Download free pdf of **Mensuration Worksheets for Class 8 with answers**. Students can easily get all the Class 8 Maths Worksheets and Exercises in Free Printable ( PDF Format) as per the latest syllabus and examination pattern in CBSE and NCERT pattern. Below you will find all the Mensuration related sums on Surface Area and Volume, that are asked in 8th Grade Mathematics Exams Across Indian Schools.

## Mensuration Worksheets for Class 8 in PDF Format

CBSE Class 8 Maths Mensuration Worksheet

CBSE Worksheet for Mensuration class 8

Worksheet for Mensurations Class 8 Maths

Mensuration Class 8 Worksheet with Answers

Class 8 Maths Mensuration Word Problems Worksheet

Mensurations Questions For CBSE Class 8 Maths

Class 8 Maths Mensurations Worksheet in PDF

Mensurations CBSE Class 8 Maths Practice Questions

CBSE Class 8 Maths Mensurations Questions

Worksheet on Mensurations for CBSE Class 8 Maths

Class 8 Maths Mensuration Worksheet (1)

Mensuration MCQ Worksheet for Class 8

Surface Area and Volume Class 8 Mensuration Worksheet

Volume Class 8 Mensuration Worksheet

Surface Area Class 8 Mensuration Worksheet

### Class 8 Mensuration Worksheet with answers

**Q.1** Mary wants to decorate her Christmas tree. She wants to place the tree on a wooden box covered with coloured paper with picture of Santa Claus on it. She must know the exact quantity of paper to buy for this purpose. If the box has length, breadth and height as 80 cm, 40 cm and 20 cm respectively how many square sheets of paper of side 40 cm would she require?

**Solution:**

First, we need to find the total surface area of the box:

Total Surface Area (TSA) = 2 * (length * breadth + length * height + breadth * height)

TSA = 2 * (80 cm * 40 cm + 80 cm * 20 cm + 40 cm * 20 cm)

TSA = 2 * (3200 cm² + 1600 cm² + 800 cm²)

TSA = 2 * 5600 cm²

TSA = 11200 cm²

Now, we need to find out how many 40 cm × 40 cm square sheets are required to cover this area:

Area of one sheet = 40 cm * 40 cm = 1600 cm²

Number of sheets required = TSA / Area of one sheet

Number of sheets required = 11200 cm² / 1600 cm² = 7 sheets

**Answer :** Mary would require 7 square sheets of paper.

**Q.2** A wall of length 10 m was to be built across an open ground. The height of the wall is 4 m and thickness of the wall is 24 cm. If this wall is to be built up with bricks whose dimensions are 24 cm × 12 cm × 8 cm, how many bricks would be required?

**Solution :**

First, we need to find the volume of the wall:

Volume of the wall = Length × Height × Thickness Volume of the wall = 10 m × 4 m × 0.24 m (converted cm to meters) Volume of the wall = 9.6 m³

Now, let’s find the volume of one brick:

Volume of one brick = Length × Breadth × Height Volume of one brick = 0.24 m × 0.12 m × 0.08 m (converted cm to meters) Volume of one brick = 0.002304 m³

Now, to find the number of bricks required, divide the volume of the wall by the volume of one brick:

Number of bricks required = Volume of the wall / Volume of one brick Number of bricks required = 9.6 m³ / 0.002304 m³ ≈ 4166.67 (rounded to the nearest whole number)

**Answer:** So, approximately 4,167 bricks would be required.

**Q.3** Find the area of trapezium whose parallel sides are 24 cm and 20 cm and the distance between them is 15 cm.

**Solution :**

To find the area of a trapezium, you can use the formula:

Area = (1/2) × (sum of parallel sides) × (distance between them)

In this case, the parallel sides are 24 cm and 20 cm, and the distance between them is 15 cm.

Area = (1/2) × (24 cm + 20 cm) × 15 cm Area = (1/2) × 44 cm × 15 cm Area = 22 cm × 15 cm Area = 330 cm²

**Answer:** So, the area of the trapezium is 330 cm².

**Q.4** The area of trapezium is 1080 cm^{2}. If the lengths of its parallel sides are 55.6 cm and 34.4 cm, find the distance between them.

**Solution :**

Let the distance between the parallel sides be “h” cm.

Using the formula for the area of a trapezium:

Area = (1/2) × (sum of parallel sides) × (distance between them)

1080 cm² = (1/2) × (55.6 cm + 34.4 cm) × h

1080 cm² = (1/2) × (90 cm) × h

Now, solve for “h”:

1080 cm² = 45 cm × h

h = 1080 cm² / 45 cm

h = 24 cm

**Answer:**So, the distance between the parallel sides is 24 cm.

**Q.5** The area of trapezium is 352 cm^{2} and the distance between its parallel sides is 16 cm. If one of the parallel sides is of length 25 cm, find the length of the other side.

**Solution :**

Let the length of the other parallel side be “x” cm.

Using the formula for the area of a trapezium:

Area = (1/2) × (sum of parallel sides) × (distance between them)

352 cm² = (1/2) × (25 cm + x) × 16 cm

Now, solve for “x”:

352 cm² = 8 cm × (25 cm + x)

44 = 25 + x

x = 44 – 25

x = 19 cm

**Answer:** So, the length of the other parallel side is 19 cm.

**Q.6 Find the lateral surface area of the cuboids whose dimensions are:**

(i) Length = 22cm, breadth = 12 cm and height = 7.5 cm

**Answer:**

Lateral Surface Area = 2 × (length × height + breadth × height)

Lateral Surface Area = 2 × (22 cm × 7.5 cm + 12 cm × 7.5 cm)

Lateral Surface Area = 2 × (165 cm² + 90 cm²)

Lateral Surface Area = 2 × 255 cm²

Lateral Surface Area = 510 cm²

(ii) Length = 15cm, breadth = 6 cm and height = 9 cm

**Answer:**

Lateral Surface Area = 2 × (length × height + breadth × height)

Lateral Surface Area = 2 × (15 cm × 9 cm + 6 cm × 9 cm)

Lateral Surface Area = 2 × (135 cm² + 54 cm²)

Lateral Surface Area = 2 × 189 cm²

Lateral Surface Area = 378 cm²

(iii) Length = 24 m, breadth = 25 cm and height = 6 m

**Answer:**

Lateral Surface Area = 2 × (length × height + breadth × height)

Lateral Surface Area = 2 × (24 m × 6 m + 25 cm × 6 m) (converted cm to meters)

Lateral Surface Area = 2 × (144 m² + 1.5 m²)

Lateral Surface Area = 2 × 145.5 m²

Lateral Surface Area = 291 m²

**Q.7** The circumference of the base of a cylinder is 176 cm and its height is 65 m. Find its curved surface area and total surface area of the cylinder.

**Solution **:

We need to find its curved surface area (CSA) and total surface area (TSA).

First, let’s find the radius of the base using the formula for the circumference of a circle:

Circumference = 2πr

176 cm = 2πr

Now, solve for “r”:

r = 176 cm / (2π) ≈ 28 cm (rounded to the nearest centimeter)

Now, let’s find the curved surface area (CSA) using the formula:

CSA = 2πrh

CSA = 2π × 28 cm × 65 m (converted cm to meters)

CSA = 2 × 3.14 × 28 cm × 65 m

CSA ≈ 114040 cm²

Next, let’s find the total surface area (TSA) of the cylinder:

TSA = 2πr² + 2πrh

TSA = 2π × (28 cm)² + 2π × 28 cm × 65 m (converted cm to meters)

TSA = 2 × 3.14 × (28 cm)² + 2 × 3.14 × 28 cm × 65 m

TSA ≈ 65936 cm² + 57140 cm²

TSA ≈ 123076 cm²

**Answer:** So, the curved surface area is approximately 114040 cm², and the total surface area is approximately 123076 cm².

**Q.8 Find the curved surface area and total surface area of the cylinders whose dimensions are:**

(i). radius = 7 cm and height = 35 cm.

**Answer: **

CSA = 2πrh

CSA = 2π × 7 cm × 35 cm

CSA = 2 × 3.14 × 7 cm × 35 cm

CSA ≈ 1547 cm²

TSA = 2πr² + 2πrh

TSA = 2π × (7 cm)² + 2π × 7 cm × 35 cm

TSA = 2 × 3.14 × (7 cm)² + 2 × 3.14 × 7 cm × 35 cm

TSA ≈ 308 cm² + 1533 cm²

TSA ≈ 1841 cm²

(ii). radius = 14 cm and height = 10 cm.

**Answer: **

CSA = 2πrh

CSA = 2π × 14 cm × 10 cm

CSA = 2 × 3.14 × 14 cm × 10 cm

CSA ≈ 879.2 cm²

TSA = 2πr² + 2πrh

TSA = 2π × (14 cm)² + 2π × 14 cm × 10 cm

TSA = 2 × 3.14 × (14 cm)² + 2 × 3.14 × 14 cm × 10 cm

TSA ≈ 1232.08 cm² + 879.2 cm²

TSA ≈ 2111.28 cm²

(iii). radius = 10 cm and height = 25 cm.

**Answer:**

CSA = 2πrh

CSA = 2π × 10 cm × 25 cm

CSA = 2 × 3.14 × 10 cm × 25 cm

CSA ≈ 1570 cm²

TSA = 2πr² + 2πrh

TSA = 2π × (10 cm)² + 2π × 10 cm × 25 cm

TSA = 2 × 3.14 × (10 cm)² + 2 × 3.14 × 10 cm × 25 cm

TSA ≈ 628 cm² + 1570 cm²

TSA ≈ 2198 cm²

**Q.9 **Find the area of a rhombus whose diagonals are of lengths 15 cm and 25 cm.

**Solution :**

Area = (1/2) × (product of diagonals)

Area = (1/2) × (15 cm × 25 cm)

Area = (1/2) × 375 cm²

Area = 187.5 cm²

**Answer:** So, the area of the rhombus is 187.5 cm².

**Q.10** The area of a rhombus is 360 cm^{2} and one of the diagonals is 18 cm. Find the other diagonal.

**Solution:**

Let the length of the other diagonal be “d” cm.

Using the formula for the area of a rhombus:

Area = (1/2) × (product of diagonals)

360 cm² = (1/2) × (18 cm × d)

Now, solve for “d”:

360 cm² = 9 cm × d

d = 360 cm² / 9 cm

d = 40 cm

**Answer:** So, the length of the other diagonal is 40 cm.

**Q.11 **The diagonal of a quadrilateral shaped field is 25 m and the perpendiculars dropped on it from the remaining opposite vertices are 18 m and 12 m. Find the area of the field.

**Solution:**

Given:

Diagonal of the field (d) = 25 m

Perpendiculars dropped on the diagonal from the remaining opposite vertices = 18 m and 12 m.

Let’s calculate the area of the field:

Area of each triangle = (1/2) × base × height

Area of the first triangle = (1/2) × 18 m × 12 m = 108 m²

Area of the second triangle = (1/2) × 18 m × 12 m = 108 m²

Now, add the areas of both triangles to find the total area of the field:

Total area of the field = Area of the first triangle + Area of the second triangle

Total area of the field = 108 m² + 108 m² = 216 m²

**Answer:** So, the area of the field is 216 square meters.

**Q.12** An aquarium is in the form of a cuboid whose external measures are 40 cm × 15 cm × 20 cm. The base, side faces and back face are to be covered with a coloured paper. Find the area of the paper needed?

**Solution:**

Given external measures of the cuboid:

Length (L) = 40 cm

Breadth (B) = 15 cm

Height (H) = 20 cm

Let’s calculate the total surface area:

Total Surface Area (TSA) = 2(LB + BH + HL)

TSA = 2(40 cm × 15 cm + 15 cm × 20 cm + 20 cm × 40 cm)

TSA = 2(600 cm² + 300 cm² + 800 cm²)

TSA = 2(1700 cm²)

TSA = 3400 cm²

**Answer:** So, the area of the paper needed to cover the aquarium is 3400 square centimeters.

**Q.13 **Hameed has built a cubical water tank with lid for his house, with each outer edge 1.5 m long. He gets the outer surface of the tank excluding the base, covered with square tiles of side 25 cm. Find how much he would spend for the tiles, if the cost of the tiles is Rs 360 per dozen.

**Solution:**

First, calculate the area of one side of the cube (excluding the base):

Area of one side = Length × Height

Area of one side = 1.5 m × 1.5 m = 2.25 m²

Now, convert the area of one side to square centimeters:

1 m² = 10,000 cm²

2.25 m² = 2.25 × 10,000 cm² = 22,500 cm²

Now, calculate the number of tiles required for one side:

Number of tiles = Area to be covered / Area of one tile

Number of tiles = 22,500 cm² / (25 cm × 25 cm) = 22,500 cm² / 625 cm² = 36 tiles

Hameed needs 36 tiles to cover one side of the tank. Since there are 6 sides on a cube, he would need a total of 6 × 36 = 216 tiles to cover the entire outer surface of the tank.

Now, let’s find out how much he would spend:

Cost of one dozen tiles = Rs 360

Cost of one tile = Rs 360 / 12 = Rs 30

Total cost = Cost per tile × Number of tiles

Total cost = Rs 30 × 216 = Rs 6,480

**Answer:** Hameed would spend Rs 6,480 for the tiles.

**Q.14 **The length, breadth and height of a room are 5 m, 4 m and 3 m respectively. Find the cost of white washing the walls of the room and the ceiling at the rate of Rs 7.50 per m^{2}.

**Solution:**

The length, breadth, and height of the room are given as:

Length (L) = 5 m

Breadth (B) = 4 m

Height (H) = 3 m

We want to find the cost of white washing the walls and the ceiling at the rate of Rs 7.50 per square meter.

First, calculate the total surface area of the room, including the walls and the ceiling:

Total Surface Area (TSA) = 2(LB + BH + HL) + 2(LB) [for the ceiling]

TSA = 2(5 m × 4 m + 4 m × 3 m + 3 m × 5 m) + 2(5 m × 4 m)

TSA = 2(20 m² + 12 m² + 15 m²) + 2(20 m²)

TSA = 2(47 m²) + 40 m²

TSA = 94 m² + 40 m²

TSA = 134 m²

Now, calculate the cost of white washing at the rate of Rs 7.50 per square meter:

Cost = Area × Rate

Cost = 134 m² × Rs 7.50/m²

Cost = Rs 1005

**Answer:** So, the cost of white washing the walls and ceiling of the room is Rs 1005.

**Q.15 **The floor of a rectangular hall has a perimeter 250 m. If the cost of painting the four walls at the rate of Rs 10 per m^{2} is Rs 15000, find the height of the hall.

**Solution:**

Let the length of the hall be “L” meters, and let the breadth of the hall be “B” meters.

Given:

Perimeter of the hall = 250 m

Cost of painting the four walls = Rs 15000

Rate of painting = Rs 10 per square meter

Perimeter of a rectangle = 2(L + B)

250 m = 2(L + B)

Now, we can express one of the variables in terms of the other:

L + B = 250 m

We also know that the cost of painting the four walls is Rs 15000, and the rate is Rs 10 per square meter. So, the area of the four walls must be 15000 / 10 = 1500 square meters.

Area of the four walls = 2Lh + 2Bh

Now, we can substitute the value of L + B from the perimeter equation:

2(L + B)h = 1500

2(250 m)h = 1500

500h = 1500

h = 1500 / 500

h = 3 meters

**Answer:** So, the height of the hall is 3 meters.

**Q.16 **The paint in a certain container is sufficient to paint an area equal to 9.375 m^{2}. How many bricks of dimensions 22.5 cm × 10 cm × 7.5 cm can be painted out of this container?

**Solution:**

First, let’s convert the area to square centimeters:

1 square meter = 10,000 square centimeters

9.375 square meters = 9.375 × 10,000 square centimeters = 93,750 square centimeters

Now, let’s calculate the volume of one brick:

Volume of one brick = Length × Breadth × Height

Volume of one brick = 22.5 cm × 10 cm × 7.5 cm = 1687.5 cubic centimeters

Now, let’s find out how many bricks can be painted with the given paint:

Number of bricks = Area to be painted / Area of one brick

Number of bricks = 93,750 square centimeters / 1687.5 cubic centimeters = 55.56 (rounded to the nearest whole number)

**Answer:** So, approximately 56 bricks can be painted out of this container.

**Q.17 **A small indoor greenhouse (herbarium) is made entirely of glass panes (including base) held together with tape. It is 30 cm long, 25 cm wide and 25 cm high.

- What is the area of the glass?
- How much of tape is needed for all the 12 edges?

**Solution:**

Length (L) = 30 cm

Width (W) = 25 cm

Height (H) = 25 cm

We need to find the area of the glass and the amount of tape needed for all the 12 edges.

Area of the Glass:

The greenhouse has 6 rectangular glass panes. Two panes for each of the 3 pairs of opposite faces.

Area of the front and back panes (L × H):

Area = 30 cm × 25 cm = 750 cm²

Area of the two side panes (W × H):

Area = 25 cm × 25 cm = 625 cm²

Area of the top and bottom panes (L × W):

Area = 30 cm × 25 cm = 750 cm²

Now, add up the areas of all 6 panes to find the total area of the glass:

Total area = 2(750 cm²) + 2(625 cm²) + 2(750 cm²) = 3000 cm² + 1250 cm² + 1500 cm² = 5750 cm²

Amount of Tape for Edges:

There are 12 edges in total.

Four edges of length L: 4 edges × 30 cm = 120 cm

Four edges of length W: 4 edges × 25 cm = 100 cm

Four edges of length H: 4 edges × 25 cm = 100 cm

Total tape needed for all edges = 120 cm + 100 cm + 100 cm = 320 cm

**Answer:** So, the area of the glass is 5750 cm², and 320 cm of tape is needed for all the edges.