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

Given:

The ∠B = 112°

Formula used:

Opposite angles of the cyclic quadrilateral = 180° 

Calculation:

In given cyclic quadrilateral ABCD

⇒ ∠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°.

Given:

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).

Given:

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

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.

Explanation:

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

add up to 180^o).

Hence option 3 is correct.

Given :

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

Calculations :

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.

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° 

Given:

∠BEC = 138° and ∠ECD = 35° 

Concept used:

In cyclic quadrilateral angles on the same arc are always same

Calculation:

∠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°  

Given:

The ∠B = 104°

Formula used:

Opposite angles of the cyclic quadrilateral = 180° 

Calculation:

In given cyclic quadrilateral ABCD

⇒ ∠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°.

Given:

AB = 16 cm

CD = 18 cm

AD = 12 cm

Concept used:

If diagonal PR bisects diagonal QS then

PQ × QR = PS × RS

In cyclic quadrilateral PQRS

PR × SQ = PQ × RS + PS × QR

Calculation:

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.

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

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

Given:

ABCD is cyclic quadrilateral. 

∠BFA = 60° and ∠AED = 30°

Calculation:

∠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 

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°

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

Given:

∠ADC = 78° and ∠BEC = 52°

Concept:

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

Calculation:

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°.