Areas of Parallelograms & Triangles — Multiple Choice Questions

Area of the rhombus whose diagonals are 16 cm and 24 cm will be :

1. 182 cm²

2. 202 cm²

3. 92 cm²

4. 192 cm²

Answer

We know that,

Area of the rhombus = $\dfrac{1}{2}$ × Product of diagonals

= $\dfrac{1}{2}$ × 16 × 24

= 8 × 24

= 192 cm²

Hence, option 4 is the correct option.

The area of the trapezium whose parallel sides are 9 cm and 6 cm respectively and distance between these sides is 8 cm, will be :

1. 50 cm²

2. 60 cm²

3. 70 cm²

4. 80 cm²

Answer

We know that,

Area of the trapezium = $\dfrac{1}{2}$ × sum of parallel sides × height

= $\dfrac{1}{2}$ × (9 + 6) × 8

= 15 × 4

= 60 cm²

Hence, option 2 is the correct option.

Area of parallelogram ABCD in the figure will be :

1. 15 cm²

2. 25 cm²

3. 35 cm²

4. 45 cm²

Answer

We know that,

Area of ∥gm = Base × Height

= AB × DB

= 5 × 7

= 35 cm².

Hence, option 3 is the correct option.

In the given figure, if AD is median on BC and AE ⟂ BC, then ar (ΔADC) =

1. ar (ΔADE)

2. ar (ΔABD)

3. ar (ΔAEC)

4. ar (ΔBCA)

Answer

Median AD divides ΔABC into two Δs of equal area.

ar (ΔADC) = ar (ΔABD)

Hence, option 2 is the correct option.

The area of trapezium PQRS will be :

1. 170 cm²

2. 180 cm²

3. 160 cm²

4. 190 cm²

Answer

In triangle RTQ,

RQ² = RT² + QT²

(17)² = RT² + (8)²

RT² = 289 - 64

RT² = 225

RT = $\sqrt{225}$ = 15 cm.

We know that,

Area of the trapezium = $\dfrac{1}{2}$ × sum of parallel sides × height

= $\dfrac{1}{2}$ × (8 + 16) × 15

= $\dfrac{1}{2}$ × (24) × 15

= 15 × 12

= 180 cm².

Hence, option 2 is the correct option.

ABCD is a parallelogram. If AB = 3.6 cm and altitude corresponding to sides AB and AD are respectively 5 cm and 4 cm, then AD will be :

1. 5.5 cm

2. 3.5 cm

3. 2.5 cm

4. 4.5 cm

Answer

We know that,

Area of ∥gm = Base × Height

Area of ∥gm ABCD using base AB = 3.6 × 5 = 18 cm²

Area of ∥gm ABCD using base AD = AD × 4

18 = AD × 4

AD = $\dfrac{18}{4}$

AD = 4.5 cm.

Hence, option 4 is the correct option.

A triangle, a parallelogram and a rectangle have the same base and are situated between the same parallels. The ratio of their areas is :

1. 2 : 1 : 2

2. 2 : 2 : 1

3. 1 : 2 : 3

4. 1 : 2 : 2

Answer

Let the common base be b and the common height be h.

Area of triangle = $\dfrac{1}{2}$ × ​b × h

Area of parallelogram = ​b × h

Area of rectangle = ​b × h

Ratio of areas of triangle, parallelogram and rectangle:

= $\dfrac{1}{2}$ × ​b × h : ​b × h : ​b × h

= 1 : 2 : 2.

Hence, option 4 is the correct option.

If the area of the parallelogram ABCD is 10 cm², then the area of ΔBCD is :

1. 10 cm²

2. 5 cm²

3. 20 cm²

4. 25 cm²

Answer

Diagonal BD divides it into triangles ABD and BCD of equal area.

ar(ΔBCD) = $\dfrac{1}{2}$ ar(∥gm ABCD)

= $\dfrac{1}{2} \times 10$

= 5 cm².

Hence, option 2 is the correct option.

A triangle and a parallelogram, both having base 5 cm, are situated between the same parallels. If the height of the triangle is 4 cm, then the area of the parallelogram is :

1. 20 cm²

2. 40 cm²

3. 10 cm²

4. 5 cm²

Answer

Area of triangle = $\dfrac{1}{2}$ × base × height

= $\dfrac{1}{2}$ × 5 × 4

= 10 cm²

A triangle and a parallelogram, both have same base and are between the same parallels.

Area of triangle = $\dfrac{1}{2}$ ar(∥gm ABCD)

10 = $\dfrac{1}{2}$ ar(∥gm ABCD)

ar(∥gm ABCD) = 10(2)

ar(∥gm ABCD) = 20 cm²

Hence, option 1 is the correct option.

The area of the parallelogram PQRS is 16 sq. units. If M is the mid-point of PQ, then the area of ΔQMR is :

1. 16 sq. units

2. 8 sq. units

3. 4 sq. units

4. 2 sq. units

Answer

ΔPQR and parallelogram PQRS, both have same base PQ and are between the same parallels PQ and SR.

Area of △PQR = $\dfrac{1}{2}$ Area of parallelogram PQRS

= $\dfrac{1}{2} \times 16$

= 8 sq.units.

M is the mid-point of PQ, QM = $\dfrac{1}{2}$ PQ

Thus, RM is the median of a triangle PQR, and divides it into two triangles of equal area.

Area of △QMR = $\dfrac{1}{2}$ Area of triangle PQR

= $\dfrac{1}{2} \times 8$

= 4 sq. units.

Hence, option 3 is the correct option.

ABCD is a rectangle and ABQC is a parallelogram. If the area of ΔABD is 5 sq. cm, then the area of the parallelogram is :

1. 5 sq. cm

2. 10 sq. cm

3. 20 sq. cm

4. 30 sq. cm

Answer

In a rectangle, a diagonal divides it into two equal triangles.

So,

Area of rectangle ABCD = 2 × Area of △ABD

= 2 × 5

= 10 cm².

Rectangle ABCD and parallelogram ABQC are on the same base AB and between the same parallel lines AB and DQ.

Area of the parallelogram ABQC = Area of rectangle ABCD = 10 cm².

Hence, option 2 is the correct option.

P and Q are two points on the side DC of a ∥ gm ABCD. If the area of ΔPAB is 10 cm², then the area of ΔQAB is :

1. 5 cm²

2. 10 cm²

3. 15 cm²

4. 20 cm²

Answer

Triangles PAB and QAB have the same base AB and lie between the same parallels AB and DC.

ar(△QAB) = ar(△PAB) = 10 cm²

Hence, option 2 is the correct option.

Two diagonals of a parallelogram ABCD intersect at O. If the area of the parallelogram is 20 cm², then the area of ΔAOB is :

1. 20 cm²

2. 15 cm²

3. 10 cm²

4. 5 cm²

Answer

The diagonals of a parallelogram divide it into four triangles of equal area.

So, Area(ΔAOB) = $\dfrac{1}{4}$ Area(∥gm ABCD)

= $\dfrac{1}{4} \times 20$

= 5 cm ².

Hence, option 4 is the correct option.

E is the mid-point of the side AB of a parallelogram ABCD. If the area of the ABCD is 60 sq. cm, then the area of ΔBDE is :

1. 60 sq. cm

2. 30 sq. cm

3. 15 sq. cm

4. 10 sq. cm

Answer

Draw diagonal BD.

A diagonal of a parallelogram divides it into two triangles of equal area :

Area (ΔABD) = $\dfrac{1}{2} \times$ Area(∥gm ABCD)

= $\dfrac{1}{2} \times$ (60)

= 30 cm².

Point E is the midpoint of AB, so:

BE = $\dfrac{1}{2}$ AB

A median of a triangle divides it into two triangles of equal area.

Area(ΔBDE) = $\dfrac{1}{2}$ Area(ΔABD)

= $\dfrac{1}{2} \times$ (30)

= 15 cm².

Hence, option 3 is the correct option.