Class 10 - Volume of Combined Solids

Understanding the volume of combined solids is an essential skill in geometry to deal with real-world objects that are made of multiple shapes. Instead of treating a complex figure as a whole, you can break it down into simple solids like cubes, cylinders, cones, or spheres, calculate their individual volumes, and then combine or subtract them as needed. In this guide, you’ll learn clear step-by-step methods, useful formulas, and practical examples to help you confidently solve problems involving 3D shapes.

Table of Contents

• What are Combined Solids
• Volume of Combined Solids Formula
• Solved Examples on Volume of Combined Solids
• Practice Questions on Volume of Combined Solids

 

What are Combined Solids

A combined solid, sometimes called a 'composite solid', is formed when two or more basic three-dimensional shapes like cones, cylinders, spheres, hemispheres, or cuboids are attached to create a single unified object. The resulting shape carries the geometric properties of each part it is assembled from.

Real Life Examples of Combined Solids:

Here are a few real-life examples of combined solids:

• Ice cream cone: A hemisphere sitting on top of a cone

• Medicine capsule: A cylinder with a hemisphere at each end

• Camping tent: A cone placed on top of a cylinder

• Sharpened pencil: A cylinder with a conical tip

• Spinning top: A hemisphere combined with a cone below it

Volume of Combined Solids Formula

The formula for the volume of a combined solid is: V=V1+V2+V3+…

where  V1,V2,V3,…are the volumes of the individual solid parts.

To find the volume of a composite solid, we use what we already know about prisms, cylinders, and other solids. We divide the solid into parts we recognise and then use the volume formulas for those parts.

• If the solid is built by combining shapes, we add the volumes.

 

• When a solid is removed from another, then you subtract the volumes.  V=Vouter−Vremoved.

 

• When a solid is melted and poured into a mould of a different shape, the volume is conserved: Volume before melting = Volume after recasting. Material is not gained or lost. It is just rearranged.

Volume formula for individual solids:

 

Solved Examples on Volume of Combined Solids 

Example 1: A solid is formed by placing a cone on top of a cylinder. Both shapes share the same base radius of 7 cm. The cylinder is 15 cm tall, and the cone is 9 cm tall. Find the total volume of the solid. (Use π = 22/7)

Solution: Given r = 7 cm , height of cone = 9 cm and height of cylinder = 15 cm.

 V_{cylinder} = πr^{2}h = (22/7) × 7^{2} × 15

= (22/7) × 49 × 15 = 22 × 7 × 15 = 2310  cm3

Vcone=(1/3)×πr2h=(1/3)×(22/7)×49×9

= (1/3) × 1386 = 462  cm3

 V=Vcylinder+Vcone= 2310 + 462 = 2772  cm3

∴ Total volume of the combined solid = 2772  cm3.

Example 2: An ice cream has a hemispherical scoop resting on a cone. The diameter of both the hemisphere and the cone's base is 14 cm. The total height of the ice cream (cone + hemisphere) is 21 cm. Find the total volume. (Use π = 22/7)

Solution: Given radius (r) = 14 ÷ 2 = 7 cm and total height = 21 cm. 

Height of cone = total height − radius of hemisphere = 21 − 7 = 14 cm

Vhemisphere=(2/3)×πr3=(2/3)×(22/7)×73

= (2/3) × (22/7) × 343 = (2/3) × 1078 ≈ 718.67  cm3

  Vcone = (1/3) × π r2 h = (1/3) × (22/7) × 49 × 14

= (1/3) × 2156 ≈ 718.67 cm3

V = 718.67 + 718.67 ≈ 1437.3  cm3

∴ Total volume of the ice cream = 1437.3  cm3.

 

Example 3: A tent is in the form of a cylinder surmounted by a cone. The base diameter of the tent is 12 m. The cylindrical part is 5 m tall, and the cone has a slant height of 10 m. Find the total volume of air inside the tent. (Use π = 3.14)

Solution: Given r = 12 ÷ 2 = 6 m, slant height l = 10 m for the cone

Height of cone =  √(l2−r2) = √(100 − 36) = √64 = 8 m

  Vcylinder =  πr2h = 3.14 × 36 × 5 = 565.2 m^{3}

  Vcone= (1/3) × πr2h= (1/3) × 3.14 × 36 × 8 = (1/3) × 904.32 = 301.44  m3

Total volume = V =  Vcylinder+Vcone = 565.2 + 301.44 = 866.64 m3

∴ Total volume of the air in the tent =  866.64  m3

 

Example 4: A wooden article is made by scooping out a hemisphere from each end of a solid cylinder. The height of the cylinder is 10 cm, and the radius is 3.5 cm. Find the volume of wood remaining. (Use π = 22/7)

Solution: Given r = 3.5 cm and h = 10 cm.

  Vcylinder=πr2h = (22/7) ×  (3.5)2× 10 = (22/7) × 12.25 × 10 = 385 cm3

  Vhemisphere = (2/3) × πr3 = (2/3) × (22/7) ×  (3.5)3 = (2/3) × (22/7) × 42.875 ≈ 89.83 cm3

The volume of two hemispheres removed = 2 × 89.83 = 179.67 cm3

Remaining wood = V = 385 − 179.67 = 205.33  cm3

∴ volume of remaining wood = 205.33 cm3

 

Example 5: A metallic sphere of radius 4.2 cm is melted and recast into a cylinder of radius 6 cm. Find the height of the cylinder. (Use π = 22/7)

Solution: When a solid is melted and recast, volume is conserved:

Volume of sphere = Volume of cylinder

Vcylinder=(4/3)×πr3

 = (4/3) × (22/7) × (4.2)3

= (4/3) × (22/7) × 74.088 ≈ 310.46  cm3
Equate the volume of sphere to volume of cylinder and solve for h

πr2h=310.46

(22/7) × 36 × h = 310.46

113.14 × h = 310.46

h = 310.46 ÷ 113.14 ≈ 2.74 cm

∴ height of the cylinder = 2.74 cm

 

Practice Questions on Volume of Combined Solids 

1. A gulab jamun is shaped roughly like a cylinder with a hemispherical end on each side. The total length of the gulab jamun is 5 cm, and its diameter is 2.8 cm. Find its approximate volume.

2. A metallic cone of radius 4 cm and height 9 cm is melted and recast into a solid sphere. Find the radius of the sphere formed.

3. Two solid cubes, each with a side of 5 cm, are joined end to end to form a cuboid. What is the total volume of the resulting cuboid?