Class 10 Mathematics — Chapter 12

Surface Areas and Volumes: Combination of Solids

1. Core Idea

Many real-life objects are formed by joining two or more basic solids, such as:

• Cuboid
• Cube
• Cylinder
• Cone
• Sphere
• Hemisphere

The key skill is to split a complicated solid into familiar solids, then calculate its surface area or volume accordingly.

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2. Surface Area of a Combination of Solids

Golden Rule

For surface area, do not simply add the total surface areas of all component solids.

Why? When two solids are joined, the surfaces touching each other become internal and are no longer exposed.

How to solve

1. Identify the individual solids.
2. Draw/visualise the combined object.
3. Identify only the exposed surfaces.
4. Add their areas.
5. Exclude surfaces hidden at the joining.

Important examples

Cylinder + two hemispheres

Only the curved surfaces are exposed:$$\text{TSA}=2\pi rh+2(2\pi r^2)$$

So,$$\boxed{\text{TSA}=2\pi rh+4\pi r^2}$$

Cone + hemisphere

If their common circular faces are joined:$$\boxed{\text{TSA}=\pi rl+2\pi r^2}$$

where $l$ is the cone’s slant height.

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3. Essential Surface-Area Formulas

Solid	Curved/Lateral Surface Area	Total Surface Area
Cube, side $a$	$4a^2$	$6a^2$
Cuboid	$2h(l+b)$	$2(lb+bh+hl)$
Cylinder	$2\pi rh$	$2\pi r(h+r)$
Cone	$\pi rl$	$\pi r(l+r)$
Sphere	—	$4\pi r^2$
Hemisphere	$2\pi r^2$	$3\pi r^2$

For a cone:$$\boxed{l=\sqrt{r^2+h^2}}$$

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4. The Most Important Surface-Area Trick

Suppose a hemisphere is attached to the top of a cube.

The circular portion of the cube covered by the hemisphere is not exposed.

Therefore:$$\boxed{\text{Required area} =\text{TSA of cube}-\text{covered circular area} +\text{CSA of hemisphere}}$$

This “subtract hidden surface + add newly exposed surface” idea is one of the most important concepts in the chapter.

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5. Different Parts May Need Different Areas

A combined solid may have different portions painted with different colours.

In that case, calculate the exposed area for each portion separately.

Example: Cone mounted on cylinder

If the cone’s base is wider than the cylinder’s base, the exposed ring between them is also painted.

Thus:$$\boxed{\text{Cone-painted area} =\text{CSA of cone}+\text{area of larger base}-\text{area of covered base}}$$

For the cylinder:$$\boxed{\text{Cylinder-painted area} =\text{CSA of cylinder}+\text{exposed base area}}$$

The textbook’s rocket example illustrates exactly this situation.

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6. Volume of a Combination of Solids

Golden Rule

Unlike surface area, volumes of joined solids are normally added, because joining does not make the volume of either component disappear.$$\boxed{\text{Volume of combined solid} =\text{sum of volumes of its parts}}$$

For example:

Cuboid + half-cylinder

$$\boxed{V=V_{\text{cuboid}}+\frac12V_{\text{cylinder}}}$$

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7. Essential Volume Formulas

Solid	Volume
Cube	$a^3$
Cuboid	$lbh$
Cylinder	$\pi r^2h$
Cone	$\frac13\pi r^2h$
Sphere	$\frac43\pi r^3$
Hemisphere	$\frac23\pi r^3$

Quick memory pair

$$\boxed{V_{\text{cone}}=\frac13\pi r^2h}$$$$\boxed{V_{\text{hemisphere}}=\frac23\pi r^3}$$

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8. Capacity Problems

Capacity is essentially the volume of the space available to hold something.

For a cylindrical container:$$V=\pi r^2h$$

But if some part of the container is occupied by a raised/depressed solid, adjust the volume.

Raised hemisphere inside a glass

$$\boxed{\text{Actual capacity} =\text{Apparent capacity}-\text{volume of raised hemisphere}}$$

The chapter demonstrates this with a cylindrical glass containing a hemispherical projection at its bottom.

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9. Empty Space / Remaining Volume

When one solid is placed inside another:$$\boxed{\text{Remaining volume} =\text{Volume of container}-\text{Volume of object}}$$

This is especially useful in:

• water displacement
• objects placed inside cylinders
• cavities
• hollowed solids
• circumscribing solids

Water displacement principle

If an object is completely immersed in water, the volume of water displaced equals the volume of the immersed object.

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10. Mass from Volume

If density/mass per unit volume is given:$$\boxed{\text{Mass}=\text{Volume}\times\text{mass per unit volume}}$$

For example, if $1\text{ cm}^3$ of iron has mass $8$ g:$$\boxed{\text{Mass}=8\times V\text{ grams}}$$

The chapter applies this idea to a composite iron pole.

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11. Unit Conversion — Don’t Lose Marks

Before calculating, make all measurements use the same unit.

Remember:$$1\,m=100\,cm$$

Therefore:$$1\,m^2=10,000\,cm^2$$

and$$1\,m^3=1,000,000\,cm^3$$

For capacity:$$1\,cm^3=1\,mL$$$$1000\,cm^3=1\,L$$

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12. Problem-Solving Method

For almost every question in this chapter, use this sequence:

IDENTIFY → SPLIT → FORMULA → ADD/SUBTRACT → UNIT

1. Identify the component solids.
2. Split the object mentally into those solids.
3. Write the appropriate formula for each part.
4. Add exposed areas or volumes; subtract hidden/removed parts when required.
5. Check units and give the final answer clearly.

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13. Surface Area vs Volume — Most Important Difference

Surface Area	Volume
Deals with exposed boundary	Deals with space occupied
Joining can hide surfaces	Joining does not normally remove volume
Consider only exposed surfaces	Add volumes of components
Hidden/covered areas must be excluded	Component volumes are generally added

One-line memory trick:

Surface area → think about what you can see.
Volume → think about how much space it occupies.

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14. High-Value Exam Points

Remember these especially:

• Joined surfaces are not counted in the external surface area.
• For a cone, calculate slant height using$$l=\sqrt{r^2+h^2}$$
• In a composite volume, add component volumes.
• In a cavity/depression problem, subtract the removed volume.
• In a container with an internal projection, subtract the projection’s volume from the apparent capacity.
• In painting problems, calculate only the surfaces actually painted.
• Always check whether a circular base is exposed, covered, or partially exposed.
• Convert all dimensions to compatible units before calculation.

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🧠 Last-Minute Revision Sheet

$$\boxed{\text{Cylinder CSA}=2\pi rh}$$$$\boxed{\text{Cone CSA}=\pi rl}$$$$\boxed{\text{Hemisphere CSA}=2\pi r^2}$$$$\boxed{\text{Sphere SA}=4\pi r^2}$$$$\boxed{\text{Cylinder Volume}=\pi r^2h}$$$$\boxed{\text{Cone Volume}=\frac13\pi r^2h}$$$$\boxed{\text{Hemisphere Volume}=\frac23\pi r^3}$$$$\boxed{\text{Sphere Volume}=\frac43\pi r^3}$$$$\boxed{\text{Combined Volume}=\text{sum of component volumes}}$$$$\boxed{\text{Required surface area}=\text{exposed surfaces only}}$$.