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

Given:

Base of the parallelogram is increased by 8%.

Height of the parallelogram is increased by 4%.

Formula Used:

Net increase percentage in area = (Percentage increase in base + Percentage increase in height + (Percentage increase in base × Percentage increase in height) / 100)

Calculation:

Let the initial area of the parallelogram be A.

After increase, base becomes 1.08 × base and height becomes 1.04 × height.

New area = 1.08 × 1.04 × A

Net increase percentage = ((1.08 × 1.04) - 1) × 100

⇒ Net increase percentage = (1.1232 - 1) × 100

⇒ Net increase percentage = 0.1232 × 100

⇒ Net increase percentage = 12.32%

The net increase percentage in its area is 12.32%.

Given:

Ratio of two sides = 3 : 5

Perimeter = 96 cm

Formula used:

Perimeter of parallelogram = 2 x (sum of adjacent sides)

Calculations:

Let the sides be 3x and 5x.

Perimeter = 2 x (3x + 5x)

96 = 2 x (8x)

96 = 16x

⇒ x = 96 / 16

⇒ x = 6

Adjacent sides are 3x and 5x.

3x = 3 x 6 = 18 cm

5x = 5 x 6 = 30 cm

∴ The adjacent sides of the parallelogram are 18 cm and 30 cm.

Given:

AE ⊥ DB

CF ⊥ DB

AE = 8 cm

BD = 15 cm

FC = 12 cm

Formula Used:

Area of the figure = Area of Triangle ADB + Area of Triangle CDB

Area of Triangle = 1/2 × base × height

Calculation:

Area of Triangle ADB:

Base (BD) = 15 cm

Height (AE) = 8 cm

Area of Triangle ADB = 1/2 × 15 × 8

⇒ Area of Triangle ADB = 60 cm²

Area of Triangle CDB:

Base (BD) = 15 cm

Height (FC) = 12 cm

Area of Triangle CDB = 1/2 × 15 × 12

⇒ Area of Triangle CDB = 90 cm²

Total Area of the figure = Area of Triangle ADB + Area of Triangle CDB

⇒ Total Area = 60 cm² + 90 cm²

⇒ Total Area = 150 cm²

The area of the figure is 150 cm².

Given:

AB = 15 cm, BC = 9 cm, and the angle ∠ABC = 60°

Formula Used:

Cosine rule: AC² = AB² + BC² − 2 × AB × BC × cos(∠ABC)

Calculation:

Applying Cosine Rule,

AC2 = AB2 + BC2 − 2 × AB × BC × cos(∠ABC)

⇒ AC2 = 15² + 9² - 2 × 15 × 9 × cos(60°)

⇒ AC2 = 225 + 81 - 270 × ½

⇒ AC2 = 306 - 135

⇒ AC2 = 171

⇒ AC = √171 = 13.08  ≈  13 cm.

∴ The correct answer is 13 cm.

Concept:

The sum of any two adjacent angles of a parallelogram is 180°

Explanation:

In a parallelogram ABCD, angle A and angle B are in the ratio 1 : 2.

Let ∠A = x, ∠B = 2x

Then x + 2x = 180

⇒ 3x = 180 ⇒ x = 60

So, ∠A = 60°

Hence Option (3) is true. 

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:

The area of a parallelogram ABCD is 300 cm2

Concept used:

Perimeter of a parallelogram = 2(a + b)

Area of a parallelogram = a × b

Where a and b is any side and height of the parallelogram from the opposite side

Calculation:

According to the concept,

CD × 20 = 300

CD = 15

Again 

AD × 30 = 300

AD = 10

So, Perimeter = 2 (15 + 10)

⇒ 50 cm

∴ The perimeter of the parallelogram 50 cm

By parallelogram Formula of diagonals,

AC² + BD² = 2 (AB² + BC²)

⇒ 8² + BD² = 2 (7² + 9²)

⇒ 64 + BD² = 2 (49 + 81) = 260

⇒ BD² = 260 – 64 = 196

⇒ BD = 14 cm

∴ The length of other diagonal = 14 cm

Explanation -

If we add both the figure then we get the new figure given below -

It is a parallelogram.

And we know that the area of Trapezium = \(\frac{1}{2}\times h \times (a+b)\)

Where h is the height and a, b is the length of the parallel sides of the parallelogram.

Now \(2(a+b)=\frac{1}{2} \times h \times (2a+2b+2a+2b)\) 

⇒ \(2(a+b)=\frac{1}{2} \times h \times (4a+4b)\)

⇒ \(2(a+b)=\frac{1}{2} \times h \times 4(a+b)\)

⇒ h = 1

Hence the option (ii) is correct.

Given:

The area of a parallelogram PQRS is 770 sq. cm

Sides RS and PS have lengths 55 cm and 28 cm respectively

Formula Used:

Area of parallelogram = base × height

Calculation:

Area = base × height

770 = 55 × height

⇒ H = 14 cm = PM

In Δ PSM

Sin ∠S = Height/hypotenuse 

Sin ∠S = 14/28 = 1/2

∠S = 30^∘ 

∴ ∠S is 30^∘ 

From the given data,

Let ABCD be the parallelogram.

We know that,

Area of parallelogram ABCD = 2 × (Area of triangle ABC)

Now let us consider a = 20 m, b = 15 m and c = 15 m

We know that,

Area of triangle ABC = √[s(s - a)(s - b)(s - c)]

Where,

s = Semi - perimeter

a, b and c = Sides of triangle ABC

⇒ s = (a + b + c)/2 = (20 + 15 + 15)/2 = 25 m

Area of triangle ABC = √[25 × (25 - 20) × (25 - 15) × (25 - 15)] = 50√5 sq.m

∴ Area of parallelogram ABCD = 2 × 50√5 = 100√5 sq.m

Extend the line QP and RS they will meet at T and they will form an equilateral triangle

∆QRT is an equilateral triangle with side = 18 cm,

⇒ Area of the ∆QRT = √3/4 × 18 × 18 = 81√3 cm²

⇒ ∆TSP is the right angle triangle with PS = 9 cm,

⇒ TS = 9 tan 30° = 9/√3 cm

⇒ Area of the ∆TSP = 1/2 × 9/√3 × 9 = (27√3)/2 cm²

Now Area of quadrilateral PQRS = Area of ∆QRT - Area of ∆TSP

⇒ Area of quadrilateral PQRS = [81√3 - (27√3)/2]

∴ Area of quadrilateral PQRS = (135√3)/2 cm²

As we know,

In quadrilateral opposite sides are parallel and equal and its diagonals bisect each other.

Go through the option (1)

In ΔADC and In ΔABD

⇒ AC = DB(Opposite sides)

⇒ AD = AD(Common)

⇒ ∠ACD = ∠ABD      (Opposite angles)

Now, we can say ΔACD ≅ ΔDBA not ΔADC ≅ ΔABD

Concept -

(1) In parallelogram ABCD, opposite angles are congruent. 
(2) Also, the angles in triangle ABC must sum to 180°
m∠ACB + m∠BAC + m∠CAB = 180°

Explanation -

ABCD is a parallelogram ∠ACB = 40° and ∠BAC = 80°



Substitute the known angle measures:
40° + 80° + m∠CBA = 180°
m∠CBA = 180° - 40° - 80° = 60°

 From (1), opposite angles are congruent. 
Therefore,  the measure of m∠ADC = 60°

Hence the correct answer is ∠ ADC = 60°

Concept used:

The opposite sides of the parallelogram are equal

Calculation:

Now,

2x + 5 = 2y + 1

⇒ 2y - 2x = 5 - 1

⇒ 2(y - x) = 4

⇒ y - x = 2      ----(1)

And, 2y - 6 = x + 3

⇒ 2y - x = 3 + 6 

⇒ 2y - x = 9      ----(2)

Subtracting eq(1) from eq(2)

2y - x - y + x = 9 - 2

⇒ y = 7

From eq (1),

x = 5

Side of the parallelogram = 2x + 5 = 2 × 5 + 5 = 15

And, x + 3 = 5 + 3 = 8

∴ The side of the parallelogram are 15 units and 8 units