iii)
Now, by arc BWC angles in same segment
∠BAC = ∠BDC
∠4 = ∠7 …..(iv)
Adding equations (iii) and (iv)
∠3 + ∠4 = ∠7 + ∠8
⇒ ∠3 + ∠4 = ∠7 + ∠8 = 90°
[Since ∆AMC and ∆BDM are right angles ∴ ∠3 + ∠4 = 90° = ∠7 + ∠8]
⇒ arc AYD + arc BWC = a semicircle

Question 6.
An equilateral ΔABC is inscribed in a circle. P is a point at minor arc which does not concides B and C then Prove that PA ¡s bisector of ∠BPC.
Solution :
Given :
∆ABC is an equilateral triangle
AB = BC = CA and P is (RBSESolutions.com) any point on minor arc BC.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 19
To Prove : PA, bisector of ∠BPC
i.e,. ∠BPA = ∠CPA
Construction : Join PB, PC and PA
Proof : AB = AC (Given)
arc AB = arc AC
∠CPA = ∠BPA [∵ angles formed by arc AB and arc AC]
Thus, PA is bisector of ∠BPC

Question 7.
In figure AB and CD arc two (RBSESolutions.com) chords of circle, intersecting at point E. Prove that ∠AEC = (Angle formed by arc CXA at center of angle formed by arc DYB at center)
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 20
Solution :
O is centre of circle and arc is AC
∠AOC is angle subtended by arc AC at center, is (RBSESolutions.com) double the angle ∠ABC, angle at remaining part by same arc
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 21
∴ ∠AOC = 2∠ABC
Similarly
∠BOD, angle subtended by arc BD at center of circle is double the angle at remaining part.
∠BOD = 2∠BCD …..(ii)
Adding equation (i) and (ii)
∠AOC + ∠BOD = 2(∠ABC + ∠BCD)
⇒ ∠AOC + ∠BOD = 2∠AEC
⇒ ∠AEC = \(\frac { 1 }{ 2 }\) [∠AOC + ∠BOD]
⇒ ∠AEC = \(\frac { 1 }{ 2 }\) [Angle (RBSESolutions.com) formed at center by CXA angle formed of center by chord DYB]

Question 8.
If in a cyclic quadrilateral ABCD, opposite angle bisectors interects at points P and Q of circumcircle of this quadrilateral, then Prove that PQ is diameter of the circle
Solution :
Given : ABCD is a cyclic quadrilateral, bisectors of ∠A and ∠C cuts the circle at Q and P
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 22
To Prove : PQ is diameter of this circle
Construction : Join PA and PD
Proof : To prove PQ is diameter of circle ∠QAP should be equal to 90°
Now, ABCD is a cyclic (RBSESolutions.com) quadrilateral, then ∠A + ∠C = 180°
= \(\frac { 1 }{ 2 }\)∠A + \(\frac { 1 }{ 2 }\)∠C = 90°
= ∠QAD + ∠PCD = 90°
∠PCD and ∠PAD are the angles subtended by arc PD is the same segment.
∴ ∠PCD = ∠PAD …(ii)
From (i) and (ii)
∠QAD + ∠PAD = 90° [∵ ∠QAD + ∠PAD = ∠QAP]
⇒ ∠QAP = 90°
⇒ ∠QAP is angle in semicircle.

Question 9.
Radius of a circle is \(\sqrt { 2 }\) cm. This circle is divides into two segment by a chord of length 2 cm. Prove that their chord subtends angle of 45° at a point in major segment.
Solution :
Given :
O is center of circle, on which (RBSESolutions.com) minor arc BC subtends ∠BAC is major segment and OB = \(\sqrt { 2 }\) cm BC = 2 cm
To Prove : ∠BAC = 45°
Construction : From center O of circle draw perpendicular to chord BC.
OM ⊥ BC
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 23
∴ ∠OMB = ∠OMC = 90°
Proof : We know that ⊥ drawn from center of the circle bisects the chord.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 24
= \(\frac { 1 }{ 2 }\) × 90°
= 45°
Thus, ∠BAC = 45°

Question 10.
AB and AC arc two chords of a circle (RBSESolutions.com) of radius r such that AB = 2 AC. If P and Q are respectively distance from center to AB and AC then Prove that 4q2 = p2 + 3r2
Solution:
Given : From the center O of circle with center r, p and q arc respectively distance to AB and AC such that AB = 2 AC
To Prove : 4q2 = p2 + 3r2
Construction : Drew perpendicular from center O to chords AB and AC which cuts chord at M and N respectively.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 25
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 26
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 27

Question 11.
In fig, O is center of (RBSESolutions.com) circle and ∠BCO = 30°, then find x and y.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 28
Solution :
In figure OM ⊥ BC
∴ ∠OMC = 90°
and ∠BCO = 30°
∠AOD = 90° (from figure)
∴ ∠OMC = ∠AOD = 90° (corresponding angles)
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 29
∴ BC || OD
∴ ∠COD = ∠BCO (alternate angles)
∴ ∠COD = 30°
By arc CD, angle subtend at center ∠COD is (RBSESolutions.com) double the angle at remaining part of circle ∠CBD.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 30

Question 12.
In figure, O is center of circle, BD = OD and CD ⊥ AB, then find ∠CAB
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 31
Solution :
We know that BD = GD
∵ GD = OB (radius of circle)
∴ BD = OD = OB
Thus, OBD will be equilateral triangle
Each angle of equilateral triangle is 60°
∴ ∠BOD = ∠ODB = ∠OBD = 60°
By arc BD, angle subtended at center ∠BOD will be (RBSESolutions.com) double the angle subtended at remaining part of circle i.e., ∠DAB
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 32
∠DAB = \(\frac { 1 }{ 2 }\) × 60°
= 30°
We know that AO, ∠CAD is of
∴ ∠CAB = ∠DAB = 30°
Thus, ∠CAB = 30°

Question 13.
Prove that the chord which passes through a point inside the circle is the smallest amongst all the chords and is perpendicular to the diameter passing through that point.
Solution :
Given :
AB is diameter passing through point R inside (RBSESolutions.com) circle of center O and required chord is PQ which is perpendicular to AB, CD is other chord passing through R.
RBSE Solutions for Class 10 Maths Chapter 12 Circle Miscellaneous Exercise 33
To Prove : Chord PQ is the smallest.
Construction : Through O, draw OM ⊥ CD
Proof : In right angled ∆OMR
Hypotenuse OR is longest side of ∆OMR
⇒ OR > OM
OR is distance of chord PQ from center O and OM is distance of chord CD from center O. Since any one of two chords of a circle, the larger chord is nearer to the center
Thus, chord PQ is the smallest.

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