Theorem on Chords MCQ Quiz - Objective Question with Answer for Theorem on Chords - Download Free PDF

Last updated on May 22, 2025

Latest Theorem on Chords MCQ Objective Questions

Theorem on Chords Question 1:

In the given figure, the chords AB and CD are parallel and are of 36 cm and 48 cm, respectively. The distance between them is 42 cm. The diameter of the circle is _______.

qImage67c2acc9672bc6a1b3178366

  1. 82 cm
  2. 60 cm
  3. 72 cm
  4. 90 cm

Answer (Detailed Solution Below)

Option 2 : 60 cm

Theorem on Chords Question 1 Detailed Solution

Given:

Parallel chords AB = 36 cm

Parallel chords CD = 48 cm

Distance between the chords = 42 cm

Formula Used:

The perpendicular from the center of a circle to a chord bisects the chord.

Pythagoras theorem: (radius)2 = (half of chord length)2 + (distance from center)2

Diameter = 2 × radius

Calculation:

qImage6829af6d7d77ec38bedde29c

Let the radius of the circle be r.

Let the distance of chord AB from the center be x cm.

Then the distance of chord CD from the center will be (42 - x) cm.

For chord AB:

r2 = (36/2)2 + x2

⇒ r2 = 182 + x2

⇒ r2 = 324 + x2 (Equation 1)

For chord CD:

r2 = (48/2)2 + (42 - x)2

⇒ r2 = 242 + (42 - x)2

⇒ r2 = 576 + (1764 - 84x + x2)

⇒ r2 = 576 + 1764 - 84x + x2

⇒ r2 = 2340 - 84x + x2 (Equation 2)

Equating Equation 1 and Equation 2:

324 + x2 = 2340 - 84x + x2

⇒ 324 = 2340 - 84x

⇒ 84x = 2340 - 324

⇒ 84x = 2016

⇒ x = 2016 / 84

⇒ x = 24 cm

Substitute the value of x in Equation 1:

r2 = 324 + (24)2

⇒ r2 = 324 + 576

⇒ r2 = 900

⇒ r = \(\sqrt{900}\)

⇒ r = 30 cm

Diameter = 2 × radius = 2 × 30 = 60 cm

∴ The diameter of the circle is 60 cm.

Theorem on Chords Question 2:

In a circle with center O, an arc ABC subtends an angle of 138º at the center of the circle. The chord AB is produced to a point P. Then, the measure of ∠CBP is:

  1. 108º
  2. 42º
  3. 111º
  4. 69º

Answer (Detailed Solution Below)

Option 4 : 69º

Theorem on Chords Question 2 Detailed Solution

Given:

The angle subtended by arc ABC at the center of the circle = 138°.

Chord AB is produced to a point P, and we need to find ∠CBP.

Formula used:

The angle subtended by an arc at the center of a circle is twice the angle subtended by the same arc at the circle's circumference.

Calculation:

F1 Priyaa SSC 14 1 25 D1

Let ∠AQC be the angle subtended by arc ABC at the circle's circumference.

⇒ ∠AOC = 2 × ∠AQC

⇒ ∠AQC = 1/2 × ∠AOC = 1/2 × 138° = 69°

Now, in the cyclic quadrilateral ABQC, the exterior angle ∠CBP is equal to the interior opposite angle ∠AQC.

⇒ ∠CBP = ∠AQC = 69°

∴ The measure of ∠CBP is 69°.

Theorem on Chords Question 3:

The two circles intersect at two points P and Q. PR and PS are diameters of the two circles. What is ∠PQR ?

  1. 90º
  2. 45º
  3. 60º

Answer (Detailed Solution Below)

Option 1 : 90º

Theorem on Chords Question 3 Detailed Solution

Given:

The two circles intersect at two points P and Q.

PR and PS are diameters of the two circles.

Formula Used:

In a circle, the angle subtended by the diameter on the circumference is 90º.

Calculation:

11-4-2025 IMG-651 -(1)

Since PR and PS are diameters of the two circles, they subtend a right angle at any point on the circumference.

Thus, ∠PQR is a right angle.

∠PQR = 90º

The correct answer is option 1.

Theorem on Chords Question 4:

A line cuts two concentric circles. The length of chords formed by this line on the circles is 6 cm and 18 cm. Find the difference in the squares of the radii of two circles.

  1. 90
  2. 120
  3. 60
  4. 72

Answer (Detailed Solution Below)

Option 4 : 72

Theorem on Chords Question 4 Detailed Solution

Given:

Length of chord on both circles is 6 cm and 18 cm.

Concept used:

Pythagoras theorem.

Calculation:

19-4-2025 IMG-816 -1

Let the radii of the smaller circle and the bigger circle be r1 and r2.

In ΔOAC 

OC2 = OA2 + AC2

⇒ r12 - AC2 = OA2    ....(1)

In ΔOAB

OB2 = OA2 + AB2

⇒ r22 - AB2 = OA2    ....(2)

From the first equation and the second equation,

r12 - 9 = r22 - 81

r22-  r12 = 81 - 9 = 72

∴ The required difference in the square is 72 cm2.

Theorem on Chords Question 5:

The radii of two concentric circles are 26 cm and 10 cm. If the chord of the greater circle is a tangent to the smaller circle, then find the length of that chord.

  1. 38.7 cm
  2. 28.7 cm
  3. 48 cm
  4. 28 cm

Answer (Detailed Solution Below)

Option 3 : 48 cm

Theorem on Chords Question 5 Detailed Solution

Given:

Radii of two concentric circles are 26 cm and 10 cm.

The chord of the greater circle is a tangent to the smaller circle.

Formula Used:

Using Pythagoras theorem in the right triangle formed by the radius of the greater circle, the radius of the smaller circle, and half of the chord length:

\(R^2 = r^2 + \left(\frac{L}{2}\right)^2\)

Where:

R = radius of the greater circle = 26 cm

r = radius of the smaller circle = 10 cm

L = length of the chord

Calculation:

Backup of F1 Ajeet  10-12-2024 d3

262 = 102\(\left(\frac{L}{2}\right)^2\)

676 = 100 + \(\left(\frac{L}{2}\right)^2\)

676 - 100 = \(\left(\frac{L}{2}\right)^2\)

576 = \(\left(\frac{L}{2}\right)^2\)

\(\left(\frac{L}{2}\right)\) = √576

\(\left(\frac{L}{2}\right)\) = 24

L = 24 × 2

L = 48 cm

The length of the chord is 48 cm.

Top Theorem on Chords MCQ Objective Questions

ΔPQR is an equilateral triangle inscribed in a circle. S is any point on the arc QR. Find the measure of ∠PSQ. 

  1. 30°
  2. 60°
  3. 90°
  4. 45°

Answer (Detailed Solution Below)

Option 2 : 60°

Theorem on Chords Question 6 Detailed Solution

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Given:

ΔPQR is an equilateral triangle inscribed in a circle.

S is any point on the arc QR.

Concept used:

Each of the angles of an equilateral triangle is equal and 60° 

The angles subtended by an arc in the same segment of a circle are equal.

Calculation:

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Both ∠PRQ and ∠PSQ are in the subtended by the arc PQ in the same segment, so

∠PRQ = ∠PSQ = 60°

 ∴ The correct option is 2

Two equal circles of radius 8 cm intersect each other in such a way that each passes through the centre of the other. The length of the common chord is:

  1. \(8\sqrt3\) cm
  2. \(\sqrt 3\) cm
  3. \(2\sqrt3\) cm
  4. \(4\sqrt3\) cm

Answer (Detailed Solution Below)

Option 1 : \(8\sqrt3\) cm

Theorem on Chords Question 7 Detailed Solution

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Given:

The radius of the circles is 8 cm

Calculation:

F1 ArunK Madhuri 11.03.2022 D6

According to the diagram,

AD = DB

O1O2 = 8

Again O1A = O2A = 8 [Radius of the circle]

∠ADO1 = 90°

O1D = O2D = 4

AD = √(64 - 16)

⇒ √48 = 4√3

AB = 2 × 4√3 = 8√3

∴ The length of the common chord is 8√3 cm

The length of the common chord of two circles of radii 15 cm and 13 cm, whose centres are 14 cm apart, is:

  1. 14 cm
  2. 12 cm
  3. 15 cm
  4. 24 cm

Answer (Detailed Solution Below)

Option 4 : 24 cm

Theorem on Chords Question 8 Detailed Solution

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Given:

Radii of circles are 13 cm and 15 cm 

Distance between centres = 14 cm 

Concept Used:

The line joining the centers of two circles is the perpendicular bisector of the common chord of the two circles.

⇒ AB = 2AM

Calculation:

F1 SSC  PriyaSS 14-2-24 D58

From the figure, P and Q are the centers of two circles of radius 15 cm and 13cm respectively and AB is the common chord

In ΔAMP, PA = 15 cm

By Pythagoras theorem,

AM2 = PA2 - PM2

⇒ AM2 = 152 - PM2

⇒ AM2 = 225 - PM2     -----(1)

In ΔAMQ, QA = 13 cm

By Pythagoras theorem,

AM2 = QA2 - MQ2

⇒ AM2 = 132 - MQ2

⇒ AM2 = 169 - MQ2     -----(2)

From (1) and (2)

225 - PM169 - MQ2 

⇒ PM - MQ2 = 56

⇒ (PM + MQ) (PM - MQ) = 56

According to the question,

(PM + MQ) = PQ = Distance between centres = 14 cm ----(3)

⇒ 14 (PM - MQ) = 56

⇒ (PM - MQ) = 56/14 = 4                                               ----(4)

From (3) and (4)

PM = 9 cm

Putting value of PM in (1)

AM2 = 225 - 92

⇒ AM = 225 - 81

⇒ AM = 144

⇒ AM = 12 cm

Thus, length of common chord = AB = 2AM = 24 cm

∴ The length of the common chord is 24 cm.

AB = 28 cm and CD = 22 cm are two parallel chords on the same side of the centre of a circle. The distance between them is 4 cm. The radius of the circle is ________. (Consider up to two decimals) 

  1. 15.20 cm
  2. 14.82 cm
  3. 15.82 cm
  4. 13.20 cm

Answer (Detailed Solution Below)

Option 3 : 15.82 cm

Theorem on Chords Question 9 Detailed Solution

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Given:

F1 Savita SSC 29-8-24 D1

AB = 28 cm and CD = 22 cm are two parallel chords on the same side of the centre of a circle.

The distance between them is 4 cm.

Calculation:

Let the OE is x, EF = 4 cm

As per the question,

OA = OC = radius of the circle

⇒ 142 + x2 = (x+4)2 + 112

⇒ 196 + x2 = x2 + 16 + 8x + 121

⇒ 8x = 59

x = 59 / 8 

The radius will be,

√(142 + (\({59\over8}\))2 = 15.82 cm

∴ The correct option is 3

Two concentric circles are of radii 10 cm and 6 cm. Find the length of the chord of the larger circle which touches the smaller circle.

  1. 8 cm
  2. 16 cm
  3. 12 cm
  4. 9 cm

Answer (Detailed Solution Below)

Option 2 : 16 cm

Theorem on Chords Question 10 Detailed Solution

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Calculation

gate SSC EE 141

The perpendicular bisector of any chord passes through the centre of the circle,

OP ⊥ AB

In OPB, ∠OPB = 900

PB2 = OB2 – OP2

⇒ PB2 = 102 – 62

⇒ PB2 = 64

⇒ PB2 = 82

⇒ PB = 8 cm

OP bisects AB

⇒ AP = PB = 8 cm

∴ AB = AP + PB = 8 + 8 = 16 cm

The length of AB is 16 cm.

In the given figure, O is the centre of the circle and ∠AOB = 130°. Find ∠APB.

F1 SSC Arbaz 19-1-24 D1

  1. 110°
  2. 115°
  3. 100°
  4. 95°

Answer (Detailed Solution Below)

Option 2 : 115°

Theorem on Chords Question 11 Detailed Solution

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Concept used:

F1 SSC Arbaz 19-1-24 D2

According to the given picture, ∠AOB = 2∠ACB

Again, ∠ACB + ∠APB = 180°

Calculation:

F1 SSC Arbaz 19-1-24 D2

According to the given concept,

∠AOB = 2∠ACB

⇒ ∠ACB = 130/2 = 65° 

Again,  ∠ACB + ∠APB = 180°

⇒ ∠APB = 180° - 65° = 115°

∴ The correct answer is 115°

Two concentric circles of radii 15 cm and 13 cm are given. Find the length of the chord of the larger circle which touches the smaller circle. 

  1. 22√7
  2. \(8\sqrt{14}\)
  3. \(4\sqrt{14}\)
  4. 12√7

Answer (Detailed Solution Below)

Option 3 : \(4\sqrt{14}\)

Theorem on Chords Question 12 Detailed Solution

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Given:

Radius of large circle (R) = 15 cm.

Radius of small circle (r) = 13 cm

Concept used:

A line drawn from the centre of the circle to a chord bisects the chord into two halves.

The tangent to the circle is perpendicular to the radius of the circle at the point of contact.

Pythagorean theorem:

H2 = P2 + B2

Where, H = hypotenuse ; P = perpendicular ; B = base

Calculation:

qImage64d3b6f625818518792cbb2a

Here, O is the centre and PQ is the chord of the circle.

In △POR

⇒ OP2 = OR2 + PR2

⇒ (15)2 = (13)2 + PR2

⇒ PR2 = 225 - 169 = 56

⇒ PR = √56 = 2√14 cm

PQ is tangent so, OR ⊥ PQ

PQ = PR + RQ

⇒ 2√14 + 2√14 (As PR = RQ)

⇒ 4√14 cm

∴ The correct answer is 4√14 cm. 

Angle subtended by the largest chord of the circle to a point on the same circle measures:

  1. < 90°
  2. 180°
  3. > 90°
  4. 90°

Answer (Detailed Solution Below)

Option 4 : 90°

Theorem on Chords Question 13 Detailed Solution

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Given:

The angle subtended by a chord at the centre = 180°

Concept Used:

The angle subtended by a chord at the centre is twice the angle subtended on the circumference.

Calculation:

F2 Savita SSC 15-6-23 Himanshu D12

⇒ Angle subtended on the circumference = 180° / 2 = 90°

Therefore, the angle subtended by the same chord on the circumference is 90°.

Two equal circles of radius 18 cm intersect each other, such that each passes through the centre of the other. The length of the common chord is _________.

  1. \(6\sqrt{27}\) cm
  2. \(9\sqrt{27}\) cm
  3. \(\sqrt{3}\)
  4. \(3\sqrt{3}\) cm

Answer (Detailed Solution Below)

Option 1 : \(6\sqrt{27}\) cm

Theorem on Chords Question 14 Detailed Solution

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Given:

The radius of the circles is 18 cm

Calculation:

F1 ArunK Madhuri 11.03.2022 D6

According to the diagram,

AD = DB

O1O2 = 18

Again O1A = O2A = 18 [Radius of the circle]

∠ADO1 = 90°

O1D = O2D = 9

AD = √(324 - 81)

⇒ (9√3) 

AB = 2 × (9√3) = 18√3 = 6√27

∴ The length of the common chord is 6√27 cm.

In the figure given below, PQ is the diameter of the circle with centre O. If ∠QOR = 100° then the measure of ∠PSR is:

F1 SSC Arbaz 6-10-23 D9

  1. 160°
  2. 80°
  3. 40°
  4. 100°

Answer (Detailed Solution Below)

Option 3 : 40°

Theorem on Chords Question 15 Detailed Solution

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Given:

PQ is a diameter.

∠QOR = 100°

Concept used:

the angle subtended by the chord at the centre of the circle is twice

the angle subtended at any point on the circle's circumference. 

Calculation:

⇒ ∠QOR + ∠ROP = 180°

⇒ ∠ROP = 180 - 100 = 80°

∠ROP = 2 × ∠PSR

⇒ ∠PSR = 80/2 = 40°

∴ The correct answer is 40°.

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