Mensuration 3D Solved Questions and Answers (original) (raw)

Last Updated : 21 Apr, 2026

3D Mensuration deals with the measurement of three-dimensional (solid) shapes, including their volume, surface area, lateral surface area, and diagonals. 3D shapes have length, width, and height/depth, making them occupy space.

**Mensuration 3D Formulas

**3D Mensuration questions and answers are provided below for you to learn and practice.

**Question 1: Find the length of the largest rod that can be kept in a cuboidal room of dimensions 12 x 16 x 9 m.

**Solution:

Largest rod would lie along the diagonal.
Length of largest rod = Length of diagonal of the room = (L2 + B2 + H2)1/2
Length of the largest rod = (122 + 162 + 92)1/2 = (144 + 256 + 81)1/2 = (481)1/2
Length of the largest rod = 21.9 m

**Question 2: Find the number of bricks of dimension 20 x 10 x 8 cm each that would be required to make a wall 10 m long, 5 m high, and 40 cm thick.

**Solution:

Volume of 1 brick = 20 x 10 x 8 = 1600 cm 3
Volume of wall = 1000 x 500 x 40 = 20000000 cm 3
Therefore, number of bricks required = 20000000 / 1600 = 12500

**Question 3: A rectangular sheet of paper measuring 44 cm x 10 cm is rolled along the longer side to make a cylinder. Find the volume of the cylinder formed.

**Solution:

Let the radius of the cylinder be 'R'.
The sheet is rolled along the longer side.
2 π R = 44
R = 7 cm
Also, height = 10 cm
Therefore, volume of the cylinder = π R2 H = π × 72 × 10 = 1540 cm3

**Question 4: If each edge of a cube is increased by 20%, what would be the percentage increase in volume?

**Solution:

Let the original edge length be 'a'
Original volume = a3
Now, new edge length = 1.2 a
New volume = (1.2 a)3 = 1.728 a3
Increase in volume = 1.728 a3 - a3 = 0.728 a3
Therefore, percentage increase in the volume = (0.728 a3/ a3) x 100 = 72.8%

**Question 5: Four metal cubes of edge lengths 2 cm, 4 cm, 6 cm, and 8 cm are melted to form a single cube. Find the edge length of the new cube.

**Solution:

Volume of new cube = Volume of metal generated on melting the cubes = Sum of volumes of the four cubes
Volume of new cube = 2 3 + 4 3 + 6 3 + 83= 800
Edge length of new cube = (800)1/3 = 9.28 cm

**Question 6: Find the length of a 1.25 m wide metal sheet required to make a conical machine of radius 7 m and height 24 m.

**Solution:

The sheet would be shaped into a cone.
Area of sheet = Area of conical machine
1.25 x Length = π x R x L
1.25 x Length = π x R x (72 + 242)1/2
1.25 x Length = π x 7 x 25
Length = 440 m
Thus, **440 m long metal sheet is required to make the conical machine.

**Question 7: From a cylindrical vessel having a radius of the base of 7 cm and a height of 6cm, water is poured into small hemispherical bowls, each of radius 3.5 cm. Find the minimum number of bowls that would be required to empty the cylindrical vessel.

**Solution:

Volume of cylindrical vessel = π R2 H = π (72) 6 = 924 cm3
Volume of each bowl = (2 / 3) π R3 = (2 / 3) π 3.53 = 269.5 / 3
Number of bowls required = (924) / (269.5 / 3) = 10.28
But since a number of bowls cannot be in fractions, we need at least **11 such bowls to empty the cylindrical vessel.

**Question 8: A cone has a radius of 5 cm and a height of 12 cm. Find the volume and surface area of the cone. Where the radius is 5cm and the height is 12cm. Use the 3.14 value of π.

**Solution:

Volume of the cone V=\frac{1}{3}​×3.14×(5)^2×12

V= \frac{1}{3}​×3.14×25×12
V=\frac{1}{3}​×3.14×300
V=\frac{1}{3}​×942
V=314cm^3

So, the volume of the cone is **314 cm³.

**Surface Area of the cone
A = \pi r ( r + \sqrt{r^2 + h^2})

the slant height l using the Pythagorean theorem:
l= \sqrt{r^2+h^2}
l= \sqrt{25+144}
l= \sqrt{169}
l= 13

A=3.14×5(5+13)
A=3.14×5×18
A=3.14×90
A=282.6cm^2

So, the surface area of the cone is **282.6 cm².