Class 10 — Mathematics (NCERT)

Surface Areas and Volumes: Combinations and Conversions of Solids

Class 10

✓By the end of this lesson students will be able to calculate the surface area of solids formed by combining two basic solids.
✓By the end of this lesson students will be able to calculate the volume of solids formed by combining two basic solids.
✓By the end of this lesson students will be able to determine the volume of a new solid formed by reshaping a given solid.
✓By the end of this lesson students will be able to solve problems involving the number of smaller solids formed from a larger solid by conversion.

Key concepts

Combinations of Solids - Surface Area

When two or more basic solids are combined to form a new solid, its surface area is the sum of the exposed surface areas of the individual constituent solids. It is crucial to remember that the surfaces where the solids are joined together are not exposed and hence are not included in the total surface area calculation. For example, if a cone is placed on a hemisphere, the base of the cone and the top circular surface of the hemisphere are hidden and thus not part of the total surface area.

Combinations of Solids - Volume

Unlike surface area, when two or more basic solids are combined to form a new solid, its volume is simply the sum of the volumes of the individual constituent solids. The volume of the combined solid does not depend on how the solids are joined, only on the total space they occupy.

V_total = V_1 + V_2 + ... + V_n

Conversion of Solids

When a solid of a certain shape is melted and recast into another shape, or when a larger solid is melted and recast into multiple smaller solids, the total volume of the material remains constant. This principle is known as the conservation of volume. The volume of the original solid (or solids) will be equal to the volume of the new solid (or solids) formed.

V_original = V_new (or V_large = n × V_small, where 'n' is the number of smaller solids)

Key facts to remember

1The surface area of a solid formed by combining basic solids is the sum of the exposed surface areas of its constituent parts.
2The volume of a solid formed by combining basic solids is the sum of the volumes of its constituent parts.
3When a solid is converted from one shape to another, its volume remains constant.
4Ensure all dimensions are in consistent units before performing calculations.
5Memorise the formulas for surface areas and volumes of basic solids: cube, cuboid, cylinder, cone, sphere, and hemisphere.
6Use the value of π (22/7 or 3.14) as specified in the problem, or 22/7 if not specified.

Worked examples

Example 1

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. (Take π = 3.14)

IGiven: Height of cone (h) = 2 cm, Diameter of base = 4 cm.

IIRadius of cone (r) = Diameter / 2 = 4 cm / 2 = 2 cm.

IIISince the cone is surmounted on a hemisphere, the radius of the hemisphere (r) will also be 2 cm.

IVVolume of cone (V_cone) = (1/3)πr²h

VV_cone = (1/3) × 3.14 × (2)² × 2

VIV_cone = (1/3) × 3.14 × 4 × 2

VIIV_cone = (1/3) × 25.12 = 8.373... cm³

VIIIVolume of hemisphere (V_hemisphere) = (2/3)πr³

9V_hemisphere = (2/3) × 3.14 × (2)³

10V_hemisphere = (2/3) × 3.14 × 8

11V_hemisphere = (2/3) × 25.12 = 16.746... cm³

12Total Volume of the toy (V_total) = V_cone + V_hemisphere

13V_total = 8.373... + 16.746...

14V_total = 25.12 cm³

Answer

The volume of the toy is 25.12 cm³.

Example 2

A decorative block is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Take π = 22/7)

IGiven: Edge of cube (a) = 5 cm, Diameter of hemisphere = 4.2 cm.

IIRadius of hemisphere (r) = Diameter / 2 = 4.2 cm / 2 = 2.1 cm.

IIITotal surface area of the cube = 6a² = 6 × (5)² = 6 × 25 = 150 cm².

IVWhen the hemisphere is fixed on the cube, the area of the base of the hemisphere is covered.

VArea of base of hemisphere = πr² = (22/7) × (2.1)² = (22/7) × 2.1 × 2.1 = 22 × 0.3 × 2.1 = 13.86 cm².

VICurved surface area of hemisphere (CSA_hemisphere) = 2πr² = 2 × (22/7) × (2.1)² = 2 × 13.86 = 27.72 cm².

VIITotal surface area of the block = (Total surface area of cube) - (Area of base of hemisphere) + (Curved surface area of hemisphere)

VIIITotal surface area = 150 - 13.86 + 27.72

9Total surface area = 136.14 + 27.72

10Total surface area = 163.86 cm².

Answer

The total surface area of the block is 163.86 cm².

Remember to subtract the area of the base of the hemisphere from the cube's surface area, as it is covered.

Example 3

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

IGiven: Radius of sphere (r_sphere) = 4.2 cm, Radius of cylinder (r_cylinder) = 6 cm.

IILet the height of the cylinder be h.

IIIAccording to the principle of conservation of volume, Volume of sphere = Volume of cylinder.

IVVolume of sphere = (4/3)π(r_sphere)³

VVolume of sphere = (4/3) × π × (4.2)³

VIVolume of cylinder = π(r_cylinder)²h

VIIVolume of cylinder = π × (6)² × h

VIIIEquating the volumes:

9(4/3) × π × (4.2)³ = π × (6)² × h

10Cancel π from both sides:

11(4/3) × (4.2)³ = (6)² × h

12(4/3) × 4.2 × 4.2 × 4.2 = 36 × h

134 × 1.4 × 4.2 × 4.2 = 36 × h

1423.52 × 4.2 = 36 × h

1598.784 = 36 × h

16h = 98.784 / 36

17h = 2.744 cm

Answer

The height of the cylinder is 2.744 cm.

It is often beneficial to keep π as a symbol until the final calculation step to simplify calculations and maintain accuracy.

Common mistakes

✗Adding the total surface areas of individual solids instead of only the exposed surfaces when calculating the surface area of a combined solid.
✗Forgetting to subtract the area of the common base/joining surface when calculating the surface area of combined solids.
✗Using incorrect formulas for the surface area or volume of basic geometric shapes.
✗Not converting all dimensions to the same unit before starting calculations, leading to incorrect results.
✗Making calculation errors, especially when dealing with fractions, decimals, and the value of π.

Exam tips

★Always draw a neat diagram of the solid(s) described in the problem to visualise the shapes and identify the exposed surfaces clearly.
★List all given dimensions and clearly identify what needs to be calculated (surface area, volume, height, number of solids, etc.).
★Write down the relevant formulas for each basic solid involved before substituting values.
★In conversion problems, equate the volumes of the original and new solids. It is often helpful to keep π as a symbol and cancel it out if it appears on both sides of the equation, simplifying calculations.
★Double-check your calculations and ensure the final answer includes appropriate units.

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