Mensuration 3D - Solved Questions and Answers
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.
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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 10 x 15 x 6 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 = (102 + 152 + 62)1/2 = (100 + 225 + 36)1/2 = (361)1/2
Length of the largest rod = 19 m
Question 2: Find the number of bricks of dimension 24 x 12 x 8 cm each that would be required to make a wall 24 m long, 8 m high, and 60 cm thick.
Solution:
Volume of 1 brick = 24 x 12 x 8 = 2304 cm 3
Volume of wall = 2400 x 800 x 60 = 115200000 cm 3
Therefore, number of bricks required = 115200000 / 2304 = 50000
Question 3: A rectangular sheet of paper measuring 22 cm x 7 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 = 22
R = 3.5 cm
Also, height = 7 cm
Therefore, volume of the cylinder = π R2 H = π (3.5)2 7 = 269.5 cm3
Question 4: If each edge of a cube is increased by 10%, what would be the percentage increase in volume?
Solution:
Let the original edge length be 'a'
Original volume = a3
Now, new edge length = 1.1 a
New volume = (1.1 a)3 = 1.331 a3
Increase in volume = 1.331 a3 - 1 a3 = 0.331 a3
Therefore, percentage increase int eh volume = (0.331 a3 / a3) x 100 = 33.1 %
Question 5: Three metal cubes of edge lengths 3 cm, 4 cm, and 5 cm are melted to form a single cube. Find the edge length of such a cube.
Solution:
Volume of new cube = Volume of metal generated on melting the cubes = Sum of volumes of the three cubes
Volume of new cube = 3 3 + 4 3 + 5 3 = 216
Edge length of new cube = (216)1/3 = 6 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.
