Area and Volume of Solids ICSE Class 10: Cylinder, Cone and Sphere

What Area and Volume of Solids – Cylinder, Cone and Sphere is about

Weightage: CISCE publishes no chapter or unit weightage for ICSE Mathematics. Area and Volume of Solids – Cylinder, Cone and Sphere is in the 4. Mensuration part of the 2027 syllabus and can be asked in both Section A and Section B of the 80-mark paper.Source: CISCE ICSE Mathematics syllabus, examination year 2027

Key concepts

Right circular cylinder

With radius $r$ and height $h$:

• Curved surface area (CSA) $=2\pi rh$ (the label of a tin, unrolled into a rectangle $2\pi r$ by $h$).
• Total surface area (TSA) $=2\pi rh+2\pi r^2=2\pi r(r+h)$.
• Volume $=\pi r^{2}h$.

An open tank (no lid) has area $2\pi rh+\pi r^2$.

Worked example

Find the CSA, TSA and volume of a cylinder of radius 7 cm and height 10 cm. ($\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. CSA $=2\times \frac{22}{7}\times 7\times 10=440$ cm².
2. TSA $=2\times \frac{22}{7}\times 7\times 17=748$ cm².
3. Volume $=\frac{22}{7}\times 49\times 10=1540$ cm³.

Answer: 440 cm², 748 cm², 1540 cm³

Hollow cylinder (pipe)

With external radius $R$, internal radius $r$ and length $h$ (thickness $=R-r$):

• Volume of material $=\pi (R^2-r^2)h$.
• Outer CSA $=2\pi Rh$; inner CSA $=2\pi rh$.
• Each ring-shaped end has area $\pi (R^2-r^2)$.
• TSA $=2\pi Rh+2\pi rh+2\pi (R^2-r^2)$.

Worked example

A pipe 14 cm long has external radius 5 cm and internal radius 4 cm. Find the volume of metal. ($\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. $R^2-r^2=25-16=9$.
2. Volume $=\frac{22}{7}\times 9\times 14=396$ cm³.

Answer: 396 cm³

Right circular cone

With radius $r$, height $h$ and slant height $l$:

• $l=\sqrt{r^2+h^2}$ (Pythagoras).
• CSA $=\pi rl$; TSA $=\pi r(l+r)$.
• Volume $=\frac{1}{3}\pi r^{2}h$, one-third of the cylinder with the same base and height.

Use the vertical height $h$ for volume and the slant height $l$ for the curved surface.

Worked example

A cone has radius 7 cm and height 24 cm. Find its slant height, CSA, TSA and volume. ($\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. $l=\sqrt{49+576}=25$ cm.
2. CSA $=\frac{22}{7}\times 7\times 25=550$ cm²; TSA $=550+154=704$ cm².
3. Volume $=\frac{1}{3}\times \frac{22}{7}\times 49\times 24=1232$ cm³.

Answer: l = 25 cm; 550 cm²; 704 cm²; 1232 cm³

Sphere, hemisphere and hollow sphere

Sphere of radius $r$: surface area $=4\pi r^2$, volume $=\frac{4}{3}\pi r^3$.

Solid hemisphere: CSA $=2\pi r^2$, TSA $=3\pi r^2$ (curved part plus the flat circle), volume $=\frac{2}{3}\pi r^3$.

Hollow sphere (external radius $R$, internal $r$): volume of material $=\frac{4}{3}\pi (R^3-r^3)$. A hemispherical bowl of thickness $R-r$ has outer CSA $2\pi R^2$, inner CSA $2\pi r^2$ and a ring-shaped rim $\pi (R^2-r^2)$.

Worked example

For radius 3.5 cm, find the surface area and volume of a sphere and the TSA of a solid hemisphere. ($\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. Sphere: $4\times \frac{22}{7}\times 12.25=154$ cm²; volume $=\frac{4}{3}\times \frac{22}{7}\times 42.875\approx 179.67$ cm³.
2. Hemisphere TSA $=3\times \frac{22}{7}\times 12.25=115.5$ cm².

Answer: 154 cm²; 179.67 cm³; 115.5 cm²

Melting and recasting

When a solid is melted and recast (or water is poured from one vessel to another), the volume stays the same. Write one equation:
$\text{volume of old solid(s)}=\text{volume of new solid(s)}.$
Cancel $\pi$ and common factors before multiplying out. For $n$ identical pieces: $n=\frac{\text{volume of big solid}}{\text{volume of one piece}}$.

Surface area is not conserved: many small pieces have more total surface than one big one.

Worked example

A metal sphere of radius 6 cm is melted and recast into a cone of radius 6 cm. Find the height of the cone.

Try it first, then show the solutionSolution

1. $\frac{4}{3}\pi \times 6^3=\frac{1}{3}\pi \times 6^2\times h$.
2. Cancel $\frac{1}{3}\pi \times 36$: $4\times 6=h$.
3. $h=24$ cm.

Answer: 24 cm

Combination of solids

For a solid made by joining shapes (a cone on a hemisphere, a cylinder with a conical top, a capsule):

• Volume: add (or subtract, if a part is hollowed out).
• Surface area: add only the surfaces you can see and touch. Joined faces are hidden; a hollowed-out cavity adds its inner surface.

When a solid is cut or separated, new flat faces appear and must be counted. Sketch the pieces and list every face before using formulas.

Worked example

A toy is a cone of radius 3.5 cm and height 12 cm on a hemisphere of the same radius. Find its total surface area. ($\pi =\frac{22}{7}$)

Try it first, then show the solutionSolution

1. $l=\sqrt{3.5^2+12^2}=\sqrt{156.25}=12.5$ cm.
2. Cone CSA $=\frac{22}{7}\times 3.5\times 12.5=137.5$ cm².
3. Hemisphere CSA $=2\times \frac{22}{7}\times 12.25=77$ cm². The joined circle is hidden.
4. Total $=214.5$ cm².

Answer: 214.5 cm²

Ratios, units and cost

• If all lengths scale by $k$: areas scale by $k^2$, volumes by $k^3$. Two spheres with radii $2:3$ have areas $4:9$ and volumes $8:27$; from volumes, take the cube root to get radii.
• Units: $1$ m³ $=1000$ L; $1$ L $=1000$ cm³.
• Cost $=$ area × rate (painting, polishing, canvas) or volume × rate (filling, material).
• Round only the final answer, to the accuracy asked.

Worked example

The volumes of two spheres are in the ratio $8:125$. Find the ratio of their surface areas.

Try it first, then show the solutionSolution

1. Radii: $\sqrt[3]{8}:\sqrt[3]{125}=2:5$.
2. Surface areas: $2^2:5^2=4:25$.

Answer: 4 : 25

Maths in real life: Area and Volume of Solids – Cylinder, Cone and Sphere

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.

Topics in this chapter

• Surface Area and Volume of a Cylinder
• Melting and Recasting of Solids
• Combination of Solids

Area and Volume of Solids – Cylinder, Cone and Sphere formula sheet

Cylinder

$\text{CSA}=2\pi rh,\text{TSA}=2\pi r(r+h),V=\pi r^{2}h$

Open tank: 2πrh + πr².

Hollow cylinder

$V=\pi (R^2-r^2)h$

R external, r internal radius.

Cone

$l=\sqrt{r^2+h^2},\text{CSA}=\pi rl,V=\frac{1}{3}\pi r^{2}h$

TSA = πr(l + r).

Sphere

$S=4\pi r^2,V=\frac{4}{3}\pi r^3$

Hollow: V = (4/3)π(R³ − r³).

Hemisphere

$\text{CSA}=2\pi r^2,\text{TSA}=3\pi r^2,V=\frac{2}{3}\pi r^3$

TSA includes the flat circle.

Melting and recasting

$V_{\text{old}}=V_{\text{new}}$

n pieces: n = V_big ÷ V_piece.

Units

$1{\text{m}}^3=1000\text{L},1\text{L}=1000{\text{cm}}^3$

Convert before comparing.

How the board asks Area and Volume of Solids – Cylinder, Cone and Sphere

The question patterns to expect, and the method that works for each.

Tricks to remember Area and Volume of Solids – Cylinder, Cone and Sphere

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

• Cone is a third

  A cone holds exactly one-third of the cylinder with the same base and height.

  Use it when: Remembering ⅓πr²h.

• Sphere four, hemisphere two (plus one lid)

  Sphere 4πr²; hemisphere curved 2πr², add the lid πr² to get 3πr².

  Use it when: Hemisphere CSA vs TSA.

• Cone 1, hemisphere 2, cylinder 3

  With equal radius r and height r, the volumes are in the ratio 1 : 2 : 3.

  Use it when: Ratio questions and checking.

• Cut it, count it

  Every cut creates new flat faces: count them.

  Use it when: Cut or separated solids.

Area and Volume of Solids – Cylinder, Cone and Sphere: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Case-study practice (Section B style)

Read the situation, then answer every part. In the board paper this is a 10-mark question split into 3 + 3 + 4 marks.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Where students lose marks in Area and Volume of Solids – Cylinder, Cone and Sphere

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

• ✗Mistake: Forgetting new faces after cutting or separating

  CISCE examiners' comments, 2024 and 2025: candidates did not include the two flat rectangular faces when a cylinder was cut in half along its height, and did not realise new flat surfaces form when a capsule is separated into a cylinder and two hemispheres.

  ✓Do this: Sketch each piece and list all its faces before writing any formula.

  Usually costs: 2 or more marks in a 4-mark part.

• ✗Mistake: Diameter used as radius

  CISCE examiners' comments, 2024 and 2025: candidates took the radius as 7 cm when the diameter was 7 cm, and took the given diameters of a hollow sphere as radii.

  ✓Do this: Underline 'diameter' in the question and halve it at once.

  Usually costs: Every answer that follows.

• ✗Mistake: Area ratio instead of volume ratio

  CISCE examiners' comments, 2025: for spheres with volumes in the ratio 27 : 64, candidates chose 9 : 16 for the radii, confusing area and volume.

  Wrong working

  Radii = √27 : √64

  Right working

  Radii = ∛27 : ∛64 = 3 : 4

  ✓Do this: Volumes use cubes; areas use squares.

  Usually costs: 1 mark.

• ✗Mistake: Hemisphere CSA and TSA confused

  CISCE examiners' comments, 2025: candidates used only the curved surface area of the hemispheres when the total was needed, and misused the formulas for a cylinder's and a hemisphere's total surface area.

  ✓Do this: Hemisphere: 2πr² curved; add πr² only if the flat face is exposed.

  Usually costs: 1 to 2 marks.

• ✗Mistake: Ignoring a condition or the rounding

  CISCE examiners' comments, 2024: candidates ignored that the oil filled only 3/4 of the sphere, and did not round 49.7 cm to the nearest cm as asked.

  ✓Do this: Circle fractions like '3/4 full' and the accuracy required; round only the final answer.

  Usually costs: 1 mark each.

• ✗Mistake: Wrong volume for a hollow solid

  CISCE examiners' comments, 2025: candidates used an incorrect formula for the volume of a hollow sphere.

  ✓Do this: Volume of a shell = outer solid − inner solid: (4/3)π(R³ − r³), not (4/3)π(R − r)³.

  Usually costs: 2 marks.

Practice questions on Area and Volume of Solids – Cylinder, Cone and Sphere

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

1. Q1 · 1 mark · MCQBoard level

   The curved surface area of a cylinder of radius 7 cm and height 5 cm is ($\pi =\frac{22}{7}$):
   (a) 110 cm²
   (b) 220 cm²
   (c) 440 cm²
   (d) 770 cm²

   Hint

   2πrh.

   Answer

   (b)

2. Q2 · 1 mark · MCQBoard level

   The total surface area of a solid hemisphere of radius 7 cm is ($\pi =\frac{22}{7}$):
   (a) 308 cm²
   (b) 154 cm²
   (c) 462 cm²
   (d) 616 cm²

   Hint

   3πr².

   Answer

   (c)

3. Q3 · 1 mark · MCQHarder board level

   A cone and a cylinder have the same radius and the same height. The ratio of their volumes is:
   (a) $1:2$
   (b) $1:3$
   (c) $2:3$
   (d) $3:1$

   Hint

   ⅓ of the cylinder.

   Answer

   (b)

4. Q4 · 1 mark · Assertion-reasonBoard level

   Assertion (A): When a metal cone is melted and recast into a sphere, the volume of the sphere equals the volume of the cone.
   Reason (R): Melting and recasting does not change the amount of metal.
   (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

   Volume is conserved.

   Answer

   (c)

5. Q5 · 4 marks · Short answerHarder board level

   A solid cone of radius 6 cm and height 24 cm is melted and recast into a sphere. Find the radius of the sphere and compare the surface areas of the cone and the sphere. ($\pi =\frac{22}{7}$, 2 d.p.)

   Hint

   4r³ = 36 × 24.

   Answer

   $\frac{4}{3}\pi r^3=\frac{1}{3}\pi (36)(24)$ gives $r^3=216$, $r=6$ cm. Sphere area $=144\pi \approx 452.57$ cm². Cone: $l=\sqrt{36+576}=\sqrt{612}\approx 24.739$, TSA $=\pi \times 6\times (l+6)\approx 579.64$ cm² (rounding only at the end). The sphere has less surface.

6. Q6 · 4 marks · Short answerHarder board level

   A cylindrical vessel of radius 14 cm contains water. A solid sphere of radius 7 cm is dropped in and is fully submerged. (a) By how much does the water level rise? (b) How many such spheres would raise the level by 14 cm?

   Hint

   Volume of sphere = π × 196 × rise.

   Answer

   (a) $\frac{4}{3}\pi (343)=\pi (196)h$ gives $h=\frac{7}{3}\approx 2.33$ cm. (b) $14÷\frac{7}{3}=6$ spheres.

7. Q7 · 3 marks · Long answerBoard level

   The curved surface area of a cylinder of radius 7 cm is 880 cm². Find its height and volume. ($\pi =\frac{22}{7}$)

   Hint

   2πrh = 880.

   Answer

   $44h=880$, $h=20$ cm; $V=\frac{22}{7}\times 49\times 20=3080$ cm³.

8. Q8 · 4 marks · Long answerHarder board level

   A metal pipe is 21 cm long, with external diameter 10 cm and thickness 1 cm. Find the volume of metal and its mass if 1 cm³ of the metal weighs 8 g. ($\pi =\frac{22}{7}$)

   Hint

   R = 5, r = 4.

   Answer

   $\frac{22}{7}\times (25-16)\times 21=594$ cm³; mass $=4752$ g $=4.752$ kg.

9. Q9 · 4 marks · Long answerHOTS / competency

   A cone, a hemisphere and a cylinder all have radius $r$, and the cone and cylinder have height $r$. Show that their volumes are in the ratio $1:2:3$, and find the ratio of the curved surface area of the hemisphere to that of the cylinder.

   Hint

   Write each volume as a multiple of πr³.

   Answer

   Cone $\frac{1}{3}\pi r^3$, hemisphere $\frac{2}{3}\pi r^3$, cylinder $\pi r^3$: ratio $1:2:3$. CSAs: $2\pi r^2$ and $2\pi r\cdot r=2\pi r^2$: ratio $1:1$.

10. Q10 · 4 marks · Long answerHarder board level

    A solid sphere of radius 3 cm is melted and recast into small spheres of radius 0.5 cm. Find the number of small spheres and how many times the total surface area increases.

    Hint

    n = (3/0.5)³.

    Answer

    $n=6^3=216$. Total small surface $=216\times 4\pi (0.25)=216\pi$; big $=36\pi$. It becomes 6 times as large.

11. Q11 · 1 mark · MCQHOTS / competency

    If the radius of a sphere is increased by 10%, its volume increases by:
    (a) 10%
    (b) 21%
    (c) 30%
    (d) 33.1%

    Hint

    1.1³.

    Answer

    (d)

12. Q12 · 5 marks · Long answerHOTS / competency

    A cylindrical vessel of radius 10 cm and height 21 cm is full of water. A solid made of a cone of radius 7 cm and height 12 cm mounted on a hemisphere of the same radius is lowered into it until fully submerged. Find (a) the volume of water that overflows, (b) the water left in the vessel, (c) the surface area of the solid. ($\pi =\frac{22}{7}$, 2 d.p.)

    Hint

    The solid is 19 cm tall and 14 cm wide, so it fits.

    Answer

    (a) $\frac{1}{3}\pi (49)(12)+\frac{2}{3}\pi (343)=196\pi +\frac{686}{3}\pi \approx 1334.67$ cm³. (b) Vessel $=\pi (100)(21)=6600$ cm³; left $\approx 5265.33$ cm³. (c) $l=\sqrt{49+144}\approx 13.89$ cm; area $=\pi (7)(13.89)+2\pi (49)\approx 305.63+308=613.63$ 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. 1.Volume of a cone?Show answer

   ⅓πr²h

2. 2.Slant height of a cone?Show answer

   l = √(r² + h²)

3. 3.TSA of a solid hemisphere?Show answer

   3πr²

4. 4.Volume of a hollow sphere?Show answer

   (4/3)π(R³ − r³)

5. 5.What stays the same in melting and recasting?Show answer

   Volume.

6. 6.1 m³ in litres?Show answer

   1000 L

7. 7.Volumes of cone, hemisphere, cylinder with radius r and height r?Show answer

   1 : 2 : 3

8. 8.Is the frustum in the ICSE 2027 syllabus?Show answer

   No.

Quick revision checklist

The night before the exam: can you do each of these without looking?

• ☐Do you halve every diameter?
• ☐Do you use l for CSA and h for volume of a cone?
• ☐Do you know when to add πr² to a hemisphere?
• ☐Do you count new faces after a cut?
• ☐Do you cancel π before multiplying in recasting?
• ☐Do you round only at the end?

Area and Volume of Solids – Cylinder, Cone and Sphere: frequently asked questions