Chapter 11 Mensuration mcqs & important questions mathematics | class 8th

MCQ Questions for Class 8 Maths: Ch 11 Mensuration

1. In a quadrilateral, half of the product of the sum of the lengths of parallel sides  and the parallel distance between them gives the area of

(a) rectangle

(b) parallelogram

(c) triangle

(d) trapezium

► (d) trapezium

2. Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm. 

(a) 215 cm³

(b) 172 cm³

(c) 150 cm³

(d) 168 cm³

►(d) 168 cm³

3. Find the volume of a cuboid whose length is 8 cm, breadth 6 cm and height 3.5 cm.
(a) 168 cm²
(b) 168 cm³
(c) 215 cm³
(d) 150 cm³
► (b) 168 cm³

4. The formula for finding total surface area of cuboid is  

(a) 2 (lb x bh x hl)

(b) 2 (lb + bh + hl)

(c) 2h (l + b)

(d) 2 lb (bh + hl)

► (b) 2 (lb + bh + hl)

5. Which of the following is an example of two dimensions  

(a) cuboid

(b) cone

(c) sphere

(d) circle

► (d) circle

6. The area of a trapezium is  

(a) 1/2 (sum of parallel sides) × h

(b) 2 (sum of parallel sides) × h

(c) (sum of parallel sides) × h

(d) 1/2 (sum of parallel sides) + h

► (a) 1/2 (sum of parallel sides) × h

7. The formula for finding lateral surface area  of cylinder is   

(a) 2πrh

(b) πr²

(c) 2πr(r+h)

(d) 2πr

► (a) 2πrh

8. Solid figures are
(a) 2 D
(b) 3 D
(c) 1 D
(d) 4 D
► (b) 3 D

9. A rectangular paper of width 7 cm is rolled along its width and a cylinder of radius 20 cm is formed. Find the volume of the cylinder.

(a) 8800 cm³
(b) 8800 cm
(c) 8800 cm²
(d) none of these
► (a) 8800 cm³

10. Find the total surface area of a cube whose volume is 343 cm3.
(a) 350 cm²
(b) 294 cm²
(c) 494 cm²
(d) 200 cm²
► (b) 294 cm2

11. Two dimensional figure is a  

(a) solid figure

(b) plane figure

(c) cylinder figure

(d) None of these

► (b) plane figure

12. A cylindrical tank has a capacity of 5632 m³. If the diameter of its base is 16 m, find its depth.

(a) 66m

(b) 30 m

(c) 26 m

(d) 28 m

► (d) 28 m

13. The length of parallel sides of trapezium is 14 cm and 6 cm and its height is 5 cm. Its area will be

(a) 50 cm²

(b) 100 cm²

(c) 210 cmsup>2

(d) 10 cm²

► (a) 50 cm²

14. The amount of space occupied by a three dimensional objects is called its
(a) area
(b) surface area
(c) volume
(d) lateral surface area
► (c) volume

15. Surface area of a cuboid = __________
(a) 2 h (l + b)
(b) 2lbh
(c) 2(lb + bh + hl)
(d) None of these
► (c) 2(lb + bh + hl)

16. Find the height of cuboid whose volume is 490 cmsup>3

and base area is 35 cmsup>3.
(a) 12 cm
(b) 14 cm
(c) 10 cm
(d) 16 cm
► (b) 14 cm

17. The cost of papering the wall of a room, 12 m long, at the rate of Rs. 1.35 per square meter is Rs. 340.20. The cost of matting the floor at Re. 0.85 per square metre is Rs. 91.80. Find the height of the room.
(a) 12 m
(b) 8 m
(c) 6 m
(d) 10 m
► (c) 6 m

18. The area of four walls of the room is  

(a) 2 (lb + bh + hl) 

(b) 2l (h + b) 

(c) 2 (lb x bh x hl) 

(d) 2h (l + b)

► (d) 2h (l + b)

19. The formula for lateral surface area of cuboid is  

(a) 2h (l + b)

(b) 2l (h + b)

(c) 2b (l + h)

(d) 2 (lb + bh + hl)

► (a) 2h (l + b)

20. Diagonals of rhombus are

(a) equal

(b) half of one diagonal

(c) of different length

(d) none of above

► (c) of different length

Short Answer Type Questions:

1. The parallel sides of a trapezium measure 12 cm and 20 cm. Calculate its area if the distance between the parallel lines is 15 cm.

Solution:

Area of trapezium = ½ × perpendicular distance between parallel sides × sum of parallel sides

= ½ × 15 × (12 + 20)

= 1/2 × 15 × 32

= 15 × 16

= 240  cm²

2. Calculate the height of a cuboid which has a base area of 180 cm² and volume is 900 cm³.

Solution:

Volume of cuboid = base area × height

900 = 180 × height

So, height = 900/180 = 5 cm

3. A square and a rectangle have the same perimeter. Calculate the area of the rectangle if the side of the square is 60 cm and the length of the rectangle is 80 cm.

Solution:

Perimeter of square formula = 4 × side of the square

Hence, P (square) = 4 × 60 = 240 cm

Perimeter of rectangle formula = 2 × (Length + Breadth)

Hence, P (rectangle) = 2 (80 + Breadth)

= 160 + 2 × Breadth

According to the given question,
160 + 2 × Breadth = 240 cm
2 × Breadth = 240 – 160
Breadth = 80/2
The breadth of the rectangle = 40 cm

Now, the area of rectangle = Length × Breadth = 80 × 40 = 3200 cm²

4. A lawnmower takes 750 complete revolutions to cut grass on a field. Calculate the area of the field if the diameter of the lawnmower is 84 cm and length is 1 m.

Solution:

Given, length of lawnmower = 1m = 100cm

Its circumference = π × D = 22/7 × 84 = 264 cm

Length of field will be = 264 × 750 = 198000 cm

Here, the width of field = length of the lawnmower i.e. 100 cm

So, area of field = 198000 × 100 = 19,800,000 cm²

Or, 1980 m²

5. The area of a rhombus is 16 cm² and the length of one of its diagonal is 4 cm. Calculate the length of other the diagonal.

Solution:

Area of rhombus = ½ × d₁ × d₂

⇒ 16 = ½ × 4 × d₂

So, d₂ = 32/4 = 8 cm

Long Answer Type Questions:

6. From a circular sheet of radius 4 cm, a circle of radius 3 cm is cut out. Calculate the area of the remaining sheet after the smaller circle is removed.

Solution:

The area of the remaining sheet after the smaller circle is removed will be = Area of the entire circle with radius 4 cm – Area of the circle with radius 3 cm

We know,

Area of circle = πr²

So,

Area of the entire circle = π(4)² = 16π cm²

And,

Area of the circle with radius 3 cm which is cut out = π(3)² = 9π cm²

Thus, the remaining area = 16π – 9π = 7π cm²

7. A cuboidal box of dimensions 1 m × 2 m × 1.5 m is to be painted except its bottom. Calculate how much area of the box has to be painted.

Solution:

Given,

Length of the box, l = 2 m,

Breadth of box, b = 1 m

Height of box, h = 1.5 m

We know that the surface area of a cuboid = 2(lb + lh + bh)

But here the bottom part is not to be painted.

So,

Surface area of box to be painted = lb + 2(bh + hl)

= 2 × 1 + 2 (1 × 1.5 + 1.5 × 2)

= 2 + 2 (1.5 + 3.0)

= 2 + 9.0

= 11

Hence, the required surface area of the cuboidal box = 11 m²

8. In a trapezium, the parallel sides measure 40 cm and 20 cm. Calculate the area of the trapezium if its non-parallel sides are equal having the lengths of 26 cm.

Solution:

From the question statement draw the diagram.

Consider a trapezium of ABCD. Let AB and DC be the parallel sides as shown in the figure.

Now, CM will be the distance between the two parallel sides or the height of the trapezium.

We know,

Area of trapezium = ½ × sum of parallel sides × height.

So, height has to be found.

In the diagram, draw CL || AD

Now, ALCD is a parallelogram ⇒ AL = CD = 20 cm and CL = AD = 26 cm

As AD = CB,

CL = CB ⇒ ΔCLB is an isosceles triangle with CB as its height.

Here, BL = AB – AL = (40 – 20) = 20 cm. So,

LM = MB = ½ BL = ½ × 20 = 10 cm

Now, in ΔCLM,

CL² = CM² + LM² (Pythagoras Theorem)

26² = CM² + 10²

CM² = 26² – 10²

Using algebraic identities, we get; 26² – 10² = (26 – 10) (26 + 10)

hence,

CM² = (26 – 10) (26 + 10) = 16 × 36 = 576

CM = √576 = 24 cm

Now, the area of trapezium can be calculated.

Area of trapezium, ABCD = ½ × (AB + CD) × CM

= ½ × (20 + 40) × 24

Or, Area of trapezium ABCD = 720 cm²