Mensuration 3D

If a shape is surrounded by a no. of surfaces or planes then it is a 3D shape.These are called Three dimensional as they have depth, breadth and length. Mensuration 3D deals with shapes like cube, cuboid, sphere etc. The problems are generally based on volume and surface area.

Cuboid
Let the length, breadth and height of the cuboid be ‘L’, ‘B’ and ‘H’ respectively.

• Volume = L x B x H
• Curved Surface area = 2 H (L + B)
• Total surface area = 2 (L B + B H + H L)
• Length of diagonal = (L² + B² + H²)^1/2

Cube
Let the side of the cube be ‘a’

• Volume = a³
• Curved Surface area = 4 a²
• Total surface area = 6 a²
• Length of diagonal =  

Cylinder (Right Circular Cylinder)
Let the radius of the base and height of the right circular cylinder be ‘R’ and ‘H’ respectively.

• Volume = π R² H
• Curved Surface area = 2 π R H
• Total surface area = 2 π R H + 2 π R²

          or

• Total surface area = 2 π R (H + R)

Hollow Cylinder (Hollow Right Circular Cylinder)
Let the inner radius of the base, outer radius of the base and height of the hollow right circular cylinder be ‘r’, ‘R’ and ‘H’ respectively.

• Volume = π H (R² – r²)
• Curved Surface area = 2 π R H + 2 π r H = 2 π H (R + r)
• Total surface area = 2 π H (R + r) + 2 π (R² – r²)

Cone
Let the radius of the base, slant height and height of the cone be ‘R’, ‘L’ and ‘H’ respectively.

• L² = R² + H²
• Volume = π R² H / 3
• Curved Surface area = π R L
• Total surface area = π R L + π R²

Sphere
Let the radius of the sphere be ‘R’

• Volume = (4 / 3) π R³
• Surface area = 4 π R²

Hemisphere
Let the radius of the hemisphere be ‘R’

• Volume = (2 / 3) π R³
• Curved Surface area = 2 π R²
• Total Surface area = 3 π R²

Please note that whenever it is mentioned to find “Surface Area”, we calculate the total surface area.

Some Examples and solutions
Example 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

Example 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 cm³
Volume of wall = 2400 x 800 x 60 = 115200000 cm³
Therefore, number of bricks required = 115200000 / 2304 = 50000

Example 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

Example 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 %

Example 5 : Three metal cubes of edge length 3 cm, 4 cm, 5 cm are melted to form a single cube. Find the edge length of such 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 = 3³ + 4³ + 5³ = 216
=> Edge length of new cube = (216)^1/3 = 6 cm