NCERT Exemplar Class 9 Maths Solutions Chapter 10 Circles 2021

Q: Can we always form a circle passing through any four points given in a plane?

No, it is not always possible to form a circle passing through four points given in any plane. For three non-Collinear points, a unique circle can be formed but it is not necessary that it will pass through the fourth point.

Q: Three vertices of a right-angle triangle with perpendicular sides a and b lie on a circle, what will be the diameter of that circle?

We know that diameter subtends 90° at any point of the circle. Therefore, the hypotenuse of the given right angle triangle will be diameter. The length of diameter will be the length of the hypotenuse.

Q: What is the longest chord length in a circle of radius R?

The longest chord of any circle will be its diameter. Hence, its length will be twice the radius that is 2R.

Q: How many questions are expected from the chapter on Circles in the final exam?

Generally, you can expect 2-3 questions under the segment of concise answer type or long answer type questions from Circles. The NCERT exemplar Class 9 Maths solutions chapter 10 can help you score well in the exam.

NCERT Exemplar Class 9 Maths Solutions Chapter 10 Circles

Edited By Ravindra Pindel | Updated on Aug 31, 2022 01:04 PM IST

NCERT exemplar Class 9 Maths solutions chapter 10 provides the circle’s properties, which are crucial and applicable in higher Mathematics. The solutions delivered for NCERT exemplar Class 9 Maths chapter 10 solutions are prepared by our experts to provide precise answers while attempting NCERT Class 9 Maths questions. These NCERT exemplar Class 9 Maths chapter 10 solutions helps the students to understand the concepts in a gradual manner leading to high proficiency in solving the problems based on circles. The CBSE Syllabus for Class 9 is followed for these NCERT exemplar Class 9 Maths solutions chapter 10.

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Exercise 10.1
Exercise 10.2
Exercise 10.3
Exercise 10.4
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Exercise 10.1

Question:1

AD is a diameter of a circle and AB is a chord. If AD = 34 cm, AB = 30 cm, the distance of AB from the center of the circle is:
(A) 17 cm
(B) 15 cm
(C) 4 cm
(D) 8 cm

Answer:

(D) 8 cm
Hint:
Solution:
Given AD = 34 cm and AB = 30 cm
In figure, draw OL $\perp$ AB

$AL = LB = \frac{1}{2} AB = 15 cm$
In Right angled $\triangle$ OLA
OA² = OL² + AL²
(By Pythagoras Theorem)
(17)² = OL² + (15)²
OL² = 289 – 225 = 64
OL = 8 cm
(Taking positive square root, because length is always positive)
Hence the distance of the chord from the centre is 8 cm.
Therefore option (D) is correct.

Question:2

In Fig. 10.3, if OA = 5 cm, AB = 8 cm and OD is perpendicular to AB, then CD is equal to:

Fig. 10.3
(A) 2 cm
(B) 3 cm
(C) 4 cm
(D) 5 cm

Answer:

(A) 2 cm
Solution:
We know that the perpendicular from the centre of a circle to a chord bisects the chord.

$AC = CB = \frac{1}{2} AB = \frac{1}{2} \times 8 = 4 cm$
Given OA = 5 cm
Using Pythagoras theorem,
$\\AO^{2} = AC^{2} + OC^{2}\\ (5)^{2} = (4)^{2}+ OC^{2}\\ 25 = 16 + OC^{2}\\ OC^{2} = 25 - 16 = 9\\ OC = 3 cm\\$
(Taking positive square root, because length is always positive)
OA = OD (same because both are radius)
OD = 5 cm
CD = OD – OC = 5 – 3 = 2 cm
Therefore option (A) is correct.

Question:3

If AB = 12 cm, BC = 16 cm and AB is perpendicular to BC, then the radius of the circle passing through the points A, B and C is:
(A) 6 cm
(B) 8 cm
(C) 10 cm
(D) 12 cm

Answer:

$AB \perp BC$
We know that every angle inscribed in a semicircle is a right angle.
So AOC is the diameter of this circle with O as centre.
Now, applying Pythagoras theorem
$\\(Hypotenuse)^{2} = (Perpendicular)^{2} + (Base)^{2}\\ AC^{2} = AB^{2} + BC^{2}\\ AC^{2} = 12^{2} + 16^{2}\\ AC^{2} = 144 + 256\\ AC^{2} = 400\\ AC =\sqrt{400} \\AC = 20 cm$
As AC is the diameter, Radius
$=\frac{AC}{2} = \frac{20}{2} = 10 cm$
Therefore option (C) is correct.

Question:4

In Fig.10.4, if $\angle ABC = 20^{\circ}$ , then $\angle AOC$ is equal to:

$\\(A) 20^{\circ}\\ (B) 40^{\circ}\\ (C) 60^{\circ}\\ (D) 10^{\circ}\\$

Answer:

(B)

Solution:

Given, $\angle ABC = 20^{\circ}$

We know that,

Angle subtended at the center by an arc is twice the angle subtended by it at any part of the circle.

Hence,

$\angle AOC = 2\angle ABC = 2 \times 20^{\circ} = 40^{\circ}$

Therefore option (B) is correct.

Question:5

In Fig.10.5, if AOB is a diameter of the circle and AC = BC, then $\angle CAB$ is equal to:

(A) 30º
(B) 60º
(C) 90º
(D) 45º

Answer:

(D)
Solution:
We know that diameter subtends a right angle to the circle.
$\therefore \angle BCA = 90^{\circ} \: \: \: \cdots (i)$
Now,
AC = BC
$\angle ABC =\angle CAB \: \: \: \cdots (ii)$ (Angles opposite to equal sides are equal in a triangle)
In $\triangle ABC$
$\angle CAB +\angle ABC +\angle BCA = 180^{\circ}$ (By angle sum property of a triangle)
$\\\angle CAB +\angle CAB +\angle 90^{\circ} = 180^{\circ} \\ 2\angle CAB = 180^{\circ}-\angle 90^{\circ} \\ \angle CAB = \frac{90^{\circ}}{2}\\ \angle CAB = 45^{\circ}$
Therefore option (D) is correct.

Question:6

In Fig. 10.6, if $\angle OAB = 40^{\circ}$ , then $\angle ACB$ is equal to:

Fig. 10.6
(A) 50º
(B) 40º
(C) 60º
(D) 70°

Answer:

(A) 50º
Solution:
AO = OB (Radius of circle)
$OBA = \angle OAB = 40^{\circ}$ (angles opposite to equal sides in a triangle are equal)
In $\triangle OAB$
$40^{\circ} + \angle AOB + 40 = 180^{\circ}$ (angle sum property of a triangle)
$\angle AOB = 100 ^{\circ}$
We know that
$2 \angle ACB = \angle AOB$ (Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)
$\angle ACB =\frac{100}{2} \\ \angle ACB = 50^{\circ}$
Hence, option A is the correct answer.

Question:7

In Fig. 10.7, if $\angle DAB = 60 ^{\circ}, \angle ABD = 50^{\circ}$ , then $\angle ACB$ is equal to:

Fig. 10.7
(A) 60º
(B) 50º
(C) 70º
(D) 80º

Answer:

(C) 70°
Solution:
In $\triangle ABD$ ,
$\angle ABD + \angle ADB +\angle DAB = 180^{\circ}$
(Angle sum property of triangle)

$\\50^o+ < ADB + 60^o = 180^o\\\Rightarrow \angle ADB = 70^o$
Now, $\angle ACB = \angle ADB = 70^{\circ}$
(Angles in the same segment are equal)
$\angle ADB = 70^{\circ} \\ \angle ACB = 70^{\circ}$
Therefore option (C) is correct.

Question:8

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and $\angle$ ADC = 140º, then $\angle$ BAC is equal to:
(A) 80º
(B) 50º
(C) 40º
(D) 30º

Answer:

(B) 50°
Solution:
Given, ABCD is cyclic Quadrilateral and $\angle$ ADC = 140°
We know that the sum the opposite angles in a cyclic quadrilateral is 180°.

$\angle$ ADC + $\angle$ ABC = 180°
140° + $\angle$ ABC = 180°
$\angle$ ABC = 180° – 140°
$\angle$ ABC = 40°
Since, $\angle$ ACB is an angle in semi circle
$\angle$ ACB = 90°
In $\triangle$ ABC
$\angle$ BAC + $\angle$ ACB + $\angle$ ABC = 180° (angle sum property of a triangle)
$\angle$ BAC + 90° + 40° = 180°
$\angle$ BAC = 180° – 130° = 50°
Therefore option (B) is correct.

Question:9

In Fig. 10.8, BC is a diameter of the circle and $\angle$ BAO = 60º. Then $\angle$ ADC is equal to:

Fig. 10.8
(A) 30º
(B) 45º
(C) 60º
(D) 120º

Answer:

$\angle$ ABO = 60° [ $\therefore$ $\angle$ BAO = 60° Given]
In $\triangle$ AOB,
$\angle$ ABO + $\angle$ OAB = $\angle$ AOC (exterior angle is equal to sum of interior opposite angles)
60° + 60° = 120° = $\angle$ AOC
$\angle ADC =\frac{1}{2} \angle AOC$
(Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)
$\angle ADC =\frac{1}{2} \left ( 120^{\circ} \right )$
$\angle$ ADC = 60°
Therefore option (C) is correct.

Question:10

In Fig. 10.9, $\angle$ AOB = 90º and $\angle$ ABC = 30º, then $\angle$ CAO is equal to:

(A) 30º
(B) 45º
(C) 90º
(D) 60º

Answer:

(D)
Solution:

In AOB,
$\angle$ OAB + $\angle$ ABO + $\angle$ BOA = 180° … (i) (angle sum property of Triangle)
OA = OB = radius
Angles opposite to equal sides are equal
$\angle$ OAB = $\angle$ ABO
Equation (i) becomes
$\angle$ OAB + $\angle$ OAB + 90° = 180°
2 $\angle$ OAB = 180° – 90°
$\angle$ OAB = 45° …(ii)
In $\triangle$ ACB,
$\angle$ ACB + $\angle$ CBA + $\angle$ CAB = 180° (angle sum property of Triangle)
$\angle ACB = \frac{1}{2} \angle BOA = 45^{\circ}$
(Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)
$\therefore$ 45° + 30° + $\angle$ CAB = 180°
$\angle$ CAB = 180° – 75° = 105°
$\angle$ CAO + $\angle$ OAB = 105°
$\angle$ CAO + 45° = 105°
$\angle$ CAO = 105° – 45° = 60°
Therefore option (D) is correct.

Exercise 10.2

Question:1

Write True or False and justify your answer in each of the following: Two chords AB and CD of a circle are each at distances 4 cm from the centre. Then AB = CD.

Answer:

True
Solution:
Given,
AB & CD are chords of a circle and they are equidistant (4 cm) from the centre O.

To find whether AB = CD
As, the two chords AB and CD are each at a distance 4 cm from the centre, this distance will be perpendicular distance.
Hence OM $\perp$ AB & ON $\perp$ CD
OM = ON = 4 cm
$\therefore$ OM bisects AB & ON bisects CD
i.e., $AM = \frac{1}{2} AB \: \: and\: \: DN = \frac{1}{2} DC$
$\angle$ OMA = $\angle$ OND = 90°
OA = OD (Both are radius)
$\triangle$ AOM $\cong$ $\triangle$ DON (RHS Congruence)
AM = DN (by CPCT)
2AM = 2DN (Multiplying both sides by 2)
AB = DC
Therefore the given statement is true.

Question:2

Write True or False and justify your answer in each of the following: Two chords AB and AC of a circle with centre O are on the opposite sides of OA. Then $\angle OAB =\angle OAC.$

Answer:

False
Solution:
Given:

AB & AC are chords of a circle with centre O.
They are given to be at the opposite sides of OA.
To find out
If $\angle$ OAB = $\angle$ OAC
Join OB and OC
Consider $\triangle$ AOC & $\triangle$ AOB we have
OB = OC (Radii of the same circle)
AO = AO (common)
But it is not sure whether AB = AC or not.
Also, we do not have any other condition to prove that the triangles are congruent.
$\therefore$ $\angle$ OAB $\neq$ $\angle$ OAC
Therefore the given statement is false.

Question:3

Write True or False and justify your answer in each of the following: Two congruent circles with centres O and ${O}'$ intersect at two points A and B. Then $\angle AOB =\angle A{O}'B.$

Answer:

True
Solution:

Two congruent circles with centres O and ${O}'$ intersecting at two points A and B are shown in the above figure.
These are congruent circles, it means that their radius is the same.
If the figure, we have joined the centres O and ${O}'$
In $\triangle OA{O}' \: \: and\: \: \triangle OB{O}'$
$OA = {O}'A$ (Radius)
So, $\angle AO{O}' =\angle A{O}'O$
(angles opposite to equal sides in a triangle are equal) …(i)
Similarly, in $\triangle OB{O}'$
$OB= {O}'B$ (Radius)
$\angle BO{O}' =\angle B{O}'O$ (angles opposite to equal sides in a triangle are equal) …(ii)
Adding (i) and (ii)
$\angle AO{O}' +\angle BO{O}' =\angle A{O}'O+\angle B{O}'O$
$\angle AOB =\angle A{O}'B$
Therefore the given statement is true.

Question:4

Write True or False and justify your answer in each of the following:Through three collinear points a circle can be drawn.

Answer: False
Hint:
Solution:
We can draw a circle from one given point. For example from point A in the figure, we can draw infinite circles passing through that point

We can draw a circle from two given points as well taking those two points as the diameter or any chord of that circle. For example, from points A and B, we can draw infinite circles as shown below:

But when we talk about three collinear points, it means that these three points lie on a straight line.
So it is not possible to draw a circle passing through three points on a straight line.

Hence the given statement is False

Question:5

Write True or False and justify your answer in each of the following: A circle of radius 3 cm can be drawn through two points A, B such that AB = 6 cm.

Answer:

True
Solution:
Yes, we can draw a circle of radius 3 cm.
The points A & B are such that AB = 6 cm.
So AB will become the diameter of the required circle as AB = 2 (Radius)
Hence, 2 × 3 = 6 cm

Therefore the given statement is true.

Question:6

Write True or False and justify your answer in each of the following: If AOB is a diameter of a circle and C is a point on the circle, then AC² + BC²= AB^2.

Answer:

True
Solution:

Given: AOB is a diameter of a circle
$\Rightarrow$ AOB is a straight line
$\Rightarrow$ $\angle$ AOB = 180°
Now, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle
$\\\angle ACB =\left ( \frac{1}{2} \right ) \angle AOB\\ \angle ACB =\left ( \frac{1}{2} \right ) 180^{\circ}\\ \angle ACB = 90^{\circ}$
$\Rightarrow$ ACB is a Right angle
So $\angle ABC$ follows Pythagoras theorem,
i.e., AB²= AC² + BC²
Therefore the given statement is true.

Question:7

Write True or False and justify your answer in each of the following: ABCD is a cyclic quadrilateral such that $\angle A = 90^{\circ}, \angle B = 70^{\circ}, \angle C = 95^{\circ} \; and \; \angle D = 105^{\circ}$ .

Answer:

False.
Solution:
In a cyclic Quadrilateral, the sum of opposite pairs of the angles in the given quadrilateral must be equal to 180°.
Here $\angle A + \angle C = 90 ^{\circ} + 95 ^{\circ} = 185 ^{\circ}$
$\angle B + \angle D = 70 ^{\circ} + 105 ^{\circ} = 175 ^{\circ}$
Hence, it is not a cyclic Quadrilateral because the sum of the measures of opposite angles is not equal to 180°
Therefore the given statement is false.

Question:8

Write True or False and justify your answer in each of the following: If A, B, C, D are four points such that $\angle BAC = 30^{\circ} \;\; and \;\;\angle BDC = 60^{\circ}$ , then D is the centre of the circle through A, B and C.

Answer:

True
Solution:
Given, ∠BAC = 30° and ∠BDC = 60°

We know that,
The angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle.
Considering BC,
At centre, $\angle BDC = 60^{\circ}$
At any other point on the circle,
$\angle BAC = 30^{\circ}=\frac{1}{2} (60^{\circ})$
Hence the above rule is justified.
Therefore the given statement is true.

Question:9

Write True or False and justify your answer in each of the following: If A, B, C and D are four points such that $\angle BAC=45^{\circ}$ and $\angle BDC=45^{\circ}$ , then A, B, C, D are concyclic.

Answer:

True
Solution:
Definition of concyclic points: The points that lie on the same circle.
We know that angles in the same segment of a circle are equal
Since $\angle BAC=45^{\circ}=\angle BDC$
We know that these angles are in the same segment of a circle.
This can be represented as shown below:

BC is the chord of the circle and hence A, B, C & D are concyclic.
Therefore the given statement is true.

Question:10

Write True or False and justify your answer in each of the following: In Fig. 10.10, if AOB is a diameter and $\angle$ ADC = 120°, then $\angle$ CAB = 30°.

Answer:

True
Solution:
Given, AOB is a diameter and $\angle$ ADC = 120°
Join CA & CB

Since ADCB is a cyclic quadrilateral
$\angle$ ADC + $\angle$ CBA = 180° (Sum of opposite angles of a cyclic quadrilateral is 180°)
120° + $\angle$ CBA = 180°
$\angle$ CBA = 60°
In $\triangle$ ACB,
$\angle$ CAB + $\angle$ CBA + $\angle$ ACB = 180° (angle sum property of a triangle)
$\angle$ ACB = 90° (Since, $\angle$ ACB is an angle in a semi-circle)
$\angle$ CAB + 60° + 90° = 180°
$\Rightarrow$ $\angle$ CAB = 180° – 150° = 30°
Therefore the given statement is true.

Exercise 10.3

Question:1

If arcs AXB and CYD of a circle are congruent, find the ratio of AB and CD.

Answer:

Given that arcs AXB and CYD are congruent arcs of a circle
To find: the ratio AB:CD
Join AB and CD

It is given that AXB and CYD are congruent arcs of a circle
Their respective chords will be equal.
So, AB = CD
Hence, the ratio of AB and CD is 1 : 1.

Question:2

If the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q, prove that arc PXA ≅ Arc PYB.

Answer:

Given that the perpendicular bisector of a chord AB of a circle PXAQBY intersects the circle at P and Q
Let AB be the chord having centre at O.
Let PQ be the perpendicular bisector of the chord AB, which intersects AB at M and it is passes through the centre O.
Join AP and BP

In $\triangle APM$ and $\triangle BPM$
AM = MB (PM is the perpendicular bisector of AB)
PM = PM (Common)
$\angle PMA = \angle PMB$ (90° each)
$\therefore \triangle APM \cong \triangle BPM$ (SAS congruence)
PA = PB (by CPCT)
If 2 chords of a circle are equal, then corresponding arcs are congruent
Hence, arc PXA $\cong$ arc PYB.
Hence proved

Question:3

A, B and C are three points on a circle. Prove that the perpendicular bisectors of AB, BC and CA are concurrent

Answer:

Given: A, B, C are 3 points on a circle
Construction $\Rightarrow$ Draw perpendicular dissector of AB and AC and they meet at a point O. Then join OA, OB and OC.

To prove: perpendicular bisector of AB, BC and CA are concurrent.
Proof: In $\triangle$ OEA and $\triangle$ OEB
AE = BE [OL is ^ar bisector of AB]
OE = OE [Common side]
$\angle$ AEO = $\angle$ BEO [each 90°]
$\therefore$ $\triangle$ OEA $\cong$ $\triangle$ DEB [By SAS congruence Rule]
OA = OC
Similarly, $\triangle$ OFA $\cong$ $\triangle$ OFC [by SAS congruence]
$\therefore$ OB = OC = OA = t [Say]
SO, we draw a perpendicular from O to the BC on join
In $\triangle$ OMB and $\triangle$ OMC
OB = OC [Proved Above]
OM = OM [Common side]
$\angle$ OMB = $\angle$ OMC [90° each]
$\therefore$ $\triangle$ OMB $\cong$ $\angle$ OMC [by RHS congruence rule]
$\Rightarrow$ BN = MC [by corresponding parts of congruent triangle]
Hence, OM is the perpendicular bisector of side BC.
Hence OL, ON, OM are concernment.
Hence proved.

Question:4

AB and AC are two equal chords of a circle. Prove that the bisector of the angle BAC passes through the centre of the circle.

Answer:

Given: AB and AC are two equal chords of the circle.
Prove: Centre lies on the bisector of the $\angle$ BAC.
Construction: Join BC

Proof: In $\triangle$ AOB and $\triangle$ AOC
AB = AC (given)
$\angle$ BAO = $\angle$ CAO (given)
AO = AO (common)
$\therefore$ $\triangle$ AOB $\cong$ $\triangle$ AOC (by SAS congruence)
$\Rightarrow$ BO = CO (CPCT) …(i)
and $\angle$ AOB = $\angle$ AOC (CPCT)
$\angle$ AOB + $\angle$ AOC = 180° (Linear pair)
$\therefore$ 2 $\angle$ AOB = 180° ( $\angle$ APB = $\angle$ APC)
$\Rightarrow$ $\angle$ AOB = 90°
$\therefore$ AO is perpendicular bisector of the chord BC and BC is the diameter as BO = CO
$\Rightarrow$ Hence, AO passes through the centre of the circle, and O is the centre of the circle.
Hence proved

Question:5

If a line segment joining mid-points of two chords of a circle passes through the centre of the circle, prove that the two chords are parallel.

Answer:

Given: M and N are midpoints of chords AB and CD respectively.
MN passes through the centre of the circle O

To prove: Chord AB and Chord CD are parallel
Construction: Join B and C.
Proof:
$\angle$ M = $\angle$ N = 90° (Segment joining the midpoint of the chord to the centre)
Consider, DOMB and DONC
$\angle$ BOM = $\angle$ CON (Vertically opposite angles)
OB = OC (Radii of the same circle)
$\angle$ OMB = $\angle$ ONC (90° each)
$\therefore$ $\triangle$ OMB $\cong$ $\angle$ ONC (AAS congruence)
$\therefore$ $\angle$ OBM = $\angle$ OCN (CPCT)
$\therefore$ $\angle$ ABC = $\angle$ BCD
We can see that $\angle$ ABC and $\angle$ BCD are alternate angles of the chords AB and CD with BC as transversal.
$\Rightarrow$ Hence, chord AB || chord CD.

Question:6

ABCD is such a quadrilateral such that A is the centre of the circle passing through B, C and D. Prove that $\angle CBD + \angle CDB =\frac{1}{2}\angle BAD.$

Answer:

Given: A circle passing through points B, C, D and ABCD is a quadrilateral having centre at A.
To prove: $\angle CBD +\angle CDB =\frac{1}{2}\angle BAD$
Construction: Join AC & BD

Proof:
We know that angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle
Arc DC subtends $\angle$ DAC at the centre and $\angle$ DBC at a point B in the remaining part of the circle.
$\therefore$ $\angle$ DAC = 2 $\angle$ CBD … (i)
Similarly, arc BC subtends ÐBAC at the centre and ÐBDC at a point D in the remaining part of the circle.
$\therefore$ $\angle$ CAB = 2 $\angle$ CDB … (ii)
On adding equations (i) and (ii)
We get
$\angle$ DAC + $\angle$ CAB = 2 $\angle$ CBD + 2 $\angle$ CDB
$\angle$ DAB = 2 ( $\angle$ CBD + $\angle$ CDB)
$\angle CBD +\angle CDB =\frac{1}{2}\angle BAD$
Hence proved

Question:7

O is the circumcentre of the triangle ABC and D is the mid-point of the base BC. Prove that $\angle$ BOD = $\angle$ A.

Answer:

Given: O is the circumcentre of the triangle ABC and D is the mid-point of the base BC.
$\therefore$ OD is perpendicular BC

In the right triangles OBD and OCD.
We have OB = OC (Radii of the same circle)
OD = OD (common)
$\angle$ ODB = $\angle$ ODC (90^o)
$\therefore$ $\triangle$ OBD $\cong$ $\triangle$ OCD (R.H.S. congruence)
$\Rightarrow$ $\angle$ BOD = $\angle$ COD (CPCT)
$\Rightarrow \angle BOD =\frac{1}{2} \angle BOC$ … (1)
Also, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle. Consider arc BC,
$\Rightarrow \angle BAC =\frac{1}{2} \angle BOC$ … (2)
From (1) and (2), we have
$\therefore$ $\angle$ BOD = $\angle$ BAC
Hence Proved.

Question:8

On a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides. Prove that $\angle$ BAC = $\angle$ BDC.

Answer:

Given that on a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides.
Let $\triangle$ ACB and $\triangle$ ADB are two right-angled triangles with common hypotenuse AB.
To prove: $\angle$ BAC = $\angle$ BDC

Proof:
Let O be the mid-point of AB
Then, OA = OB = OC = OD
Mid-point of the hypotenuse of a right triangle is equidistant from both vertices.
We can draw a circle passing through the points A, B, C and D with O as centre and radius equal to OA.
We know that angles in the same segment of a circle are equal
From the figure, $\angle$ BAC and $\angle$ BDC are angles of same segment BC.
$\therefore$ $\angle$ BAC = $\angle$ BDC
Hence proved.

Question:9

Two chords AB and AC of a circle subtends angles equal to 90º and 150º, respectively at the centre. Find $\angle$ BAC, if AB and AC lie on the opposite sides of the centre.

Answer:

60°
Solution:
Given: Two chords AB and AC of a circle subtends angles equal to 90º and 150º, respectively at the centre.
AB and AC lie on the opposite sides of the centre
In $\triangle$ BOA
OB = OA (Both equal to radius)
$\angle$ OAB = $\angle$ OBA … (i) (angles opposite to equal sides in a triangle are equal)

In $\triangle$ OAB,
$\angle$ OBA + $\angle$ BAO + $\angle$ AOB = 180° (angle sum property of a triangle)
$\Rightarrow$ $\angle$ OAB + $\angle$ OAB + 90° = 180° [From equation (1)]
$\Rightarrow$ 2 $\angle$ OAB = 180° – 90°
$\Rightarrow$ $\angle$ OAB = $\frac{90^{\circ}}{2}$ = 45°
Now in $\triangle$ AOC, AO = OC (Both equal to radius)
$\angle$ OCA = $\angle$ OAC … (ii) (angles opposite to equal sides in a triangle are equal)
Also,
$\angle$ AOC + $\angle$ OAC + $\angle$ OCA = 180° (angle sum property of a triangle)
150° + 2 $\angle$ OAC = 180°
2 $\angle$ OAC = 180° – 150°
2 $\angle$ OAC = 30°
$\angle$ OAC = 15°
$\angle$ BAC = $\angle$ OAB + $\angle$ OAC = 45° + 15° = 60°
Hence the answer is 60°

Question:10

If BM and CN are the perpendiculars drawn on the sides AC and AB of the triangle ABC, prove that the points B, C, M and N are concyclic.

Answer:

Given: In DABC, MB $\perp$ AC and CN $\perp$ AB.
To Prove: Points B, C, M and N are Concyclic (lie on the same circle)

Proof:
Assume, BC is the diameter of a circle.
So BC will subtend an angle of 90° at any point on the circle.
Now, MB $\perp$ AC and CN $\perp$ AB. So these angles lie on the circle and the points N and M lie on this circle.
Hence, BCMN is concyclic.
Hence Proved.

Question:11

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.

Answer:

Given: A line is drawn parallel to the base of an isosceles triangle to intersect its equal sides
Let $\triangle$ PQR be an isosceles triangle such that PR = PQ
Also, RQ || MN
To prove: Quadrilateral MNQR is a cyclic quadrilateral.
Proof:

In DPQR
PR = PQ [Given]
$\Rightarrow$ $\angle$ PQR = $\angle$ PRQ … (1) (angles opposite to equal sides in a triangle are equal)
Now, RQ || MN and PQ is a transversal
$\Rightarrow$ $\angle$ PNM = $\angle$ PQR … (2) (Corresponding angles)
On adding $\angle$ MNQ to both sides in equation (2), we get:
$\angle$ PNM + $\angle$ MNQ = $\angle$ PQR + $\angle$ MNQ
180° = $\angle$ PQR + $\angle$ MNQ ( $\angle$ PNM + $\angle$ MNQ forms a linear pair)
$\Rightarrow$ $\angle$ MNQ + $\angle$ PRQ = 180° (From equation (1))
Hence, MNRQ is a cyclic quadrilateral as the sum of opposite angles is 180^o
Hence proved

Question:12

If a pair of opposite sides of a cyclic quadrilateral are equal, prove that its diagonals are also equal.

Answer:

Given: A pair of opposite sides of a cyclic quadrilateral are equal.
Let ABCD be a cyclic quadrilateral with AD = BC

To Prove: Diagonals are equal, i.e., AC = BD
Proof:
We know that angles is same segment are equal.
Consider segment CD, $\angle$ DAC = $\angle$ DBC
So we can write, $\angle$ DAO = $\angle$ CBO …(i)
Consider segment AB, $\angle$ ADB = $\angle$ ACD
So we can write, $\angle$ ADO = $\angle$ BCO…(ii)
In $\triangle$ AOD & $\triangle$ BOC,
$\angle$ DAO = $\angle$ CBO (from (i))
$\angle$ ADO = $\angle$ BCO (from (ii))
$\angle$ AOD = $\angle$ BOC (vertically opposite angles)
$\triangle$ AOD $\cong$ $\triangle$ BOC [AAA congruence rule]
AO = BO (CPCT)…(iii)
DO = CO (CPCT)…(iv)
Adding (iii) and (iv),
AO + OC = BO + OD
Hence, AC = BD
Hence proved

Question:13

The circumcenter of the triangle ABC is O. Prove that $\angle$ OBC + $\angle$ BAC = 90º.

Answer:

Given that the circumcenter of the triangle ABC is O. So O will be the radius of the circle passing through the points A, B, C

Now, OB = OC = radius of the circle
We know that angles opposite to equal sides in a triangle are equal
$\Rightarrow$ $\angle$ OBC = $\angle$ OCB = Let x
In DOBC,
$\therefore$ x + x + $\angle$ BOC = 180° (angle sum property of a triangle)
2x + $\angle$ BOC = 180°
$\angle$ BOC = 180° – 2x
Now, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle
So considering the arc BC,
$\angle$ BOC = 2 $\angle$ BAC
Putting the value of $\angle$ BOC
180° – 2x = 2 $\angle$ BAC
90° – x = $\angle$ BAC
We have to find $\angle$ OBC + $\angle$ BAC
$\therefore$ $\angle$ ÐBAC + $\angle$ OBC = (90° – x) + x
$\angle$ BAC + $\angle$ OBC = 90°
Hence proved

Question:14

A chord of a circle is equal to its radius. Find the angle subtended by this chord at a point in major segment

Answer:

30^o
Solution:
Given that a chord of a circle is equal to its radius
Let the chord be AB as shown in the figure.
AB = Radius of Circle.

Now, AB = OA = OB
In $\triangle$ OAB,
AB = OA = OB = radius
Hence it is an equilateral triangle. So all angles are 60o
$\angle$ OAB = $\angle$ ABO = $\angle$ BOA = 60^o
Let point D be a point on the major arc.
Considering the arc AB, we know that angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle
So,
$\angle ADB = \frac{1}{2} \angle AOB = \frac{1}{2} (60^{\circ}) = 30^{\circ}$

Question:16

In Fig.10.14, $\angle$ ACB = 40º. Find $\angle$ OAB.

Answer:

50°
Solution:
Given, $\angle$ ACB = 40°
As we know that angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle
$\\\Rightarrow \angle AOB = 2 \angle ACB.\\ \\ \Rightarrow \angle ACB = \frac{ \angle AOB}{2}\\ \\ \Rightarrow 40^{\circ} =\frac{1}{2} \angle AOB\\ \\ \Rightarrow \angle AOB = 80^{\circ}$
In $\triangle$ AOB, AO = BO (Both are the radius of the same circle)
$\angle$ OBA = $\angle$ OAB (angles opposite to equal sides in a triangle are equal)
As we know, the sum of all three angles in a triangle AOB is 180°.
In $\triangle$ AOB
$\\ \angle AOB + \angle OBA + \angle OAB = 180^{\circ}\\ \Rightarrow 80^0 + \angle OAB + \angle OAB = 180^{\circ}\\ \Rightarrow 2\angle OAB = 180^{\circ} -80^{\circ}\\ \Rightarrow 2\angle OAB = 100^0\\\\ \Rightarrow \angle OAB = \frac{100^{\circ}}{2}=50^{\circ}$

Question:17

A quadrilateral ABCD is inscribed in a circle such that AB is diameter and $\angle$ ADC = 130º. Find $\angle$ BAC.

Answer:

40°
Solution:
Given: Quadrilateral ABCD inscribed in a circle
$\angle$ ADC = 130º
$\therefore$ Quadrilateral ABCD is a cyclic Quadrilateral. Sum of opposite angles in a cyclic quadrilateral is 180°

$\Rightarrow$ $\angle$ ADC + $\angle$ ABC = 180°
$\Rightarrow$ 130° + $\angle$ ABC = 180°
$\Rightarrow$ $\angle$ ABC = 180° – 130°
$\Rightarrow$ $\angle$ ABC = 50° …(i)
AB is a diameter of a circle (given)
$\therefore$ ACB = 90° (angle in a semicircle is 90º)
In $\triangle$ ABC,
$\Rightarrow$ $\angle$ ABC + $\angle$ ACB + $\angle$ BAC = 180°(angle sum property of a triangle)
Putting all the values,
$\Rightarrow$ 50° + 90° + $\angle$ BAC = 180°
$\Rightarrow$ $\angle$ BAC = 180° –50° – 90°
$\therefore$ $\angle$ BAC = 40°

Question:18

Two circles with centres O and O′ intersect at two points A and B. A line PQ is drawn parallel to OO′ through A(or B) intersecting the circles at P and Q. Prove that PQ = 2 OO′.

Answer:

Two circles with centres O and O′ intersect at two points A and B.
PQ || OO'

To Prove that: PQ = 2OO'
Proof:
Draw OM perpendicular to PB and O'N perpendicular to BQ
From the figure, we have:
OP = OB (radius of same circle)
O'B = O'Q (radius of the same circle)
In $\triangle$ OPB,
BM = MP … (i) (perpendicular from the centre of the circle bisects the chord)
Similarly in $\triangle$ O'BQ,
BN = NQ … (ii) (perpendicular from the centre of the circle bisects the chord)
Adding (i) and (ii),
BM + BN = PM + NQ
Adding BM + BN to both the sides.
BM + BN + BM + BN = BM + PM + NQ + BN.
2BM + 2BN = PQ
2(BM + BN) = PQ … (iii)
OO' = MN (As OO' NM is a rectangle)
OO' = BM + BN ... (iv)
Using equation (iv) in (iii)
2 OO'= PQ
Hence proved.

Question:19

In Fig.10.15, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of $\angle$ ACD + $\angle$ BED.

Answer:

270°
Solution:
Join AE

ACDE is a cyclic quadrilateral and sum of opposite angles in a cyclic quadrilateral is 180°
$\therefore$ $\angle$ ACD + $\angle$ AED = 180° … (i)
Now, we know that the angle in a semi-circle is 90°
So, $\angle$ AEB = 90° … (ii)
An adding equation (i) & (ii), we get.
$\angle$ ACD + $\angle$ AED + $\angle$ AEB = 180° + 90°
$\angle$ ACD + $\angle$ BED = 270°
Hence the value of $\angle$ ACD + $\angle$ BED is 270°

Question:20

In Fig. 10.16, $\angle$ OAB = 30º and $\angle$ OCB = 57º. Find $\angle$ BOC and $\angle$ AOC.

Answer:

$\angle$ AOC = 54° and $\angle$ BOC = 66°
Solution:
Given: $\angle$ OAB = 30°, $\angle$ OCB = 57°C
In $\triangle$ OAB,
AO = BO (radius of the same circle)
$\angle$ OAB = $\angle$ OBA = 30° (Angles opposite to equal sides are equal)
In $\triangle$ AOB, the sum of all angles is 180°.
$\Rightarrow$ $\angle$ OAB + $\angle$ OBA + $\angle$ AOB = 180°.
$\Rightarrow$ 30° + 30° + $\angle$ AOB = 180°
$\Rightarrow$ $\angle$ AOB = 180° – 30° – 30°
$\Rightarrow$ $\angle$ AOB = 120° … (i)
Now, in $\triangle$ OBC,
OC = OB (radius of the same circle)
$\angle$ OBC = $\angle$ OCB = 57° (Angles Opposite to equal sides are equal).
In $\triangle$ OBC, the sum of all angles is 180°.
$\angle$ OBC + $\angle$ OCB + $\angle$ BOC = 180°
$\Rightarrow$ 57° + 57° + $\angle$ BOC = 180°
$\angle$ BOC = 180° – 57° – 57°
$\Rightarrow$ $\angle$ BOC = 66° … (ii)
Now, form equation (i) we have
$\Rightarrow$ $\angle$ AOB = 120°.
$\Rightarrow$ $\angle$ AOC + $\angle$ COB = 120°
$\Rightarrow$ $\angle$ AOC + 66° = 120° (from ii)
$\Rightarrow$ $\angle$ AOC = 120° – 66°
$\Rightarrow$ $\angle$ AOC = 54° and $\angle$ BOC = 66°.

Exercise 10.4

Question:1

If two equal chords of a circle intersect, prove that the parts of one chord are separately equal to the parts of the other chord.

Answer:

Given: Two equal chords of a circle intersect
Let us construct a circle with centre O.
Its two equal chords are AB and CD which intersect at E as shown in the figure.

To prove: AE = DE and CE = BE
Proof:
Draw OM $\perp$ AB and ON $\perp$ CD
Join OE
In $\triangle$ OME and $\triangle$ ONE,
OM = ON (Equals chords of a circle are equidistant from the centre.)
OE = OE (Common)
$\angle$ OME = $\angle$ ONE (90^o)
$\because$ $\triangle$ OME $\cong$ $\triangle$ ONE (RHS congruence)
ME = NE (C.P.C.T.) ..…(i)
Now as AB = CD
$\frac{1}{2}AB =\frac{1}{2}CD$
AM = DN..…(ii)
Adding (i) and (ii)
AM + ME = DN + NE
Þ AE = DE
Also, AB – AE = CD – DE ( $\because$ AB = CD)
BE = CE
Hence proved

Question:2

If non-parallel sides of a trapezium are equal, prove that it is cyclic.

Answer:

Given: non-parallel sides of a trapezium are equal
Let ABCD is a trapezium where AB || CD & non-parallel sides are equal
i.e., AD = BC

To Prove: ABCD is a cyclic quadrilateral
Proof:
Draw DE $\perp$ AB
And CF $\perp$ AB
In $\triangle$ ADE & $\triangle$ BCF
$\angle$ AED = $\angle$ BFC (Both are 90°)
AD = BC (Given)
DE = CF (perpendicular distance between parallel sides will remain the same)
$\because$ $\triangle$ ADE $\cong$ $\triangle$ BCF (RHS congruence rule)
So $\angle$ DAE = $\angle$ CBF (CPCT)
i.e. $\angle$ A = $\angle$ B ....…(i)
Now, AB || DC, AD is a transversal line
$\angle$ A + $\angle$ D = 180° [Interior angles on the same side of the transversal are supplementary]
$\angle$ B + $\angle$ D = 180° [From (i)]
So, in ABCD, the sum of one pair of opposite angle is 180°.
Therefore, ABCD is a cyclic quadrilateral
Hence Proved.

Question:3

If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC, prove that P, Q, R and D are concyclic.

Answer:

In $\triangle$ ABC, P, Q and R are the midpoints of the sides BC, CA and AB respectively
Also AD $\perp$ BC

To prove: P, Q, R and D are concyclic
Constructions: Join PR, RD, QD
Proof:
By midpoint theorem
RP || AC
QP || AB
So, ARPQ is a parallelogram
$\angle$ RAQ = $\angle$ RPQ … (1) (opposite angles of a ||gm are equal)
Now, in right-angled $\triangle$ ADB, R is the midpoint of AB.
The median on the hypotenuse of a right triangle divides the triangle into two isosceles triangles because the median equals one-half the hypotenuse.
$\because$ RD = RA
$\angle$ RAD = $\angle$ RDA … (2) (angles opposite to equal sides in a triangle are equal)
Similarly, in $\triangle$ ADQ
$\angle$ DAQ = $\angle$ ADQ … (3)
Adding (2) and (3)
$\angle$ RAD + $\angle$ DAQ = $\angle$ RDA + $\angle$ ADQ
$\angle$ RAQ = vRDQ
From (1), $\angle$ RAQ = $\angle$ RPQ
Hence, we can see if we consider arc RQ of a circle then $\angle$ RAQ and $\angle$ RPQ are the angles subtended by it on a circle as these angles are equal.
So, P, Q, R and D are concyclic
Hence proved

Question:4

ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.

Answer:

Given: ABCD is a parallelogram. A circle whose centre O passes through A, B has intersected AD at P and BC at Q.
To Prove: Points P, Q, C and D are concyclic.

Proof:
Join PQ
We know that sum of opposite angles of a cyclic quadrilateral is 180°
$\angle$ A + $\angle$ PQB = 180°
Also, $\angle$ PQB + $\angle$ PQC = 180° (linear pair)
Comparing the above two equations we get
$\angle$ PQC = $\angle$ A
Let $\angle$ PQC = $\angle$ 1
Then, $\angle$ A = $\angle$ 1
Also, $\angle$ A = $\angle$ C (Opposite angles of a parallelogram are equal)
So, $\angle$ 1 = $\angle$ C
But, $\angle$ C + $\angle$ D = 180° (As ABCD is a parallelogram)
Hence, $\angle$ 1 + $\angle$ D = 180°
Therefore P, Q, C and D are concyclic
Hence proved

Question:5

Prove that angle bisector of any angle of a triangle and perpendicular bisector of the opposite side if intersect, they will intersect on the circumcircle of the triangle.

Answer:

Given: Angle bisector of any angle of a triangle and perpendicular bisector of the opposite side intersect on the circumcircle of the triangle.

Consider DABC
Let the angle bisector of angle A intersect circumcircle of triangle ABC at point D.
Join DC and DB
$\angle$ BCD = $\angle$ BAD (Angle in the same segment are equal)
$\angle$ BCD = $\angle$ BAD = $\frac{1}{2}$ $\angle$ A … (1) (AD is bisector of ÐCAB)
Similarly,
$\angle$ DAC = $\angle$ DBC (Angle in the same segment are equal)
$\angle$ DBC = $\angle$ DAC = $\frac{1}{2}$ $\angle$ ÐA … (2) (AD is bisector of ÐCAB)
From (1) and (2) we have
$\angle$ DBC = $\angle$ BCD
BD = DC (sides opposite to equal angles in a triangle are equal)
Hence, D also lies on the perpendicular bisector of BC.
Hence, proved

Question:6

If two chords AB and CD of a circle AYDZBWCX intersect at right angles (see Fig.10.18), prove that arc CXA + arc DZB = arc AYD + arc BWC = semicircle.

Answer:

Given: In circle AYDZBWCX two chords AB and CD intersect at right angles
To Prove: Arc CXA + Arc DZB = arc AYD + arc BWC = Semicircle
Proof:
Draw Diameter EF || CD having centre M
Since CD || EF
Arc EC = DF … (i)
Arc ECXA = arc EWB (Symmetrical about diameter of circle)
Arc AF = arc FB … (ii) (Symmetrical about diameter of circle)
We know that arc ECXAYDF = Semicircle

Arc EA + arc AF = Semicircle
arc EC + arc CXA + arc FB = semicircle (from ii, Arc AF = arc FB)
arc DF + arc CXA + arc FB = semicircle (from i, Arc EC = arc DF)
=> arc DF + arc FB + arc CXA = semicircle
Arc DZB + arc CXA = semicircle. (First statement is proved)
Now, we know that a circle is divided into two semi-circles.
Therefore the remaining portion of the circle (other than, Arc DZB + arc CXA) is also equal to a semi-circle
So, Arc AYD + arc BWC = semicircle
So we get: Arc CXA + Arc DZB = arc AYD + arc BWC = Semicircle
Hence proved

Question:7

If ABC is an equilateral triangle inscribed in a circle and P be any point on the minor arc BC which does not coincide with B or C, prove that PA is angle bisector of $\angle$ BPC.

Answer:

Let DABC is an equilateral triangle inscribed in a circle with centre O.
So all sides of this equilateral triangle are equal by definition
AB = AC = BC

Now we know that equal chords subtend and equal angles at the centre. So,
$\angle$ AOB = $\angle$ AOC = $\angle$ BOC … (i)
Consider the arc AB, $\angle$ AOB and $\angle$ APB are the angles subtended by AB at the centre and at a remaining part of the circle
Therefore
$\angle APB = \frac{1}{2} \angle AOB$ … (ii)
Similarly,
$\angle APC = \frac{1}{2} \angle AOC$ … (iii)
Using (i), (ii) and (iii), we have
$\angle$ APB = $\angle$ APC
Hence, PA is the angle bisector of $\angle$ BPC.
Hence proved

Question:8

In Fig. 10.19, AB and CD are two chords of a circle intersecting each other at point E. Prove that $\angle$ AEC = $\frac{1}{2}$ (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre).

Answer:

Given: AB and CD are two chords of a circle intersecting each other at point E
To prove:
$\angle$ AEC =0.5 (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre)
$\angle AEC=0.5\left ( \angle AOC+\angle DOB \right )$

Proof:
Extend the lines DO to L and BO to point X on the circle. Join AC.
Now we know that angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle.
For arc LC,
$\angle$ 1 = 2 $\angle$ 6 … (i)
For arc XA,
$\angle$ 3 = 2 $\angle$ 7 … (ii)
Consider $\triangle$ OAC
OC = OA (Radius of the same circle)
$\angle$ OCA = $\angle$ 4 … (iii) (angles opposite to equal sides in a triangle are equal)
$\angle$ AOC + $\angle$ OCA + $\angle$ 4 = 180° (angle sum property of a triangle)
$\angle$ AOC + $\angle$ 4 + $\angle$ 4 = 180°
$\angle$ AOC = 180° – 2 $\angle$ 4 …(iv)
Consider $\triangle$ DCO
OD = OC (Radius of the same circle)
$\angle$ OCD = $\angle$ 6 … (v) (angles opposite to equal sides in a triangle are equal)
Consider $\triangle$ AEC
$\angle$ AEC + $\angle$ ECA + $\angle$ CAE = 180° (angle sum property of a triangle)
$\angle$ AEC = 180° – ( $\angle$ ECA + $\angle$ CAE)
$\angle$ AEC = 180° – [( $\angle$ ECO + $\angle$ OCA) + ( $\angle$ CAO + OAE)]
$\angle$ AEC = 180° – ( $\angle$ 6 + $\angle$ 4 + $\angle$ 4 + $\angle$ 5) [from (iii) and (v)]
$\angle$ AEC = 180° – (2 $\angle$ 4 + $\angle$ 5 + $\angle$ 6)
$\angle$ AEC = 180° – 2 $\angle$ 4 - $\angle$ 5 - $\angle$ 6
From (iv), $\angle$ AOC = 180° – 2 $\angle$ 4
$\angle$ AEC = $\angle$ AOC - $\angle$ 5 - $\angle$ 6
Also, $\angle$ AOC = $\angle$ 1 + $\angle$ 2 + $\angle$ 3
So, $\angle$ AEC = $\angle$ 1 + $\angle$ 2 + $\angle$ 3 - $\angle$ 5 - $\angle$ 6
In $\triangle$ AOB,
OA = OB (Radius of the same circle)
$\angle$ 5 = $\angle$ 7 (angles opposite to equal sides in a triangle are equal)
$\angle$ AEC = $\angle$ 1 + $\angle$ 2 + $\angle$ 3 - $\angle$ 7 - $\angle$ 6
From (ii), $\angle$ 3 = 2 $\angle$ 7
$\angle$ AEC = $\angle$ 1 + $\angle$ 2 + $\angle$ 3 - 0.5<3 - $\angle$ 6
From (i), $\angle$ 1 = 2 $\angle$ 6
$\angle$ AEC = $\angle$ 1 + $\angle$ 2 + $\angle$ 3 - 0.5 $\angle$ 3 - 0.5 $\angle$ 1
$\angle$ AEC = 0.5 $\angle$ 1 + 0.5 $\angle$ 3 + $\angle$ 2
$\angle AEC = \frac{1}{2}\angle 1 + \frac{1}{2}\angle 3 + \frac{1}{2}\angle 2 + \frac{1}{2}\angle 2$
$\angle$ 2 = $\angle$ 8 (vertically opposite angles)
$\angle AEC = \frac{1}{2}\angle 1 + \frac{1}{2}\angle 3 + \frac{1}{2}\angle 2 + \frac{1}{2}\angle 8$
$\angle$ AEC = 0.5 ( $\angle$ 1 + $\angle$ 2 + $\angle$ 3 + $\angle$ 8)
$\angle AEC=0.5\left ( \angle AOC+\angle DOB \right )$
$\angle$ AEC =0.5 (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre)
Hence proved

Question:9

If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.

Answer:

Given: Bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q
Let ABCD be a cyclic quadrilateral
DP and QB are the bisectors of $\angle$ ADC and $\angle$ ABC respectively.
To Prove: PQ is the diameter of circle

Proof:
Joint QD and QC.
As, DP and QB are the bisectors of $\angle$ ADC and $\angle$ ABC respectively.
$\angle 1 =\frac{1}{2}\angle CDA\; and \; \;\angle2 =\frac{1}{2} \angle CBA$
Since ABCD is cyclic quadrilateral, sum of opposite angles is 180°
$\angle$ CDA + $\angle$ CBA = 180°
Divide both sides by 2,
$\frac{1}{2} \angle CDA +\frac{1}{2} \angle CBA = 90^{\circ}$
$\angle$ 1 + $\angle$ 2 = 90° … (i)
Now consider the segment QC. We know that angles in the same segment of a circle are equal. So,
$\angle$ 2 = $\angle$ 3
Putting the value in equation (i),
$\angle$ 1 + $\angle$ 3 = 90°
$\angle$ POQ = 90°
We know that angle in a semicircle is 90°
Hence PQ is a diameter of a circle.

Question:10

A circle has radius $\sqrt{2}$ cm. It is divided into two segments by a chord of length 2 cm. Prove that the angle subtended by the chord at a point in major segment is 45º.

Answer:

A circle with radius $\sqrt{2}$ cm is divided into two segments by a chord of length 2 cm.
We have drawn a circle having centre O
Let AB = 2 cm be the required chord which divides the circle into two segment.

To Prove: the angle subtended by the chord at a point in major segment is 45º.
Which means we have to prove that $\angle$ APB = 45°
We have, radius = $\sqrt{2}$ cm
OA = OB = $\sqrt{2}$ cm (given)
OA² = OB² = 2 cm
Also, chord AB = 2 cm (given)
AB² = 4cm
Now, we can see that in DAOB,
OA² + OB² = AB²
As, 2 + 2 = 4
So this follows Pythagoras theorem.
Hence $\angle$ AOB = 90°
Now we know that angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle
$\\\because \angle APB = \angle AOB\\ \Rightarrow \angle APB = 45^{\circ}$
Hence proved.

Question:11

Two equal chords AB and CD of a circle when produced intersect at a point P. Prove that PB = PD

Answer:

Given: Two equals chords AB and CD of a circle intersect at a point P.
To Prove : PB = PD

Proof: Given,
Join OP draw OL $\perp$ AB and OM $\perp$ CD
AB = CD
OL = OM (equal chords are equidistant from the centre)
Consider $\triangle$ OLP and $\triangle$ OMP
OL = OM
$\angle$ OLP = $\angle$ OMP (both are 90o)
OP = OP
$\triangle$ OLP $\cong$ $\triangle$ OMP (SAS congruence)
LP = MP …(i) (CPCT)
AB = CD (Given)
$\frac{1}{2}\left ( AB \right )=\frac{1}{2}\left ( CD \right )$
BL = DM …(ii)
Subtracting equation (ii) from equation (i) we get
LP – BL = MP – DM
$\Rightarrow$ PB = PD
Hence proved.

Question:12

AB and AC are two chords of a circle of radius r such that AB = 2AC. If p and q are the distances of AB and AC from the centre, prove that 4q² = p² + 3r²

Answer:

AB and AC are two chords of a circle of radius r such that AB = 2AC.
The distance of AB and AC from the centre are p and q, respectively
To Prove: 4q²= p² + 3r²
Proof:

Let AB = 2x then AC = x (Given, AB = 2AC)
Draw ON perpendicular to AB and OM perpendicular to AC
AM = MC = x/2 (perpendicular from the centre bisects the chord)
AN = NB = x (perpendicular from the centre bisects the chord)
In DOAM, applying Pythagoras theorem
AO² = AM²+ MO²
AO² = (x/2)² + q² ….(i)
In $\triangle$ OAN, applying Pythagoras theorem
AO² = (AN)² + (NO)²
AO² = (x)² + p² ….(ii)
From equation (i) and (ii)
$\\\left ( \frac{x}{2} \right )^{2}+\left ( q \right )^{2}=x^{2}+\left ( p \right )^{2}\\ \frac{x^{2}}{4}+q^{2}=x^{2}+p^{2}$
x² + 4q² = 4x² + 4p² [Multiply both sides by 4]
4q² = 3x² + 4p²
4q²= p² + 3 (x² + p²)
4q²= p²+ 3r² (In right angle $\triangle$ OAN, r² = x² + p²)
Hence Proved.

Question:13

In Fig. 10.20, O is the centre of the circle, $\angle$ BCO = 30°. Find x and y.

Answer:

x = 30° and y = 15°
Solution:
Given: O is the centre of the circle
$\angle$ BCO = 30°
Join OB and AC.
In $\triangle$ BOC,

CO = BO (Radius of the same circle)
$\angle$ OBC = $\angle$ OCB = 30° (angles opposite to equal sides in a triangle are equal)
In $\triangle$ OBC,
$\angle$ OBC + $\angle$ OCB + $\angle$ BOC = 180° (angle sum property of a triangle)
$\angle$ BOC = 180° – (30° + 30°) = 120°
$\angle$ BOC = 2 $\angle$ BAC (Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)
$\angle BAC = \frac{120^{\circ}}{2}=60^{\circ}$
Consider $\triangle$ AEB, $\triangle$ AEC
AE = AE (common)
$\angle$ AEB = $\angle$ AEC = 90°
BE = EC (given)
$\triangle$ AEB $\cong$ $\triangle$ AEC (SAS congruence)
$\angle$ BAE = $\angle$ CAE (CPCT)
$\angle$ BAE + $\angle$ CAE = $\angle$ BAC = 60°
2 $\angle$ BAE = 60°
$\angle$ BAE = 30° = x
Now, OD is perpendicular to AE
and, CE is perpendicular to AE
Two lines perpendicular to the same line are parallel to each other
So, OD || CE and OC is transversal
$\angle$ DOC = $\angle$ ECO = 30° (alternate interior angles)
Now, the angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle.
Consider the arc CD,
$\angle$ DOC = 2 $\angle$ DBC = 2y
30° = y
y = 15°
Hence, x = 30° and y = 15°

Question:14

In Fig. 10.21, O is the centre of the circle, BD = OD and CD $\perp$ AB. Find $\angle$ CAB.

Answer:

30°
Solution:
Given: In the figure BD = OD, CD $\perp$ AB
In $\triangle$ OBD,
BD = OD (given)
OD = OB (radius of the same circle)
OB = OD = BD
Hence the triangle is equilateral

$\angle$ BOD = $\angle$ OBD = $\angle$ ODB = 60°
Consider $\triangle$ MBC and MBD
MB = MB (common)
$\angle$ CMB = $\angle$ BMD = 90° (given)
CM = MD (perpendicular from the centre on the chord bisects the chord]
$\triangle$ MBC = $\triangle$ MBD (SAS rule)
$\angle$ MBC = $\angle$ MBD (CPCT)
$\angle$ MBC = $\angle$ OBD = 60° ( $\because$ OBD = 60°)
Since AB is the diameter of the circle
$\angle$ ACB = 90° (angle is a semi-circle)
$\angle$ CAB + $\angle$ CBA + $\angle$ ACB = 180° (By angle sum property of a triangle)
Putting the values,
$\angle$ CAB + 60° + 90° = 180°
$\angle$ CAB = 180° – (60° + 90°) = 30°

Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 10:

Topics covered in NCERT exemplar Class 9 Maths solutions chapter 10 deals with the understanding of:

Definition of Circles: It can be considered a regular polygon of infinite sides, where each side length tends to 0.
The related terms of circles such as radius, diameter, chords, arc length, segments, and sectors are discussed in detail.
The three-point property states, “if three points are not Collinear, then a unique circle can be drawn passing through them.”
NCERT exemplar Class 9 Maths solutions chapter 10 explains theorems of a circle such as:
The angle subtended by any chord at the centre will be twice the angle subtended at any point of the circle.
The perpendicular bisector of any chord will pass through the centre of the circle.
If two chords are equidistant from the centre, then their length will be equal.
This chapter discusses cycling quadrilaterals in detail.

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Chapter 1 Number System

Chapter 2 Polynomials

Chapter 3 Coordinate geometry

Chapter 4 Linear equations in Two Variable

Chapter 5 Introduction to Euclid’s Geometry

Chapter 6 Lines and Angles

Chapter 7 Triangles

Chapter 8 Quadrilaterals

Chapter 9 Area of Parallelograms and Triangles

Chapter 11 Constructions

Chapter 12 Heron’s Formula

Chapter 13 Surface Areas and Volumes

Chapter 14 Statistics and Probability

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 10:

These Class 9 Maths NCERT exemplar chapter 10 solutions provide a basic knowledge of circles and are useful for Class 10 and 11 with Mathematics as a subject. The Class 9 Maths NCERT exemplar solutions chapter 10 Circles are highly detailed and provide a step-by-step solution to the problems, which helps clarify the queries of students in an extensive manner. These problems and solutions are enough to solve other books such as Mathematics Pearson Class 9, NCERT Class 9 Maths, RD Sharma Class 9 Maths, RS Aggarwal Class 9 Maths, et cetera.

Students can download/view these solutions by clicking on NCERT exemplar Class 9 Maths solutions chapter 10 pdf download to access the pdf version of the chapter’s solutions while attempting NCERT exemplar Class 9 Maths chapter 10 in an offline environment.

Check the Solutions of Questions Given in the Book

Chapter No.	Chapter Name
Chapter 1	Number Systems
Chapter 2	Polynomials
Chapter 3	Coordinate Geometry
Chapter 4	Linear Equations In Two Variables
Chapter 5	Introduction to Euclid's Geometry
Chapter 6	Lines And Angles
Chapter 7	Triangles
Chapter 8	Quadrilaterals
Chapter 9	Areas of Parallelograms and Triangles
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Heron’s Formula
Chapter 13	Surface Area and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Also, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Can we always form a circle passing through any four points given in a plane?

No, it is not always possible to form a circle passing through four points given in any plane.

For three non-Collinear points, a unique circle can be formed but it is not necessary that it will pass through the fourth point.

2. Three vertices of a right-angle triangle with perpendicular sides a and b lie on a circle, what will be the diameter of that circle?

We know that diameter subtends 90° at any point of the circle.

Therefore, the hypotenuse of the given right angle triangle will be diameter. The length of diameter will be the length of the hypotenuse.

3. What is the longest chord length in a circle of radius R?

The longest chord of any circle will be its diameter. Hence, its length will be twice the radius that is 2R.

4. How many questions are expected from the chapter on Circles in the final exam?

Generally, you can expect 2-3 questions under the segment of concise answer type or long answer type questions from Circles. The NCERT exemplar Class 9 Maths solutions chapter 10 can help you score well in the exam.

Also Read

NCERT Solutions for Class 9 - Subject-wise PDFs Updated for 2024-25

NCERT Solutions for Class 9 Science

NCERT Solutions for Class 9 Maths

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NCERT Class 9 Maths Notes - Download Chapter Wise PDFs

NCERT Class 9 Maths Chapter 8 Notes Quadrilaterals - Download PDF

NCERT Class 9 Maths Chapter 2 Notes Polynomials - Download PDF

NCERT Class 9 Maths Chapter 1 Notes Number System - Download Free PDF

NCERT Class 9 Maths Chapter 5 Notes Introduction To Euclid’s Geometry - Download PDF

NCERT Class 9 Maths Chapter 6 Notes Lines and Angles - Download PDF

NCERT Exemplar Class 9 Maths Solutions - Chapter Wise Problems with Solutions

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 9 Maths Solutions Chapter 10 Circles

Exercise 10.1

Exercise 10.2

Exercise 10.3

Exercise 10.4

Important Topics for NCERT Exemplar Solutions Class 9 Maths Chapter 10:

NCERT Class 9 Exemplar Solutions for Other Subjects:

NCERT Class 9 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 9 Maths Solutions Chapter 10:

Check the Solutions of Questions Given in the Book

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Check NCERT Notes Subject Wise

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