Mensuration — Multiple Choice Questions

Area of a triangle is 30 cm². If its base is 10 cm, then its height is

1. 5 cm

2. 6 cm

3. 7 cm

4. 8 cm

Answer

By formula,

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

Substituting values we get,

30 = $\dfrac{1}{2}$ × 10 × height

height = $\dfrac{30 \times 2}{10} = \dfrac{60}{10}$ = 6 cm.

Hence, Option 2 is the correct option.

If the perimeter of a square is 80 cm, then its area is

1. 800 cm²

2. 600 cm²

3. 400 cm²

4. 200 cm²

Answer

By formula,

Perimeter of square = 4 × side

Substituting values we get,

80 = 4 × side

side = $\dfrac{80}{4}$ = 20 cm.

By formula,

Area of square = side × side = 20 × 20 = 400 cm².

Hence, Option 3 is the correct option.

Area of a parallelogram is 48 cm². If its height is 6 cm then its base is

1. 8 cm

2. 4 cm

3. 16 cm

4. None of these

Answer

By formula,

Area of parallelogram = base × height

Substituting values we get,

⇒ 48 = base × 6

⇒ base = $\dfrac{48}{6}$ = 8 cm.

Hence, Option 1 is the correct option.

If d is the diameter of a circle, then its area is

1. πd²

2. $\dfrac{πd^2}{2}$

3. $\dfrac{πd^2}{4}$

4. 2πd²

Answer

r = $\dfrac{\text{Diameter}}{2} = \dfrac{d}{2}$.

By formula,

Area of circle = πr² = $π \times \Big(\dfrac{d}{2}\Big)^2 = \dfrac{πd^2}{4}.$

Hence, Option 3 is the correct option.

If the area of trapezium is 64 cm² and the distance between parallel sides is 8 cm, then sum of its parallel sides is

1. 8 cm

2. 4 cm

3. 32 cm

4. 16 cm

Answer

By formula,

Area of trapezium = $\dfrac{1}{2}$ × sum of parallel sides × distance between them

Substituting values we get,

⇒ 64 = $\dfrac{1}{2}$ × sum of parallel sides × 8

⇒ 64 = 4 × sum of parallel sides

⇒ sum of parallel sides = $\dfrac{64}{4}$ = 16 cm.

Hence, Option 4 is the correct option.

Area of a rhombus whose diagonals are 8 cm and 6 cm is

1. 48 cm²

2. 24 cm²

3. 12 cm²

4. 96 cm²

Answer

By formula,

Area of rhombus = $\dfrac{1}{2}$ × d₁ × d₂

Substituting values we get,

⇒ Area of rhombus = $\dfrac{1}{2}$ × 8 × 6 = 24 cm².

Hence, Option 2 is the correct option.

If the lengths of diagonals of a rhombus is doubled, then area of rhombus will be

1. doubled

2. tripled

3. four times

4. remains same

Answer

Let diagonals be d₁ and d₂.

By formula,

Area of rhombus = $\dfrac{1}{2}$ × d₁ × d₂

If doubled, diagonals = 2d₁ and 2d₂.

Area of new rhombus = $\dfrac{1}{2}$ × 2d₁ × 2d₂

= 4 × $\dfrac{1}{2}$ × d₁ × d₂

= 4 × Area of rhombus.

Hence, Option 3 is the correct option.

If the length of a diagonal of a quadrilateral is 10 cm and lengths of the perpendiculars on it from opposite vertices are 4 cm and 6 cm, then area of quadrilateral is

1. 100 cm²

2. 200 cm²

3. 50 cm²

4. None of these.

Answer

Let ABCD be the quadrilateral with diagonal BD.

Let AM and CN be the perpendiculars from A and C on diagonal BD.

From figure,

BD divides quadrilateral in two triangles.

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

= $\dfrac{1}{2}$ × BD × AM

= $\dfrac{1}{2}$ × 10 × 4

= 20 cm².

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

= $\dfrac{1}{2}$ × BD × CN

= $\dfrac{1}{2}$ × 10 × 6

= 30 cm².

Area of quadrilateral = Area of △ABD + Area of △BCD

= 20 + 30 = 50 cm².

Hence, Option 3 is the correct option.

Area of a rhombus is 90 cm². If the length of one diagonal is 10 cm then the length of other diagonal is

1. 18 cm

2. 9 cm

3. 36 cm

4. 4.5 cm

Answer

Let diagonals be d₁ and d₂.

By formula,

Area of rhombus = $\dfrac{1}{2}$ × d₁ × d₂

Substituting values we get,

⇒ 90 = $\dfrac{1}{2}$ × 10 × d₂

⇒ d₂ = $\dfrac{90 \times 2}{10}$

⇒ d₂ = 18 cm.

Hence, Option 1 is the correct option.

In the adjoining figure, OACB is a quadrant of a circle of radius 7 cm. The perimeter of the quadrant is

1. 11 cm

2. 18 cm

3. 25 cm

4. 36 cm

Answer

Perimeter of quadrant = $\dfrac{2πr}{4} + 2r$

$= \dfrac{2 \times \dfrac{22}{7} \times 7}{4} + 2 \times 7 \\[1em] = \dfrac{44}{4} + 14 \\[1em] = 11 + 14 \\[1em] = 25 \text{ cm}.$

Hence, Option 3 is the correct option.

In the adjoining figure, OABC is a square of side 7 cm. OAC is a quadrant of a circle with O as center. The area of the shaded region is

1. 10.5 cm²

2. 38.5 cm²

3. 49 cm²

4. 11.5 cm²

Answer

Area of square OABC = (7)² = 49 cm².

Area of quadrant OAC = $\dfrac{πr^2}{4}$

$= \dfrac{\dfrac{22}{7} \times (7)^2}{4} \\[1em] = \dfrac{22 \times 7}{4} \\[1em] = \dfrac{154}{4}\\[1em] = 38.5 \text{ cm}^2.$

Area of shaded region = Area of square OABC - Area of quadrant OAC

= 49 - 38.5 = 10.5 cm².

Hence, Option 1 is the correct option.

The adjoining figure shows a rectangle and a semicircle. The perimeter of the shaded region is

1. 70 cm

2. 56 cm

3. 78 cm

4. 46 cm

Answer

From figure,

Let length = 14 cm and breadth = 10 cm.

Diameter of semi-circle = 14 cm and radius = 7 cm.

Perimeter of shaded region = length + breadth + breadth + πr.

= 14 + 10 + 10 + $\dfrac{22}{7} \times 7$

= 34 + 22 = 56 cm.

Hence, Option 2 is the correct option.

The area of the shaded region shown in the below figure is

1. 140 cm²

2. 77 cm²

3. 294 cm²

4. 217 cm²

Answer

Area of shaded region = Area of rectangle + Area of semi-circle

= length × breadth + $\dfrac{πr^2}{2}$

= 14 × 10 + $\dfrac{\dfrac{22}{7} \times (7)^2}{2}$

= 140 + $\dfrac{154}{2}$

= 140 + 77

= 217 cm².

Hence, Option 4 is the correct option.

In the adjoining figure, the boundary of the shaded region consists of semicircular arcs. The area of the shaded region is equal to

1. 616 cm²

2. 385 cm²

3. 231 cm²

4. 308 cm²

Answer

From figure,

Radius of larger circle (R) = 14 cm.

Area of shaded region = Area of larger circle - Area of 1st smaller circle + Area of 2nd smaller circle .......(1)

From figure,

Diameter of both the smaller semi-circles = 14 cm.

∴ Radius = 7 cm and area of both the circle are equal.

∴ Area of shaded region = Area of larger circle = $\dfrac{πR^2}{2}$

$= \dfrac{1}{2} \times \dfrac{22}{7} \times (14)^2 \\[1em] = \dfrac{11}{7} \times 196 \\[1em] = 11 \times 28 \\[1em] = 308 \text{ cm}^2.$

Hence, Option 4 is the correct option.

The perimeter of the shaded region shown in the below figure is

1. 44 cm

2. 88 cm

3. 66 cm

4. 132 cm

Answer

Radius of larger semi-circle (R) = 14 cm and radius of smaller semi-circle (r) = 7 cm.

From figure,

Perimeter of shaded region = Circumference of larger semi-circle + 2 × Circumference of smaller semi-circle

= πR + 2πr

= $\dfrac{22}{7} \times 14 + 2 \times \dfrac{22}{7} \times 7$

= 44 + 44 = 88 cm.

Hence, Option 2 is the correct option.

In the adjoining figure, ABC is a right angled triangle at B. A semicircle is drawn on AB as diameter. If AB = 12 cm and BC = 5 cm, then the area of the shaded region is

1. (60 + 18π) cm²

2. (30 + 36π) cm²

3. (30 + 18π) cm²

4. (30 + 9π) cm²

Answer

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

= $\dfrac{1}{2}$ × AB × BC

= $\dfrac{1}{2}$ × 12 × 5

= 30 cm².

From figure,

AB = 12 cm is the diameter of circle.

Radius = $\dfrac{AB}{2}$ = 6 cm.

Area of semi-circle = $\dfrac{πr^2}{2} = \dfrac{π(6)^2}{2} = \dfrac{36π}{2}$ = 18π cm².

Area of shaded region = Area of right angle triangle ABC + Area of semi-circle

= (30 + 18π) cm².

Hence, Option 3 is the correct option.

The perimeter of the shaded region shown in the below figure is

1. (30 + 6π) cm

2. (30 + 12π) cm

3. (18 + 12π) cm

4. (18 + 6π) cm

Answer

In right angle triangle ABC,

⇒ AC² = AB² + BC²

⇒ AC² = (12)² + (5)²

⇒ AC² = 144 + 25

⇒ AC² = 169

⇒ AC = $\sqrt{169}$ = 13 cm.

From figure,

radius of semi-circle (r) = $\dfrac{AB}{2} = \dfrac{12}{2}$ = 6 cm.

Perimeter of shaded region = AC + CB + Circumference of semi-circle

= 13 + 5 + πr

= (18 + 6π) cm.

Hence, Option 4 is the correct option.

If the volume of a cube is 729 m³, then its surface area is

1. 486 cm²

2. 324 cm²

3. 162 cm²

4. None of these

Answer

By formula,

Volume of cube = (Side)³

∴ (Side)³ = 729

Side = $\sqrt[3]{729}$ = 9 cm.

⇒ Surface area of cube = 6(side)²

= 6(9)²

= 6 × 81 = 486 cm².

Hence, Option 1 is the correct option.

If the total surface area of a cube is 96 cm², then the volume of cube is

1. 8 cm³

2. 512 cm³

3. 64 cm³

4. 27 cm³

Answer

Let side of cube = x cm.

Given,

Surface area of cube = 96 cm²

By formula,

Surface area of a cube = 6x²

∴ 6x² = 96

⇒ x² = 16

⇒ x = $\sqrt{16}$ = 4 cm.

Volume of cube = (side)³

= (x)³

= 4³ = 64 cm³.

Hence, Option 3 is the correct option.

The length of the longest pole that can be put in a room of dimensions (10 m × 10 m × 5 m) is

1. 15 m

2. 16 m

3. 10 m

4. 12 m

Answer

The longest pole in a cuboid is equal to the diagonal of cuboid.

Diagonal of cuboid = $\sqrt{l^2 + b^2 + h^2}$

$= \sqrt{10^2 + 10^2 + 5^2} \\[1em] = \sqrt{100 + 100 + 25} \\[1em] = \sqrt{225} \\[1em] = 15 \text{ m}.$

Hence, Option 1 is the correct option.

The lateral surface area of a cube is 256 m². The volume of the cube is

1. 512 m³

2. 64 m³

3. 216 m³

4. 256 m³

Answer

By formula,

Lateral surface area of cube = 4(side)²

⇒ 4(side)² = 256

⇒ (side)² = 64

⇒ side = $\sqrt{64}$ = 8 m.

Volume of cube = (side)³

= 8³ = 512 m³.

Hence, Option 1 is the correct option.

If the perimeter of one face of a cube is 40 cm, then the sum of lengths of its edge is

1. 80 cm

2. 120 cm

3. 160 cm

4. 240 cm

Answer

Each face of cube is a square. Let length of each side = x cm.

Given, perimeter = 40 cm.

∴ 4x = 40 cm

⇒ x = 10 cm.

There are 12 edges in a cube.

Sum of edges = 12 × 10 = 120 cm.

Hence, Option 2 is the correct option.

A cuboid container has the capacity to hold 50 small boxes. If all the dimensions of the container are doubled, then it can hold (small boxes of same size)

1. 100 boxes

2. 200 boxes

3. 400 boxes

4. 800 boxes

Answer

Let l, b and h be the length, breadth and height of the cuboid container.

Volume = l × b × h = lbh.

If they are doubled then,

New Volume = 2l × 2b × 2h = 8lbh

Hence, volume becomes 8 times.

So, capacity becomes 8 times.

So, container can hold 50 × 8 = 400 boxes.

Hence, Option 3 is the correct option.

The number of planks of dimensions (4 m × 50 cm × 20 cm) that can be stored in a pit which is 16 m long, 12 m wide and 4 m deep is

1. 1900

2. 1920

3. 1800

4. 1840

Answer

Volume of plank = 4 m × 50 cm × 20 cm

= 4 m × 0.50 m × 0.20 m

= 0.4 m³.

Volume of pit = 16 m × 12 m × 4 m

= 768 m³.

No. of planks that can be stored in pit = $\dfrac{\text{Volume of pit}}{\text{Volume of plank}} = \dfrac{768}{0.4}$ = 1920.

Hence, Option 2 is the correct option.

Consider the following two statements:

Statement 1: If the circumference of a circle is 10π cm, then its area is 25π cm².

Statement 2: The area of a circle is π times its circumference.

Which of the following is valid?

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and Statement 2 is false.

4. Statement 1 is false, and Statement 2 is true.

Answer

Given that the circumference is 10π cm.

⇒ 2πr = 10π

⇒ r = $\dfrac{10π}{2π}$

⇒ r = 5 cm

Now, calculate the area using this radius,

⇒ A = πr²

= π.5²

= 25π cm².

∴ Statement 1 is true.

We know that,

Circumference of the circle = 2πr

2πr × π = 2π²r ≠ πr².

∴ Statement 2 is false.

∴ Statement 1 is true, and Statement 2 is false.

Hence, option 3 is the correct option.