NCERT Exemplar Class 9 Maths Solutions Chapter 10 Circles

Albert Einstein once said, “The circle is the symbol of eternity. It has no beginning and no end.” In the NCERT Exemplar Class 9 Maths Chapter 10 solutions, students will learn about circles in detail and recall the concepts they have learnt in previous classes, such as radius, diameter, chords, tangents, and arcs. After completing the textbook exercises, students can practice questions from class 9 math NCERT exemplar books to deepen their understanding.

Circles are one of the most important and common shapes in our day-to-day lives. NCERT exemplar class 9 maths problems are prepared to test students' higher thinking skills and build strong conceptual understanding, as it is a significant chapter not only in this class but also in higher classes and other competitive exams. These maths exemplar problems class 9 solutions are prepared by experienced Careers360 teachers following the latest CBSE Syllabus. Students can also click on this link to check NCERT book solutions.

Exercise: 10.1 Total Questions: 10 Page number: 99-101

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
Solution:
Given AD = 34 cm and AB = 30 cm
In the figure, draw OL $\perp$ AB

$AL=LB=\frac{1}{2}AB=15cm$
In Right angled $\mathrm{△}$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.

$AC=CB=\frac{1}{2}AB=\frac{1}{2}\times 8=4cm$
Given OA = 5 cm
Using Pythagoras theorem,
$A{O}^2=A{C}^2+O{C}^2$
$\Rightarrow (5{)}^2=(4{)}^2+O{C}^2$
$\Rightarrow 25=16+O{C}^2$
$\Rightarrow O{C}^2=25-16=9$
$\Rightarrow OC=3$ 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:

(C) 10 cm
Solution:
Given: AB = 12 cm, BC = 16 cm and AB is perpendicular to BC

$AB\perp BC$
We know that every angle inscribed in a semicircle is a right angle.
So AOC is the diameter of this circle with O as the centre.
Now, applying Pythagoras theorem
(Hypotenuse)²= (Perpendicular² + (Base)²
$\Rightarrow A{C}^2=A{B}^2+B{C}^2$
$\Rightarrow A{C}^2={12}^2+{16}^2$
$\Rightarrow A{C}^2=144+256$
$\Rightarrow A{C}^2=400$
$\Rightarrow AC=\sqrt{400}$
$\Rightarrow AC=20$ cm
As AC is the diameter, Radius
$=\frac{AC}{2}=\frac{20}{2}=10cm$
Therefore, option (C) is correct.

Question:7

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

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

Answer:

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

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

Question:8

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

Answer:

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

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

Question:9

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

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

Answer:

(C)
Solution:
In $\mathrm{△}$AOB, AO = OB (Radius)
$\mathrm{\angle }$ABO = $\mathrm{\angle }$BAO [Angle opposite to equal sides are equal]

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

Question:10

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

(A) 30º
(B) 45º
(C) 90º
(D) 60º

Answer:

(D)
Solution:

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

Exercise: 10.2 Total Questions: 10 Page number: 101-102

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 a 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 $DN=\frac{1}{2}DC$
$\mathrm{\angle }$ OMA = $\mathrm{\angle }$OND = 90°
OA = OD (Both are radius)
$\mathrm{△}$AOM $\cong$ $\mathrm{△}$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 $\mathrm{\angle }OAB=\mathrm{\angle }OAC.$

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 $\mathrm{\angle }$OAB =$\mathrm{\angle }$OAC
Join OB and OC
Consider $\mathrm{△}$AOC & $\mathrm{△}$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.
$\therefore$ $\mathrm{\angle }$OAB $\ne$ $\mathrm{\angle }$OAC
Therefore, the given statement is false.

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

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.
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
$\Rightarrow$ $\mathrm{\angle }$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
$\mathrm{\angle }ACB=\left(\frac{1}{2}\right)\mathrm{\angle }AOB$
$\Rightarrow \mathrm{\angle }ACB=\left(\frac{1}{2}\right){180}^{\circ }$
$\Rightarrow \mathrm{\angle }ACB={90}^{\circ }$
$\Rightarrow$ ACB is a Right angle.
So, $\mathrm{\angle }ABC$ follows Pythagoras theorem,
i.e., AB² = AC² + BC²
Therefore, the given statement is true.

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 $\mathrm{\angle }BAC={30}^{\circ }and\mathrm{\angle }BDC={60}^{\circ }$, 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, $\mathrm{\angle }BDC={60}^{\circ }$
At any other point on the circle,
$\mathrm{\angle }BAC={30}^{\circ }=\frac{1}{2}({60}^{\circ })$
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 $\mathrm{\angle }BAC={45}^{\circ }$ and $\mathrm{\angle }BDC={45}^{\circ }$, then A, B, C, and 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 $\mathrm{\angle }BAC={45}^{\circ }=\mathrm{\angle }BDC$
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 $\mathrm{\angle }$ADC = 120°, then $\mathrm{\angle }$CAB = 30°.

Answer:

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

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

Exercise: 10.3 Total Questions: 20 Page number: 103-105

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 its 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 $\mathrm{△}APM$ and $\mathrm{△}BPM$
AM = MB (PM is the perpendicular bisector of AB)
PM = PM (Common)
$\mathrm{\angle }PMA=\mathrm{\angle }PMB$ (90° each)
$\therefore \mathrm{△}APM\cong \mathrm{△}BPM$ (SAS congruence)
PA = PB (by CPCT)
If 2 chords of a circle are equal, then the 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, and C are 3 points on a circle
Construction $\Rightarrow$ Draw a 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 $\mathrm{△}$OEA and $\mathrm{△}$OEB
AE = BE [OL is ^ar bisector of AB]
OE = OE [Common side]
$\mathrm{\angle }$AEO = $\mathrm{\angle }$BEO [each 90°]
$\therefore$ $\mathrm{△}$OEA $\cong$ $\mathrm{△}$DEB [By SAS congruence Rule]
OA = OC
Similarly, $\mathrm{△}$OFA $\cong$ $\mathrm{△}$OFC [by SAS congruence]
$\therefore$ OB = OC = OA = t [Say]
SO, we draw a perpendicular from O to the BC on join
In $\mathrm{△}$OMB and $\mathrm{△}$OMC
OB = OC [Proved Above]
OM = OM [Common side]
$\mathrm{\angle }$OMB = $\mathrm{\angle }$OMC [90° each]
$\therefore$ $\mathrm{△}$OMB $\cong$ $\mathrm{\angle }$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, and 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: The Centre lies on the bisector of the $\mathrm{\angle }$BAC.
Construction: Join BC

Proof: In $\mathrm{△}$AOB and $\mathrm{△}$AOC
AB = AC (given)
$\mathrm{\angle }$BAO = $\mathrm{\angle }$CAO (given)
AO = AO (common)
$\therefore$ $\mathrm{△}$AOB $\cong$ $\mathrm{△}$AOC (by SAS congruence)
$\Rightarrow$ BO = CO (CPCT) …(i)
and $\mathrm{\angle }$AOB = $\mathrm{\angle }$AOC (CPCT)
$\mathrm{\angle }$AOB + $\mathrm{\angle }$AOC = 180° (Linear pair)
$\therefore$ 2 $\mathrm{\angle }$AOB = 180° ($\mathrm{\angle }$APB = $\mathrm{\angle }$APC)
$\Rightarrow$ $\mathrm{\angle }$AOB = 90°
$\therefore$ AO is a 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:
$\mathrm{\angle }$M = $\mathrm{\angle }$N = 90° (Segment joining the midpoint of the chord to the centre)
Consider, DOMB and DONC
$\mathrm{\angle }$BOM = $\mathrm{\angle }$CON (Vertically opposite angles)
OB = OC (Radii of the same circle)
$\mathrm{\angle }$OMB = $\mathrm{\angle }$ONC (90° each)
$\therefore$ $\mathrm{△}$OMB$\cong$ $\mathrm{\angle }$ONC (AAS congruence)
$\therefore$ $\mathrm{\angle }$OBM = $\mathrm{\angle }$OCN (CPCT)
$\therefore$ $\mathrm{\angle }$ABC = $\mathrm{\angle }$BCD
We can see that $\mathrm{\angle }$ABC and $\mathrm{\angle }$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 $\mathrm{\angle }CBD+\mathrm{\angle }CDB=\frac{1}{2}\mathrm{\angle }BAD.$

Answer:

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

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

Question:7

O is the circumcentre of the triangle ABC, and D is the mid-point of the base BC. Prove that $\mathrm{\angle }$BOD = $\mathrm{\angle }$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 to BC.

In the right triangles OBD and OCD.
We have OB = OC (Radii of the same circle)
OD = OD (common)
$\mathrm{\angle }$ODB = $\mathrm{\angle }$ODC (90^o)
$\therefore$ $\mathrm{△}$OBD $\cong$ $\mathrm{△}$OCD (R.H.S. congruence)
$\Rightarrow$ $\mathrm{\angle }$BOD = $\mathrm{\angle }$COD (CPCT)
$\Rightarrow \mathrm{\angle }BOD=\frac{1}{2}\mathrm{\angle }BOC$ … (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,
$\Rightarrow \mathrm{\angle }BAC=\frac{1}{2}\mathrm{\angle }BOC$ … (2)
From (1) and (2), we have
$\therefore$ $\mathrm{\angle }$BOD = $\mathrm{\angle }$BAC
Hence Proved.

Question:8

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

Answer:

Given that on a common hypotenuse AB, two right triangles ACB and ADB are situated on opposite sides.
Let $\mathrm{△}$ACB and $\mathrm{△}$ADB be two right-angled triangles with common hypotenuse AB.
To prove: $\mathrm{\angle }$BAC = $\mathrm{\angle }$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 the centre and radius equal to OA.
We know that angles in the same segment of a circle are equal
From the figure, $\mathrm{\angle }$BAC and $\mathrm{\angle }$BDC are angles of the same segment BC.
$\therefore$ $\mathrm{\angle }$BAC = $\mathrm{\angle }$BDC
Hence proved.

Question:9

Two chords AB and AC of a circle subtend angles equal to 90º and 150º, respectively, at the centre. Find $\mathrm{\angle }$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 subtend angles equal to 90º and 150º, respectively, at the centre.
AB and AC lie on the opposite sides of the centre
In $\mathrm{△}$BOA
OB = OA (Both equal to radius)
$\mathrm{\angle }$OAB = $\mathrm{\angle }$OBA … (i) (angles opposite to equal sides in a triangle are equal)

In $\mathrm{△}$OAB,
$\mathrm{\angle }$OBA + $\mathrm{\angle }$BAO + $\mathrm{\angle }$AOB = 180° (angle sum property of a triangle)
$\Rightarrow$ $\mathrm{\angle }$OAB + $\mathrm{\angle }$OAB + 90° = 180° [From equation (1)]
$\Rightarrow$ 2$\mathrm{\angle }$OAB = 180° – 90°
$\Rightarrow$ $\mathrm{\angle }$OAB = $\frac{{90}^{\circ }}{2}$= 45°
Now in $\mathrm{△}$AOC, AO = OC (Both equal to radius)
$\mathrm{\angle }$OCA = $\mathrm{\angle }$OAC … (ii) (angles opposite to equal sides in a triangle are equal)
Also,
$\mathrm{\angle }$AOC + $\mathrm{\angle }$OAC + $\mathrm{\angle }$OCA = 180° (angle sum property of a triangle)
150° + 2 $\mathrm{\angle }$OAC = 180°
2$\mathrm{\angle }$OAC = 180° – 150°
2$\mathrm{\angle }$OAC = 30°
$\mathrm{\angle }$OAC = 15°
$\mathrm{\angle }$BAC = $\mathrm{\angle }$OAB +$\mathrm{\angle }$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 $\mathrm{△}$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]
$\Rightarrow$ $\mathrm{\angle }$PQR = $\mathrm{\angle }$PRQ … (1) (angles opposite to equal sides in a triangle are equal)
Now, RQ || MN and PQ is a transversal
$\Rightarrow$ $\mathrm{\angle }$PNM = $\mathrm{\angle }$PQR … (2) (Corresponding angles)
On adding $\mathrm{\angle }$MNQ to both sides in equation (2), we get:
$\mathrm{\angle }$PNM + $\mathrm{\angle }$MNQ = $\mathrm{\angle }$PQR + $\mathrm{\angle }$MNQ
180° = $\mathrm{\angle }$PQR + $\mathrm{\angle }$MNQ ($\mathrm{\angle }$PNM + $\mathrm{\angle }$MNQ forms a linear pair)
$\Rightarrow$ $\mathrm{\angle }$MNQ + $\mathrm{\angle }$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 in the same segment are equal.
Consider segment CD, $\mathrm{\angle }$DAC = $\mathrm{\angle }$DBC
So we can write, $\mathrm{\angle }$DAO = $\mathrm{\angle }$CBO …(i)
Consider segment AB, $\mathrm{\angle }$ADB = $\mathrm{\angle }$ACD
So we can write, $\mathrm{\angle }$ADO = $\mathrm{\angle }$BCO…(ii)
In $\mathrm{△}$AOD & $\mathrm{△}$BOC,
$\mathrm{\angle }$DAO = $\mathrm{\angle }$CBO (from (i))
$\mathrm{\angle }$ADO = $\mathrm{\angle }$BCO (from (ii))
$\mathrm{\angle }$AOD = $\mathrm{\angle }$BOC (vertically opposite angles)
$\mathrm{△}$AOD $\cong$ $\mathrm{△}$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 $\mathrm{\angle }$OBC + $\mathrm{\angle }$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
$\Rightarrow$ $\mathrm{\angle }$OBC = $\mathrm{\angle }$OCB = Let x
In DOBC,
$\therefore$ x + x + $\mathrm{\angle }$BOC = 180° (angle sum property of a triangle)
2x + $\mathrm{\angle }$BOC = 180°
$\mathrm{\angle }$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,
$\mathrm{\angle }$BOC = 2 $\mathrm{\angle }$BAC
Putting the value of $\mathrm{\angle }$BOC
180° – 2x = 2 $\mathrm{\angle }$BAC
90° – x = $\mathrm{\angle }$BAC
We have to find $\mathrm{\angle }$OBC + $\mathrm{\angle }$BAC
$\therefore$ $\mathrm{\angle }$ÐBAC +$\mathrm{\angle }$OBC = (90° – x) + x
$\mathrm{\angle }$BAC + $\mathrm{\angle }$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 a 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 $\mathrm{△}$OAB,
AB = OA = OB = radius
Hence, it is an equilateral triangle. So all angles are 60o
$\mathrm{\angle }$OAB =$\mathrm{\angle }$ABO = $\mathrm{\angle }$BOA = 60^o
Let point D be a point on the major arc.
Considering the arc AB, 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
So,
$\mathrm{\angle }ADB=\frac{1}{2}\mathrm{\angle }AOB=\frac{1}{2}({60}^{\circ })={30}^{\circ }$

Question:17

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

Answer:

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

$\Rightarrow$ $\mathrm{\angle }$ADC + $\mathrm{\angle }$ABC = 180°
$\Rightarrow$ 130° + $\mathrm{\angle }$ABC = 180°
$\Rightarrow$ $\mathrm{\angle }$ABC = 180° – 130°
$\Rightarrow$ $\mathrm{\angle }$ABC = 50° …(i)
AB is the diameter of a circle (given)
$\therefore$ ACB = 90° (angle in a semicircle is 90º)
In $\mathrm{△}$ABC,
$\Rightarrow$ $\mathrm{\angle }$ABC + $\mathrm{\angle }$ACB + $\mathrm{\angle }$BAC = 180°(angle sum property of a triangle)
Putting all the values,
$\Rightarrow$ 50° + 90° + $\mathrm{\angle }$BAC = 180°
$\Rightarrow$ $\mathrm{\angle }$BAC = 180° –50° – 90°
$\therefore$ $\mathrm{\angle }$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 $\mathrm{△}$OPB,
BM = MP … (i) (perpendicular from the centre of the circle bisects the chord)
Similarly in $\mathrm{△}$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 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 $\mathrm{\angle }$ACD + $\mathrm{\angle }$BED.

Answer:

270°
Solution:
Join AE

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

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

Question:3

If P, Q and R are the midpoints 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 $\mathrm{△}$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
$\mathrm{\angle }$RAQ = $\mathrm{\angle }$RPQ … (1) (opposite angles of a ||gm are equal)
Now, in right-angled $\mathrm{△}$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
$\mathrm{\angle }$RAD = $\mathrm{\angle }$RDA … (2) (angles opposite to equal sides in a triangle are equal)
Similarly, in$\mathrm{△}$ADQ
$\mathrm{\angle }$DAQ = $\mathrm{\angle }$ADQ … (3)
Adding (2) and (3),
$\mathrm{\angle }$RAD + $\mathrm{\angle }$DAQ = $\mathrm{\angle }$RDA +$\mathrm{\angle }$ADQ
$\mathrm{\angle }$RAQ = vRDQ
From (1), $\mathrm{\angle }$RAQ =$\mathrm{\angle }$RPQ
Hence, we can see if we consider arc RQ of a circle then $\mathrm{\angle }$RAQ and$\mathrm{\angle }$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 the sum of opposite angles of a cyclic quadrilateral is 180°
$\mathrm{\angle }$A + $\mathrm{\angle }$PQB = 180°
Also, $\mathrm{\angle }$PQB + $\mathrm{\angle }$PQC = 180° (linear pair)
Comparing the above two equations, we get
$\mathrm{\angle }$PQC = $\mathrm{\angle }$A
Let $\mathrm{\angle }$PQC = $\mathrm{\angle }$1
Then, $\mathrm{\angle }$A = $\mathrm{\angle }$1
Also, $\mathrm{\angle }$A = $\mathrm{\angle }$C (Opposite angles of a parallelogram are equal)
So, $\mathrm{\angle }$1 = $\mathrm{\angle }$C
But, $\mathrm{\angle }$C + $\mathrm{\angle }$D = 180° (As ABCD is a parallelogram)
Hence, $\mathrm{\angle }$1 + $\mathrm{\angle }$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 the 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 the perpendicular bisector of the opposite side intersect on the circumcircle of the triangle.

Consider DABC
Let the angle bisector of angle A intersect the circumcircle of triangle ABC at point D.
Join DC and DB
$\mathrm{\angle }$BCD = $\mathrm{\angle }$BAD (Angle in the same segment are equal)
$\mathrm{\angle }$BCD = $\mathrm{\angle }$BAD = $\frac{1}{2}$ $\mathrm{\angle }$A … (1) (AD is bisector of ÐCAB)
Similarly,
$\mathrm{\angle }$DAC = $\mathrm{\angle }$DBC (Angle in the same segment are equal)
$\mathrm{\angle }$DBC =$\mathrm{\angle }$DAC = $\frac{1}{2}$ $\mathrm{\angle }$ÐA … (2) (AD is bisector of ÐCAB)
From (1) and (2), we have
$\mathrm{\angle }$DBC =$\mathrm{\angle }$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 = a 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 the diameter of the 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 $\mathrm{\angle }$BPC.

Answer:

Let DABC be 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,
$\mathrm{\angle }$AOB = $\mathrm{\angle }$AOC =$\mathrm{\angle }$BOC … (i)
Consider the arc AB, $\mathrm{\angle }$AOB and $\mathrm{\angle }$APB are the angles subtended by AB at the centre and at a remaining part of the circle
Therefore
$\mathrm{\angle }APB=\frac{1}{2}\mathrm{\angle }AOB$ … (ii)
Similarly,
$\mathrm{\angle }APC=\frac{1}{2}\mathrm{\angle }AOC$ … (iii)
Using (i), (ii) and (iii), we have
$\mathrm{\angle }$APB = $\mathrm{\angle }$APC
Hence, PA is the angle bisector of $\mathrm{\angle }$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 $\mathrm{\angle }$AEC =$\frac{1}{2}$ (Angle subtended by arc CXA at the 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:
$\mathrm{\angle }$AEC =0.5 (Angle subtended by arc CXA at centre + angle subtended by arc DYB at the centre)
$\mathrm{\angle }AEC=0.5(\mathrm{\angle }AOC+\mathrm{\angle }DOB)$

Proof:
Extend the lines DO to L and BO to point X on the circle. Join AC.
Now, 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.
For arc LC,
$\mathrm{\angle }$1 = 2$\mathrm{\angle }$6 … (i)
For arc XA,
$\mathrm{\angle }$3 = 2$\mathrm{\angle }$7 … (ii)
Consider $\mathrm{△}$OAC
OC = OA (Radius of the same circle)
$\mathrm{\angle }$OCA = $\mathrm{\angle }$4 … (iii) (angles opposite to equal sides in a triangle are equal)
$\mathrm{\angle }$AOC + $\mathrm{\angle }$OCA + $\mathrm{\angle }$4 = 180° (angle sum property of a triangle)
$\mathrm{\angle }$AOC + $\mathrm{\angle }$4 + $\mathrm{\angle }$4 = 180°
$\mathrm{\angle }$AOC = 180° – 2 $\mathrm{\angle }$4 …(iv)
Consider $\mathrm{△}$DCO
OD = OC (Radius of the same circle)
$\mathrm{\angle }$OCD = $\mathrm{\angle }$6 … (v) (angles opposite to equal sides in a triangle are equal)
Consider $\mathrm{△}$AEC
$\mathrm{\angle }$AEC + $\mathrm{\angle }$ECA + $\mathrm{\angle }$CAE = 180° (angle sum property of a triangle)
$\mathrm{\angle }$AEC = 180° – ($\mathrm{\angle }$ECA + $\mathrm{\angle }$CAE)
$\mathrm{\angle }$AEC = 180° – [($\mathrm{\angle }$ECO + $\mathrm{\angle }$OCA) + ($\mathrm{\angle }$CAO + OAE)]
$\mathrm{\angle }$AEC = 180° – ($\mathrm{\angle }$6 + $\mathrm{\angle }$4 + $\mathrm{\angle }$4 + $\mathrm{\angle }$5) [from (iii) and (v)]
$\mathrm{\angle }$AEC = 180° – (2$\mathrm{\angle }$4 + $\mathrm{\angle }$5 + $\mathrm{\angle }$6)
$\mathrm{\angle }$AEC = 180° – 2$\mathrm{\angle }$4 - $\mathrm{\angle }$5 - $\mathrm{\angle }$6
From (iv), $\mathrm{\angle }$AOC = 180° – 2 $\mathrm{\angle }$4
$\mathrm{\angle }$AEC = $\mathrm{\angle }$AOC - $\mathrm{\angle }$5 - $\mathrm{\angle }$6
Also, $\mathrm{\angle }$AOC = $\mathrm{\angle }$1 + $\mathrm{\angle }$2 + $\mathrm{\angle }$3
So, $\mathrm{\angle }$AEC = $\mathrm{\angle }$1 + $\mathrm{\angle }$2 + $\mathrm{\angle }$3 - $\mathrm{\angle }$5 - $\mathrm{\angle }$6
In $\mathrm{△}$AOB,
OA = OB (Radius of the same circle)
$\mathrm{\angle }$5 =$\mathrm{\angle }$7 (angles opposite to equal sides in a triangle are equal)
$\mathrm{\angle }$AEC = $\mathrm{\angle }$1 +$\mathrm{\angle }$2 + $\mathrm{\angle }$3 - $\mathrm{\angle }$7 - $\mathrm{\angle }$6
From (ii), $\mathrm{\angle }$3 = 2$\mathrm{\angle }$7
$\mathrm{\angle }$AEC = $\mathrm{\angle }$1 + $\mathrm{\angle }$2 + $\mathrm{\angle }$3 - 0.5<3 - $\mathrm{\angle }$6
From (i), $\mathrm{\angle }$1 = 2$\mathrm{\angle }$6
$\mathrm{\angle }$AEC = $\mathrm{\angle }$1 + $\mathrm{\angle }$2 + $\mathrm{\angle }$3 - 0.5$\mathrm{\angle }$3 - 0.5$\mathrm{\angle }$1
$\mathrm{\angle }$AEC = 0.5$\mathrm{\angle }$1 + 0.5$\mathrm{\angle }$3 + $\mathrm{\angle }$2
$\mathrm{\angle }AEC=\frac{1}{2}\mathrm{\angle }1+\frac{1}{2}\mathrm{\angle }3+\frac{1}{2}\mathrm{\angle }2+\frac{1}{2}\mathrm{\angle }2$
$\mathrm{\angle }$2 = $\mathrm{\angle }$8 (vertically opposite angles)
$\mathrm{\angle }AEC=\frac{1}{2}\mathrm{\angle }1+\frac{1}{2}\mathrm{\angle }3+\frac{1}{2}\mathrm{\angle }2+\frac{1}{2}\mathrm{\angle }8$
$\mathrm{\angle }$AEC = 0.5 ($\mathrm{\angle }$1 + $\mathrm{\angle }$2 + $\mathrm{\angle }$3 + $\mathrm{\angle }$8)
$\mathrm{\angle }AEC=0.5(\mathrm{\angle }AOC+\mathrm{\angle }DOB)$
$\mathrm{\angle }$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 $\mathrm{\angle }$ADC and $\mathrm{\angle }$ABC, respectively.
To Prove: PQ is the diameter of the circle

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

Question:10

A circle has a radius of $\sqrt{2}$ 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 the 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 segments.

To Prove: The angle subtended by the chord at a point in the major segment is 45º.
This means we have to prove that $\mathrm{\angle }$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 $\mathrm{\angle }$AOB = 90°
Now 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
$$\begin{gathered} \because \mathrm{\angle }APB=\mathrm{\angle }AOB \\ \Rightarrow \mathrm{\angle }APB={45}^{\circ } \end{gathered}$$
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 equal 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 $\mathrm{△}$OLP and $\mathrm{△}$OMP
OL = OM
$\mathrm{\angle }$OLP = $\mathrm{\angle }$OMP (both are 90o)
OP = OP
$\mathrm{△}$OLP $\cong$ $\mathrm{△}$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 distances 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$\mathrm{△}$OAN, applying Pythagoras theorem
AO² = (AN)² + (NO)²
AO² = (x)² + p² ….(ii)
From equations (i) and (ii)
$$\begin{gathered} {\left(\frac{x}{2}\right)}^2+{\left(q\right)}^2=x^2+{\left(p\right)}^2 \\ \frac{x^2}{4}+q^2=x^2+p^2 \end{gathered}$$
x² + 4q² = 4x² + 4p² [Multiply both sides by 4]
4q² = 3x² + 4p²
4q² = p² + 3 (x² + p²)
4q² = p² + 3r² (In right angle $\mathrm{△}$OAN, r² = x² + p²)
Hence Proved.

Question:13

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

Answer:

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

CO = BO (Radius of the same circle)
$\mathrm{\angle }$OBC =$\mathrm{\angle }$OCB = 30° (angles opposite to equal sides in a triangle are equal)
In $\mathrm{△}$OBC,
$\mathrm{\angle }$OBC + $\mathrm{\angle }$OCB + $\mathrm{\angle }$BOC = 180° (angle sum property of a triangle)
$\mathrm{\angle }$BOC = 180° – (30° + 30°) = 120°
$\mathrm{\angle }$BOC = 2$\mathrm{\angle }$BAC (Angle subtended at the centre by an arc is twice the angle subtended by it at any part of the circle)
$\mathrm{\angle }BAC=\frac{{120}^{\circ }}{2}={60}^{\circ }$
Consider $\mathrm{△}$AEB, $\mathrm{△}$AEC
AE = AE (common)
$\mathrm{\angle }$AEB =$\mathrm{\angle }$AEC = 90°
BE = EC (given)
$\mathrm{△}$AEB $\cong$ $\mathrm{△}$AEC (SAS congruence)
$\mathrm{\angle }$BAE =$\mathrm{\angle }$CAE (CPCT)
$\mathrm{\angle }$BAE + $\mathrm{\angle }$CAE = $\mathrm{\angle }$BAC = 60°
2$\mathrm{\angle }$BAE = 60°
$\mathrm{\angle }$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
$\mathrm{\angle }$DOC = $\mathrm{\angle }$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,
$\mathrm{\angle }$DOC = 2$\mathrm{\angle }$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 $\mathrm{\angle }$CAB.

Answer:

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

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

NCERT Class 9 Exemplar Solutions for Other Subjects

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Importance of NCERT Exemplar Solutions Class 9 Maths Chapter 10

NCERT Exemplar Class 9 Maths chapter 10 solutions are highly detailed and give a step-by-step solution to the problems, extensively clarifying students' queries. Here are some more important features of these solutions.

• These NCERT exemplar problems contain MCQ, Short answer, long answer and assertion type questions, that will help students build a strong core in this chapter.
• These solutions are not only ideal for class 9 board exams but also in exams like NTSE, Olympiads, JEE, NEET, etc.
• Some of the questions include real-life scenarios and practical uses of circle geometry. Students can relate these questions to real-life situations.
• Finding solutions to these questions will give students well-deserved confidence and boost their morale before the exam.
• Consistent practice of these Exemplar problems will increase the problem-solving speed and reduce the errors.

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