MathsICSE Class 10 · Area and Volume of Solids – Cylinder, Cone and Sphere

Combination of Solids: Surface Area and Volume of Cylinders, Cones and Hemispheres Joined Together (ICSE Class 10)

Updated 15 October 2026 · Fact-checked

For a solid made of two or more shapes, the volume is the sum of the parts (minus any part removed). The surface area is only what you can touch from outside: add the exposed curved and flat faces, and leave out the faces where the parts are joined. When parts are separated, the hidden faces reappear.

In a joined solid the dashed joins are inside and not counted; once separated, every piece shows its own flat face and must be counted.

Illustration drawn for QuestionBuddy.

Understand Combination of Solids

Combined solids use the cylinder, cone, sphere and hemisphere formulas from the CISCE syllabus (no frustum). The skill is deciding which faces are on the outside.

Volume: add or subtract.

Joined solids: $V = V_{1} + V_{2} + \dots$
A part scooped or hollowed out: $V = V_{outer} - V_{removed}$ .

Surface area: count the visible faces.

Cone on a cylinder (same $r$ ): cone CSA + cylinder CSA + one base.
Toy, cone on a hemisphere: $π r l + 2 π r^{2}$ .
Capsule, a cylinder with a hemisphere at each end: $2 π r h + 4 π r^{2}$ .
Hemisphere scooped from each end of a cylinder: $2 π r h + 4 π r^{2}$ .
Cone removed from one end of a cylinder: $2 π r h + π r^{2} + π r l$ .

Where two parts meet, the circle of contact is inside the solid and is not counted. When something is scooped out, the flat circle disappears but the new curved surface (inside the hollow) is added.

Separating a solid. If a capsule is pulled apart, each piece is a complete solid with its own flat faces: cylinder $2 π r (r + h)$ and each hemisphere $3 π r^{2}$ . In 2025 Q5(iii) the total for the three parts was $308 + 308 = 616$ cm²; the examiners report candidates who used only the curved surface of the hemispheres or missed the new flat faces.

Useful results.

Slant height $l = r^{2} + h^{2}$ .
Hemisphere: CSA $2 π r^{2}$ , TSA $3 π r^{2}$ , volume $\frac{2}{3} π r^{3}$ .
Work with the coefficient of $π$ (for example $216 π$ ) and multiply by $\frac{22}{7}$ once at the end.

Combination of Solids: formulas to remember

Volume of a combination: $V = V_{1} + V_{2} (joined), V = V_{1} - V_{2} (removed)$; Add parts; subtract hollows.
Toy: cone on hemisphere: $TSA = π r l + 2 π r^{2}$; Same radius $r$ ; $l$ slant height.
Capsule: $TSA = 2 π r h + 4 π r^{2}, V = π r^{2} h + \frac{4}{3} π r^{3}$; $h$ = length of the cylindrical part.
Hemisphere: $CSA = 2 π r^{2}, TSA = 3 π r^{2}, V = \frac{2}{3} π r^{3}$; Use TSA only for a separate hemisphere.

How to solve Combination of Solids questions

Sketch the solid, label each part, and tick the faces you can see.

1Split the solid into standard parts and write $r$ , $h$ and $l$ for each.
2Volume: add the parts, subtract any hollow or scooped part.
3Surface area: list each exposed face; leave out the joining circles; add any new inner surfaces.
4If the parts are separated, use the total surface area of each part.
5Keep answers as multiples of $π$ until the end, then substitute and round.
6Finish with cost, capacity or mass if asked.

Combination of Solids in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Cylinder plus cone

Suppose a sharpened pencil is a cylinder of radius 0.35 cm and length 14 cm, with a conical tip of the same radius and height 1.2 cm.

Find the volume of the pencil to 2 decimal places. ( $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

Cylinder: $\frac{22}{7} \times 0.1225 \times 14 = 5.39$ cm³.
Cone: $\frac{1}{3} \times \frac{22}{7} \times 0.1225 \times 1.2 = 0.154$ cm³.
Total $= 5.544$ cm³.

Answer: About 5.54 cm³

Memory hook: Volume of a combination = sum of the pieces.

Two cones base to base

Imagine a wooden bead made of two cones of radius 3 mm and height 4 mm joined at their bases.

Find the bead's surface area and volume in terms of $π$ .

Try it first, then show the solutionSolution

$l = 3^{2} + 4^{2} = 5$ mm. The joined bases are hidden, so area $= 2 \times π \times 3 \times 5 = 30 π$ mm².
Volume $= 2 \times \frac{1}{3} π \times 9 \times 4 = 24 π$ mm³.

Answer: $30 π \approx 94.29$ mm²; $24 π \approx 75.43$ mm³

Memory hook: Joined faces hide; they are not painted.

Tricks to remember Combination of Solids

Short memory aids. Say each one aloud once, then use it on a question the same day.

Paint test
Would a painter's brush reach this face? If not, do not count it.
Use it when: Deciding which faces go into the TSA.
Pull it apart, faces appear
Separated parts each get their own flat faces back.
Use it when: Questions like 2025 Q5(iii).
Scoop swaps flat for curved
Scooping a hemisphere removes $π r^{2}$ of flat face and adds $2 π r^{2}$ of curved face.
Use it when: Surface area after hollowing.

Combination of Solids: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Toy formula · 1 mark · MCQBoard level

A solid toy is a cone of radius $r$ and slant height $l$ mounted on a hemisphere of the same radius. Its total surface area is:
(a) $π r l + 2 π r^{2}$
(b) $π r l + 3 π r^{2}$
(c) $π r (l + r)$
(d) $2 π r (l + r)$

Try it first, then show the solutionSolution

Visible faces: the cone's curved surface $π r l$ and the hemisphere's curved surface $2 π r^{2}$ ; the joining circle is hidden.

Answer: (a)

Example 2: A capsule · 3 marks · Short answerBoard level

A solid is a cylinder of radius 2.1 cm and length 6 cm with a hemisphere of the same radius at each end. Find its total surface area and volume. ( $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

TSA $= 2 π r h + 4 π r^{2} = 2 \times \frac{22}{7} \times 2.1 \times 6 + 4 \times \frac{22}{7} \times 4.41 = 79.2 + 55.44 = 134.64$ cm².
Volume $= π r^{2} h + \frac{4}{3} π r^{3} = \frac{22}{7} \times 4.41 \times 6 + \frac{4}{3} \times \frac{22}{7} \times 9.261 = 83.16 + 38.808 = 121.968$ cm³.

Answer: TSA 134.64 cm²; volume about 121.97 cm³

Example 3: Cone on a hemisphere · 4 marks · Long answerHarder board level

A toy is a cone of radius 5 cm and height 12 cm mounted on a hemisphere of the same radius. Find its total surface area and its volume, to 2 decimal places. ( $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

$l = 5^{2} + 1 2^{2} = 169 = 13$ cm.
TSA $= π r l + 2 π r^{2} = π (5 \times 13) + 2 π (25) = 115 π = 115 \times \frac{22}{7} \approx 361.43$ cm².
Volume $= \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π (25) (12) + \frac{2}{3} π (125) = 100 π + \frac{250}{3} π = \frac{550}{3} π \approx 576.19$ cm³.

Answer: TSA $115 π \approx 361.43$ cm²; volume $\frac{550}{3} π \approx 576.19$ cm³

Example 4: Hemispheres scooped from a cylinder · 4 marks · Long answerHarder board level

A wooden article is made by scooping out a hemisphere of radius 3.5 cm from each end of a solid cylinder of radius 3.5 cm and height 10 cm. Find (a) its total surface area, (b) the volume of wood left, to 2 decimal places. ( $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

(a) Faces: the cylinder's curved surface and two hollow hemispheres. TSA $= 2 π r h + 2 \times 2 π r^{2} = 220 + 154 = 374$ cm².
(b) Cylinder $= \frac{22}{7} \times 12.25 \times 10 = 385$ cm³.
Two hemispheres $= \frac{4}{3} \times \frac{22}{7} \times 42.875 \approx 179.67$ cm³.
Wood left $= 385 - 179.67 = 205.33$ cm³.

Answer: (a) 374 cm² (b) 205.33 cm³

Example 5: Joined versus separated · 4 marks · Long answerHOTS / competency

A solid consists of a cylinder of radius 6 cm and height 10 cm with a cone of the same radius and height 8 cm on top.
(a) Find the total surface area of the solid in terms of $π$ .
(b) The cone is cut off along the joint. Find the sum of the total surface areas of the two pieces, in terms of $π$ and to 2 decimal places. ( $π = \frac{22}{7}$ )
(c) Explain the difference.

Try it first, then show the solutionSolution

$l = 36 + 64 = 10$ cm.
(a) $2 π r h + π r^{2} + π r l = 120 π + 36 π + 60 π = 216 π$ cm².
(b) Cylinder $2 π r (r + h) = 192 π$ ; cone $π r (l + r) = 96 π$ . Sum $= 288 π \approx 905.14$ cm².
(c) $288 π - 216 π = 72 π = 2 \times π (6)^{2}$ : the two circular faces at the cut are now exposed.

Answer: (a) $216 π$ cm² (b) $288 π \approx 905.14$ cm² (c) two new circular faces

Example 6: Case study: a grain silo · 4 marks · Case-basedHOTS / competency

A grain silo is a cylinder of radius 1.5 m and height 7 m with a conical roof of the same radius and height 2 m.
(a) Find the volume of the silo to 2 decimal places.
(b) Find the area of its outer curved surfaces (walls and roof).
(c) Find the cost of painting them at ₹40 per m². ( $π = \frac{22}{7}$ )

Try it first, then show the solutionSolution

(a) Cylinder $= \frac{22}{7} \times 2.25 \times 7 = 49.5$ m³; cone $= \frac{1}{3} \times \frac{22}{7} \times 2.25 \times 2 \approx 4.71$ m³. Total $\approx 54.21$ m³.
(b) $l = 1. 5^{2} + 2^{2} = 2.5$ m. Walls $= 2 \times \frac{22}{7} \times 1.5 \times 7 = 66$ m²; roof $= \frac{22}{7} \times 1.5 \times 2.5 \approx 11.79$ m². Total $\approx 77.79$ m².
(c) Cost $= 77.786 \times 40 \approx$ ₹3111.43 (using the unrounded area).

Answer: (a) 54.21 m³ (b) 77.79 m² (c) ₹3111.43

Where students lose marks in Combination of Solids

The slips examiners see most often, and what to write instead.

Mistake: Missing the new flat faces when solids are separated
Each separated part has its own flat circular faces.
Wrong working
Capsule parts: cylinder TSA + two hemisphere CSAs only.
Right working
Cylinder $2 π r (r + h)$ + two hemispheres at $3 π r^{2}$ each.
Do this: Draw the separated pieces and shade every face.
Usually costs: Reported in 2025 Q5(iii): candidates did not realise new flat surfaces form, or used only the CSA of the hemispheres.
Mistake: Counting a hidden joining face
Where two parts meet, the circle is inside the solid.
Wrong working
Toy TSA $= π r (l + r) + 3 π r^{2}$ .
Right working
Toy TSA $= π r l + 2 π r^{2}$ .
Do this: Use curved surfaces for joined parts and add only the bases that are still showing.
Usually costs: Usually 2 of the 4 marks.
Mistake: Hemisphere CSA and TSA confused
$2 π r^{2}$ is curved only; $3 π r^{2}$ includes the flat circle.
Wrong working
Using $3 π r^{2}$ for a dome on a building.
Right working
A dome is curved only: $2 π r^{2}$ .
Do this: Ask whether the flat face is visible.
Usually costs: Reported in 2025 Q5(iii): formulae for the hemisphere were confused.

Exam tips

2025 Q5(iii) (4 marks): capsule with radius 3.5 cm and cylinder 14 cm, separated into three parts: $2 π r (r + h) + 2 \times 3 π r^{2} = 308 + 308 = 616$ cm².
2024 Q2(ii) (4 marks) also tested new faces made by cutting a solid.
Keep the answer as a multiple of $π$ until the last line; it avoids rounding errors and makes checking easy.
State clearly which faces you are adding; the method marks depend on it.

Practice questions on Combination of Solids

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
A solid is a cylinder of radius 3 cm and height 5 cm with a hemisphere of radius 3 cm on top. Its volume is:
(a) $54 π$ cm³
(b) $63 π$ cm³
(c) $72 π$ cm³
(d) $81 π$ cm³
Hint
$45 π + 18 π$ .
Answer
(b)
Q2 · 2 marks · Short answerBoard level
A solid is a hemisphere of radius 7 cm surmounted by a cone of the same radius and height 7 cm. Find its volume. ( $π = \frac{22}{7}$ )
Hint
$\frac{2}{3} π (343) + \frac{1}{3} π (49) (7)$ .
Answer
$\frac{686}{3} π + \frac{343}{3} π = 343 π = 1078$ cm³.
Q3 · 1 mark · Assertion-reasonHarder board level
Assertion (A): The total surface area of a solid made of a cylinder with a hemisphere on top equals the total surface area of the cylinder plus the total surface area of the hemisphere.
Reason (R): The faces where two solids are joined are not part of the outer surface.
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
Two circles disappear at the joint.
Answer
(b)
Q4 · 4 marks · Long answerHarder board level
From a solid cylinder of radius 3 cm and height 10 cm, a cone of radius 3 cm and height 4 cm is hollowed out from each end. Find the volume and the total surface area of the remaining solid, to 2 decimal places. ( $π = \frac{22}{7}$ )
Hint
Each cone has slant height 5 cm.
Answer
Volume $= 90 π - 2 \times 12 π = 66 π \approx 207.43$ cm³. TSA $= 2 π (3) (10) + 2 π (3) (5) = 90 π \approx 282.86$ cm².
Q5 · 3 marks · Short answerHarder board level
A building is a cylinder of radius 7 m and height 4 m topped by a hemispherical dome of the same radius. Find the inner area of the walls and dome, and the cost of painting it at ₹10 per m². ( $π = \frac{22}{7}$ )
Hint
Curved wall $2 π r h$ , dome $2 π r^{2}$ ; no floor.
Answer
$56 π + 98 π = 154 π = 484$ m²; cost ₹4840.
Q6 · 3 marks · Short answerHarder board level
A capsule is 16 cm long overall and 4 cm in diameter: a cylinder with a hemisphere at each end. Find its volume to 2 decimal places. ( $π = \frac{22}{7}$ )
Hint
The cylindrical part is $16 - 2 \times 2 = 12$ cm long.
Answer
$π (4) (12) + \frac{4}{3} π (8) = \frac{176}{3} π \approx 184.38$ cm³.
Q7 · 4 marks · Long answerHOTS / competency
A hemisphere of radius $r$ is scooped out of one end of a solid cylinder of radius $r$ and height $h$ . The volume left equals the volume removed.
(a) Show that $h = \frac{4 r}{3}$ .
(b) If $r = 6$ cm, find $h$ and the total surface area of the remaining solid, to 2 decimal places. ( $π = \frac{22}{7}$ )
Hint
Remaining TSA: curved side, one flat end, one hollow hemisphere.
Answer
(a) $π r^{2} h - \frac{2}{3} π r^{3} = \frac{2}{3} π r^{3}$ gives $h = \frac{4 r}{3}$ . (b) $h = 8$ cm; TSA $= 96 π + 36 π + 72 π = 204 π \approx 641.14$ cm².

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.TSA of a toy (cone on hemisphere)?Show answer
$π r l + 2 π r^{2}$
2.TSA of a separate solid hemisphere?Show answer
$3 π r^{2}$
3.Volume of a capsule with cylinder length $h$ ?Show answer
$π r^{2} h + \frac{4}{3} π r^{3}$
4.Are joining faces counted in the TSA?Show answer
No.
5.Slant height for $r = 6$ , $h = 8$ ?Show answer
10

Combination of Solids: frequently asked questions

How do I find the surface area of a combined solid?

Add only the faces you can see from outside. Leave out the circles where the parts are joined, and add any inner surface left by hollowing out a part.

Is the volume of a combined solid just the sum of the volumes?

Yes, for solids joined together. If a part is removed or scooped out, subtract its volume.

What changes when the parts of a combined solid are separated?

The volume is the same, but the total surface area increases, because the faces that were joined are now exposed on each piece.