Cyclic Quadrilateral MCQ Quiz - Objective Question with Answer for Cyclic Quadrilateral - Download Free PDF

Last updated on Jun 12, 2025

The trick to solving Cyclic Quadrilateral MCQs Quiz is to understand its properties and then practice applying them to Cyclic Quadrilateral objective questions. Here is a collection of questions for the practice of candidates preparing the Mensuration Topic for competitive exams. Mensuration is a commonly featured Topic in most competitive exams and hence candidates should be well prepared for it. Therefore, we also have a few tips and tricks for candidates to save time and increase the accuracy of Cyclic Quadrilateral question answers.

Latest Cyclic Quadrilateral MCQ Objective Questions

Cyclic Quadrilateral Question 1:

ABCD is a cyclic quadrilateral such that ∠B = 112°. The tangents at A and C meet at a point P. What is the measure of ∠APC?  

  1. 44°
  2. 68°
  3. 38°
  4. 42°

Answer (Detailed Solution Below)

Option 1 : 44°

Cyclic Quadrilateral Question 1 Detailed Solution

Given:

The ∠B = 112°

Formula used:

Opposite angles of the cyclic quadrilateral = 180° 

Calculation:

In given cyclic quadrilateral ABCD

Task Id 1150 Daman

⇒ ∠ABC + ∠ADC = 180° 

⇒ ∠ADC = 180° - 112° = 68° 

Since PA is tangent on the circle at point A and AC is the cord which is subtending the angle ∠D = 68°

The angle between a tangent and a chord of a circle is equal to the angle subtended by the chord in the alternate segment of the circle.

In this case, the tangent at point A (line segment PA) and chord AC subtend ∠PAC in the alternate segment, which is equal to the angle ∠D subtended by the same chord AC. Therefore, ∠PAC = ∠D = 68°.

⇒ ∠PAC = ∠D = 68°

Also,

⇒ ∠PAC  =  ∠PCA, (As PA and PC are the tangents of A and C) 

⇒ ∠PAC = ∠PCA = ∠ADC = 68°    

In ΔPAC

⇒ ∠PAC + ∠PCA + ∠APC = 180° 

⇒ 68° + 68° + ∠APC = 180° 

⇒ ∠APC = 180° - 136° = 44° 

∴ The required result will be 44°.

Cyclic Quadrilateral Question 2:

In a quadrilateral ABCD, angles A, B, C and D are respectively (3x - 10°), (x + 30°), (2x + 30°) and (2x - 10°). Then, this quadrilateral is a

  1. kite
  2. rectangle
  3. trapezium
  4. parallelogram

Answer (Detailed Solution Below)

Option 4 : parallelogram

Cyclic Quadrilateral Question 2 Detailed Solution

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Given:

The angles of quadrilateral ABCD are:

  • Angle A = 3x-10° 
  • Angle B = x+30° 
  • Angle C = 2x+30° 
  • Angle D = 2x-10° 
Concept:

The sum of the interior angles of any quadrilateral is \( 360^\circ \).

Calculation:
  1. Set up the equation for the sum of angles:
    3x-10 + x+30 + 2x+30 + 2x-10 = 360°
  2. Simplify the equation:  8x + 40 = 360° 
  3. Solve for x" id="MathJax-Element-363-Frame" role="presentation" style="position: relative;" tabindex="0"> :
    8x = 320° 
    x = 40° 
  4. Find each angle:
    • Angle A = 110° 
    • Angle B = 70° 
    • Angle C = 110° 
    • Angle D = 70° 
Analysis of the Quadrilateral:
  • Adjacent Angles:
    • Angle A (110°)  + Angle B (70°)  = 180° 
    • Angle B (70°) + Angle C (110°) = 180° 
    • Angle C (110°) + Angle D (70°) = 180° 
    • Angle D (70°) + Angle A (110°) = 180° 
Conclusion:
The quadrilateral has:
  1. Two distinct pairs of adjacent sides where one pair is equal
  2. One pair of equal opposite angles (110° and 110°)
  3. The other pair of equal opposite angles (70° and 70°)
  4. Adjacent angles are supplementary (add up to 180°)

These properties are characteristic of a Parallelogram.

NOTE:

As per the official answer key, Option 3 and 4 both are correct.

Cyclic Quadrilateral Question 3:

The angles of cyclic quadrilaterals ABCD are: A = (6x + 10), B = (5x)°, C = (x + y)° and D =(3y - 10)°. The value of x and y is:

  1. x = 20° and y = 10°
  2. x = 20° and y = 30°
  3. x = 44° and y = 15°
  4. x = 15° and y = 15°

Answer (Detailed Solution Below)

Option 2 : x = 20° and y = 30°

Cyclic Quadrilateral Question 3 Detailed Solution

Formula used:

In a cyclic quadrilateral, opposite angles sum up to 180º.

A + C = 180º and B + D = 180º

Calculations:

⇒ (6x + 10) + (x + y) = 180

⇒ 7x + y = 170   .........(1)

⇒ (5x) + (3y - 10) = 180

⇒ 5x + 3y = 190  .........(2)

Now, solving the equations 1 and 2:

7x + y = 170    

5x + 3y = 190 

____________

21x + 3y = 510    ......(3)

5x + 3y = 190    .......(4)

Subtracting (4) from (3):

16x = 320 ⇒ x = 20º

Substitute x in (4):

5 x 20 + 3y = 190

⇒ 100 + 3y = 190

⇒ y = 30º

⇒ x = 20º and y = 30º

∴ The correct answer is option (2).

Cyclic Quadrilateral Question 4:

A quadrilateral PQRS is inscribed in a circle of centre O, such that PQ is a diameter and ∠ PSR = 120°. Find the value of ∠QPR.

  1. 30°
  2. 60°
  3. 40°
  4. 50°

Answer (Detailed Solution Below)

Option 1 : 30°

Cyclic Quadrilateral Question 4 Detailed Solution

Given:

PQRS is a cyclic quadrilateral, and ∠PSR = 120°

14-4-2025 IMG-673 -4

Calculation:

As we know, the sum of the either pair of the opposite angles of a cyclic quadrilateral is 180°

⇒ ∠PSR + ∠PQR = 180°

⇒ ∠PQR = 180° - 120° = 60°

In ΔPQR,

if PQ is diameter, then ∠PRQ = 90°

⇒ ∠PQR + ∠PRQ + ∠QPR = 180°

⇒ 60° + 90° + ∠QPR = 180°

⇒ ∠QPR = 180° – 150° = 30°

The correct answer is option 1.

Cyclic Quadrilateral Question 5:

In a cyclic quadrilateral the sum of opposite angles is:

  1. Equal
  2. Complimentary
  3. Supplementary
  4. Unequal

Answer (Detailed Solution Below)

Option 3 : Supplementary

Cyclic Quadrilateral Question 5 Detailed Solution

Explanation:

In a cyclic quadrilateral, the sum of opposite angles is supplementary (i.e., they

add up to 180o).

Hence option 3 is correct.

Top Cyclic Quadrilateral MCQ Objective Questions

A circle touches all four sides of a quadrilateral PQRS. If PQ = 11 cm. QR = 12 cm and PS = 8 cm. then what is the length of RS ?

  1. 7 cm
  2. 15 cm
  3. 9 cm
  4. 7.3 cm

Answer (Detailed Solution Below)

Option 3 : 9 cm

Cyclic Quadrilateral Question 6 Detailed Solution

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Given :

A circle touches all four sides of a quadrilateral PQRS. If PQ = 11 cm. QR = 12 cm and PS = 8 cm

Calculations :

F1 Ashish S 25-10-21 Savita D1

If a circle touches all four sides of quadrilateral PQRS then, 

PQ + RS = SP + RQ

So,

⇒ 11 + RS = 8 + 12

⇒ RS = 20 - 11

⇒ RS = 9

∴ The correct choice is option 3.

PQRS is a cyclic trapezium where PQ is parallel to SR and PQ is the diameter. If ∠QPR = 40° then the ∠PSR is equal to:

  1. 130°
  2. 120°
  3. 140°
  4. 110°

Answer (Detailed Solution Below)

Option 1 : 130°

Cyclic Quadrilateral Question 7 Detailed Solution

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F1 Arun Madhuri 26.10.2021 D6

Given:

PQRS is a cyclic trapezium where PQ is parallel to RS.

PQ is the diameter & ∠QPR = 40° 

Concept:

Angle made in a semicircle is a right angle. 

The sum of the opposite angles of a cyclic trapezium is 180°.

Calculation:

In triangle PQR, 

∠RPQ + ∠RQP + ∠QRP = 180°  [Angle sum property] 

⇒ 40° + ∠RQP + 90° = 180° 

⇒ ∠RQP = 180° - 130° = 50° 

∠RQP + ∠PSR = 180° [Supplementary Angles]

∴ ∠PSR = 180° - 50° = 130° 

ABCD is a cyclic quadrilateral. Diagonals BD and AC intersect each other at E. If ∠BEC = 138° and ∠ECD = 35°, then what is the measure of ∠BAC?

  1. 133°
  2. 123°
  3. 113°
  4. 103°

Answer (Detailed Solution Below)

Option 4 : 103°

Cyclic Quadrilateral Question 8 Detailed Solution

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Given:

∠BEC = 138° and ∠ECD = 35° 

Concept used:

In cyclic quadrilateral angles on the same arc are always same

Calculation:

F1 HimanshuC Madhuri 14.03.2022 D1

∠BEC and ∠CED are on the same straight lines 

∠BEC  =138°

∠CED = 180° – 138° 

⇒ ∠CED = 42° 

In ΔCDE, ∠CED = 42° and ∠DCE = 35°

∠CDE = 180° - (42° + 35°)

∠CDE = 103° 

 ∠BAC and ∠BDC are on the same arc BC

We know that In cyclic quadrilateral angles on the same arc are always same.

∠BAC = 103°

∴  The measure of ∠BAC is 103°  

ABCD is a cyclic quadrilateral such that B = 104°. The tangents at A and C meet at a point P. What is the measure of ∠APC?

  1. 24°
  2. 38°
  3. 28°
  4. 26°

Answer (Detailed Solution Below)

Option 3 : 28°

Cyclic Quadrilateral Question 9 Detailed Solution

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Given:

The ∠B = 104°

Formula used:

Opposite angles of the cyclic quadrilateral = 180° 

Calculation:

In given cyclic quadrilateral ABCD

F2 Madhuri Engineering 26.04.2022 D3

⇒ ∠ABC + ∠ADC = 180° 

⇒ ∠ADC = 180° - 104° = 76° 

Since PA is tangent on the circle at point A and AC is the cord which is subtending the angle ∠D = 76°
 

The angle between a tangent and a chord of a circle is equal to the angle subtended by the chord in the alternate segment of the circle.

In this case, the tangent at point A (line segment PA) and chord AC subtend ∠PAC in the alternate segment, which is equal to the angle ∠D subtended by the same chord AC. Therefore, ∠PAC = ∠D = 76°.

⇒ ∠PAC = ∠D = 76°

Also,

⇒ ∠PAC  =  ∠PCA, (As PA and PC are the tangents of A and C) 

⇒ ∠PAC = ∠PCA = ∠ADC = 76°    

In ΔPAC

⇒ ∠PAC + ∠PCA + ∠APC = 180° 

⇒ 76° + 76° + ∠APC = 180° 

⇒ ∠APC = 180° - 152° = 28° 

∴ The required result will be 28°.

ABCD is a cyclic quadrilateral in which AB = 16 cm, CD = 18 cm and AD = 12 cm, and AC bisects BD. What is the value of AC.BD?

  1. 450
  2. 360
  3. 300
  4. 825

Answer (Detailed Solution Below)

Option 1 : 450

Cyclic Quadrilateral Question 10 Detailed Solution

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Given:

AB = 16 cm

CD = 18 cm

AD = 12 cm

Concept used:

F1 Vinanti SSC 26.10.22 D1

If diagonal PR bisects diagonal QS then

PQ × QR = PS × RS

In cyclic quadrilateral PQRS

PR × SQ = PQ × RS + PS × QR

Calculation:

F1 Vinanti SSC 26.10.22 D2

According to the concept,

AB × BC = CD × AD

⇒ 16BC = 18 × 12

⇒ 16BC = 216

⇒ BC = 13.5 cm

Now,

Again according to the concept,

AC.DB = AB × CD + AD × BC

⇒ AC.DB = 16 × 18 + 12 × 13.5

⇒ AC.DB = 288 + 162

⇒ AC.DB = 450

∴ The value of AC.BD is 450.

In the given figure, PQ is a chord passing through the centre 'O' of the circle. Calculate ∠PQS.

F1 SSC Arbaz 6-10-23 D2

  1. 40°
  2. 60°
  3. 20°
  4. 80°

Answer (Detailed Solution Below)

Option 3 : 20°

Cyclic Quadrilateral Question 11 Detailed Solution

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Concept used:

The sum of the opposite angles of the cyclic quadrilateral is 180°.

∠PSQ = 90° because the angle is in a semi-circle.

Calculation:

∠SPQ = 180° - 110°  = 70° (The sum of the opposite angles of the cyclic quadrilateral is 180°)

Now, In triangle PQS.

∠PSQ = 90° (Angle in semi-circle)

∠PQS = 180° - ( ∠PSQ +∠QPS)
 

⇒ 180° - (90° + 70°) = 20°

∴ The correct option is 3

PQRS is a cyclic quadrilateral in which PQ = x cm, QR = 16.8 cm, RS = 14 cm, PS = 25.2 cm, and PR bisects QS. What is the value of x?

  1. 24
  2. 21
  3. 28
  4. 18

Answer (Detailed Solution Below)

Option 2 : 21

Cyclic Quadrilateral Question 12 Detailed Solution

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     AAV

From the following figure

∠SQP = ∠OQP = ∠SRO

OQ = OS (Given)

∠QPR = ∠QSR = ∠OSR 

∴ We can say that,

ΔPOQ∼ΔSOR

PQ/SR = OQ/OR

OQ/OR = x/14

Given, OQ = OS

OS/OR = x/14     --- (1)

As we know,

ΔPOS∼ΔQOR

OS/OR = PS/QR

OS/OR = 25.2/16.8      --- (2)

From equation (1) and equation (2)

x/14 = 25.2/16.8

x = (25.2 × 14)/16.8

∴ x = 21 cm

ABCD is cyclic quadrilateral. Sides AB and DC, when produced, meet at E, and sides BC and AD, when produced, meet at F. If ∠BFA = 60° and ∠AED = 30°, then the measure of ∠ABC is:

  1. 65°
  2. 75°
  3. 70°
  4. 80°

Answer (Detailed Solution Below)

Option 2 : 75°

Cyclic Quadrilateral Question 13 Detailed Solution

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Given:

ABCD is cyclic quadrilateral. 

∠BFA = 60° and ∠AED = 30°

Calculation:

F1 Sharad.C 27-02-2020 Savita D12

∠BFA = 60 & ∠BED = 30

Let ∠ABC = θ

∵ ABCD is a cyclic quadrilateral

∴ ∠ADC + ∠ABC = 180

∠ADC = 180 – θ ---- (1)

Now,

∠ADC + ∠FDC = 180 [straight line]

From Equation (1);

180 – θ + ∠FDC = 180

∠FDC = θ

In ΔDFC

∠FDC + ∠DFC + FCD = 180

θ + 60 + ∠FCD = 180

∠FCD = 180 – 60 – θ = 120 – θ

∠FCD = ∠BCE = (120 – θ) [Vertical opposite angle]

∠ABC + ∠CBE = 180 [straight line]

∠CBE = 180 – θ

In ΔBEC

∠CBE + ∠BCE + ∠BEC = 180

180 – θ + 120 – θ + 30 = 180

2θ = 330 – 180 = 150

θ = 150/2 = 75

Shortcut Trick F3 Madhuri SSC 23.01.2023 D1 V2

In triangle ABF

α + θ + 60 = 180 

⇒ α + θ = 120 → (1)

In triangle ADE

α + π - θ + 30 = 180

⇒ α - θ + 30 = 0 → (2)

From Eqn (1) and (2), we have

α = 45° and θ = 75°

∴ ∠ABC = 75°

The length of two parallel sides of a trapezium are 53 cm and 68 cm respectively, and the distance between the parallel sides is 16 cm. Find the area of the trapezium. 

  1. 968 cm2
  2. 972 cm2
  3. 988 cm2
  4. 1024 cm2

Answer (Detailed Solution Below)

Option 1 : 968 cm2

Cyclic Quadrilateral Question 14 Detailed Solution

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Area of the Trapezium = 1/2 × (Sum of the parallel sides) × (Distance between parallel sides)

⇒ 1/2 × (53 + 68) × 16

⇒ 1/2 × 121 × 16

∴ Area of the Trapezium = 968 cm2

Sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E and sides AD and BC are produced to meet at F. If ∠ADC = 78° and ∠BEC = 52°, then the measure of ∠AFB is:

  1. 30°
  2. 28°
  3. 26°
  4. 32°

Answer (Detailed Solution Below)

Option 2 : 28°

Cyclic Quadrilateral Question 15 Detailed Solution

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Given:

∠ADC = 78° and ∠BEC = 52°

Concept:

The sum of a pair of opposite angles is 180° of a cyclic quadrilateral. 

Calculation:

F1 Ashish Madhu 19.10.21 D5

Now, In quadrilateral ABCD,

∠ADC + ∠ABC = 180° (Sum of opposite angles are 180°)

⇒ ∠ABC = 180° - 78° = 102° 

∠EBC = 180° - 102° = 78° [Linear Pair]

In ΔBEC,

∠BEC + ∠EBC + ∠BCE = 180° 

⇒ 52° + 78° + ∠BCE = 180° 

⇒ ∠BCE = 180° - 130° = 50°

⇒ ∠BCE = ∠DCF = 50° (Vertically opposite angle)

∠ADE + ∠FDC = 180° [Linear Pair]

⇒ ∠FDC = 180° - ∠ADE = 180° - 78° = 102°

In ΔDCF,

∠FDC + ∠DCF + ∠DFC = 180° 

102 + 50 + ∠DFC = 180° 

∠DFC = 180° - 152° = 28° 

∠DFC = ∠AFB = 28° 

∴ The measure of ∠AFB is 28°.

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