Quadrilaterals MCQ Quiz - Objective Question with Answer for Quadrilaterals - Download Free PDF

Given:

ABCD is a trapezium (trapezoid) with BC parallel to AD (BC || AD).

AC = CD (This means triangle ACD is an isosceles triangle).

Angle ABC (∠ABC) = 69°

Angle BAC (∠BAC) = 23°

Find: The measure of Angle ACD (∠ACD).

Given

AB = 17 cm, BC = 8 cm, CD = 9 cm, AD = 12 cm, and AC = 15 cm

Calculation

In the above figure:

Area of ΔACD 

= 1/2 × 12 × 9 = 54 

Area of ΔABD 

= 1/2 × 8 × 15 = 60

Area of quadrilateral = 60 + 54 = 114 cm2

The correct answer is 114. 

Given:

Measure of an exterior angle of a regular polygon: 45°

Calculation:

Let's denote the number of sides of the regular polygon as "n".

According to the given information,

The measure of an exterior angle is 45°.

Using the formula mentioned above,

We can write the equation: 360° / n = 45°

To solve for "n", we can cross-multiply and simplify:

⇒ 360° = 45n

Dividing both sides by 45°: 360° / 45° = n

⇒ 8 = n

Therefore, the number of sides of the regular polygon is 8.

Given:

ABCD is a cyclic quadrilateral

∠A = 67°

∠B = 92°

Formula used:

In a cyclic quadrilateral, the opposite angles are supplementary:

∠A + ∠C = 180°

∠B + ∠D = 180°

Difference between ∠C and ∠D = |∠C - ∠D|

Calculations:

∠A + ∠C = 180°

⇒ ∠C = 180° - 67° = 113°

∠B + ∠D = 180°

⇒ ∠D = 180° - 92° = 88°

Difference between ∠C and ∠D = |∠C - ∠D|

⇒ Difference = |113° - 88°|

⇒ Difference = 25°

∴ The correct answer is option (4).

Given:

Angles of a pentagon are in the ratio 1 : 3 : 6 : 7 : 10.

Sum of interior angles of a pentagon = 540º.

Formula Used:

Sum of angles in a polygon = (n - 2) × 180, where n is the number of sides.

Individual angle = (Ratio of angle / Total ratio) × Sum of angles.

Calculation:

Total ratio = 1 + 3 + 6 + 7 + 10 = 27.

Smallest angle corresponds to ratio 1.

Smallest angle = (1 / 27) × 540.

⇒ Smallest angle = 540 / 27.

⇒ Smallest angle = 20º.

The smallest angle is 20º.

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.

Concept:

Octagon has eight sides.

Dodecagon has twelve sides.

Formula:

Interior angle of polygon = [(n – 2) × 180°] /n

Calculation:

Interior angle of octagon = [(8 – 2)/8] × 180° = 1080°/8 = 135°

Interior angle of dodecagon = [(12 – 2)/12] × 180° = 1800°/12 = 150°

∴ The ratio of the measures of the interior angles for octagon and dodecagon is 9 : 10

Given:

In parallelogram ABCD, AL and CM are perpendicular to CD and AD respectively. 

AL = 20 cm, CD = 18 cm and CM = 15 cm

Formula used:

Area of  parallelogram = Base × Height 

Perimeter of parallelogram = 2 × (Sum of parallel sides) 

Calculation:

Area of ABCD with base DC = AL × DC = 20 × 18 

⇒ 360 cm²

Again, Area of ABCD with base AD = CM × AD = 15 × AD 

⇒ 360 cm2 =  15 × AD 

⇒ AD = 24 cm 

∴ AD = BC = 24 cm, DC = AB = 18 cm 

Perimeter of ABCD = 2 × (24 + 18) 

⇒ 2 × 42

⇒ 84 cm 

∴ The required result = 84 cm 

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° 

Figure:

Calculation:

As the diagonals of a rectangle intersect each other,

⇒ AO = OB

⇒ ∠OBA = ∠OAB = 25° [∵ Angle opposite to equal side are equal]

By angle sum property in ΔAOB,

⇒ ∠AOB + ∠OAB + ∠OBA = 180°

⇒ ∠AOB + 25° + 25° = 180°

⇒ ∠AOB = 130°

By linear pair property,

⇒ ∠DOA + ∠AOB = 180°

⇒ ∠DOA + 130° = 180°

⇒ ∠DOA = 50°

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.

Given:

Radius of semicircle = 10 cm

Formula Used:

Area of semicircle = (1/2)πr²

Area of rectangle = length × breadth

Calculation:

Area of semicircle = (1/2)πr²

⇒ Area of semicircle = (1/2) × (22/7) × (10)²

⇒ Area of semicircle = 157.14 cm²

∵ A is the mid point of OP

OA = 5 cm

In a ΔAOD,

OD² = OA² + AD²

⇒ (10)² = (5)² + AD²

⇒ AD² = 100 - 25

⇒ AD2 = 75

⇒ AD = 5√3

Area of rectangle ABCD = AB × AD

⇒ Area of rectangle ABCD = 10 × 5√3

⇒ Area of rectangle ABCD = 50√3

⇒ Area of rectangle ABCD = 86.60 cm²

∴ Area of shaded portion = Area of semicircle - Area of rectangle

⇒ Area of shaded portion = 157.14 - 86.60

⇒ Area of shaded portion = 70.54 cm²

∴ The area of shaded portion is 70.54 cm².

Given:

External angle = 45° 

Formula used:

External angle = (360°/n)

Number of diagonal of a n side polygon = (n² - 3n)/2

Where, n = Equal to the number of side of a polygon

Calculation:

External angle = (360°/n)

⇒ 45° = (360°/n)

⇒ n = 8 

Now, Number of diagonal of a 'n' side polygon

⇒ (n2 - 3n)/2

⇒ (64 - 24)/2

⇒ 20

∴ The number of diagonal is 20.