MathsCBSE Class 10 · Surface Areas and Volumes

Cone on a Hemisphere and Similar Shapes: Toys and Capsules

Updated 14 October 2026 · Fact-checked

A cone on a hemisphere is a toy-like solid made by joining a cone and a hemisphere of the same radius at their flat circular faces. Similar shapes include capsules and cylinders with hemispherical ends. The board asks you to find a hidden height or slant height first, then the total surface area, volume or cost.

Both solids share radius r; the red dashed joined circles are hidden, so only curved parts count, with H = h + r and L = h + 2r.

Illustration drawn for QuestionBuddy.

Understand Cone on a Hemisphere and Similar Shapes

Many everyday objects are two solids joined along a common circle. A spinning top is a cone on a hemisphere. A medicine capsule is a cylinder with a hemisphere at each end. A bottle or vessel can be a cylinder with a hemisphere at one end. In every case the two solids have the same radius $r$ , so their flat circles match exactly and fit together.

The key idea is the hidden face. Where the cone meets the hemisphere, the cone's base and the hemisphere's flat face press against each other. Neither can be seen or painted, so neither is part of the surface area. Only curved surfaces remain: TSA $= π r l + 2 π r^{2} = π r (l + 2 r)$ . For a capsule, both ends of the cylinder are covered, so TSA $= 2 π r h + 4 π r^{2} = 2 π r (h + 2 r)$ .

Volume behaves differently. Joining does not lose any space, so you simply add: for the toy $V = \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π r^{2} (h + 2 r)$ , and for the capsule $V = π r^{2} h + \frac{4}{3} π r^{3}$ , since two hemispheres make one full sphere.

The lengths are where marks are lost. In a toy, the total height is the cone's height plus one radius, so $h = H - r$ . In a capsule, each end sticks out by one radius, so the cylinder length is $h = L - 2 r$ . A cone also needs its slant height $l = r^{2} + h^{2}$ from Pythagoras before you can find its curved surface.

A neat result for a capsule: $h + 2 r$ is the total length $L$ , so its surface area is just $2 π r L$ , the same as an open cylinder of the full length. This builds directly on the Class 9 formulas and on the general rule of the chapter: count only exposed surfaces, but add all volumes. These formulas hold only when both solids share the same radius; if the radii differ, a ring of flat face stays exposed and must be added.

Cone on a Hemisphere and Similar Shapes: formulas to remember

Cone on hemisphere: cone height: $h = H - r$; $H$ is the total height of the toy, from the tip of the cone to the bottom of the hemisphere.
Slant height of the cone: $l = r^{2} + h^{2}$; Find this before any cone curved surface area.
Cone on hemisphere: total surface area: $TSA = π r l + 2 π r^{2} = π r (l + 2 r)$; Both flat circles are hidden; only the two curved surfaces count.
Cone on hemisphere: volume: $V = \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = \frac{1}{3} π r^{2} (h + 2 r)$; Also gives the ratio cone : hemisphere $= h : 2 r$ .
Capsule: cylinder length: $h = L - 2 r$; $L$ is the total length, end to end.
Capsule: total surface area: $TSA = 2 π r h + 4 π r^{2} = 2 π r (h + 2 r) = 2 π r L$; Same radius for the cylinder and both hemispheres.
Capsule: volume: $V = π r^{2} h + \frac{4}{3} π r^{3} = π r^{2} (h + \frac{4 r}{3})$; The two hemispherical ends together make one full sphere.
Solid cylinder with hemisphere on one end: $TSA = π r^{2} + 2 π r H + 2 π r^{2}, V = π r^{2} (H + \frac{2 r}{3})$; $H$ is the cylinder height. The other flat end of the cylinder is still exposed.
Open vessel: hemisphere with cylinder on top: $Inner area = 2 π r H + 2 π r^{2} = 2 π r (H + r)$; No lid, so no top circle. $H$ is total height minus $r$ .
Cone on a cylinder (solid): $TSA = 2 π r h + π r^{2} + π r l, V = π r^{2} h + \frac{1}{3} π r^{2} h_{c}$; h is the cylinder height, h_c the cone height and l its slant height. The joined circle is hidden; the base circle of the cylinder is outside.
Cavity scooped out of a solid: $TSA = outer surfaces left + inner surface of the cavity, V = V_{solid} - V_{cavity}$; Scooping removes the flat circle on top but adds the curved inner surface of the cavity.

How to solve Cone on a Hemisphere and Similar Shapes questions

This method works for any toy, capsule or vessel made from a cone, cylinder and hemisphere of the same radius.

1Read the question and name the two solids. Check that they share the same radius and note what is given: total height, total length, diameter, volume or area.
2Write $r$ as a fraction such as $\frac{7}{2}$ or $\frac{21}{2}$ so that $π = \frac{22}{7}$ cancels neatly.
3Find the hidden length first: cone height $h = H - r$ , or cylinder length $h = L - 2 r$ . Write it on its own line with units.
4If a cone is involved and surface area is needed, find $l = r^{2} + h^{2}$ .
5For surface area, write a word line such as 'TSA = CSA of cone + CSA of hemisphere', leaving out every hidden circle. Add a flat face only if it is really outside.
6For volume, add the volumes of the two parts. Take $π r^{2}$ or $\frac{1}{3} π r^{2}$ out as a common factor before multiplying.
7If the volume or area is given and a length is unknown, set up the equation from the formula and solve it.
8Finish with the unit asked: cm² or cm³, litres (divide cm³ by 1000), or rupees for a cost.

Cone on a Hemisphere and Similar Shapes 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.

Capsule: a hemisphere at each end

Suppose your grandmother is stitching a new cover for a round bolster pillow before a family function. The bolster is a cylinder with a hemispherical end at each side, of diameter 21 cm and total length 84 cm. She wants to know how much cloth the cover needs and how much filling the pillow holds.

Find the area of cloth needed for the cover (ignore seams) and its cost at ₹2 per 100 cm². Find the volume of filling inside the bolster in cm³ and in litres. Filling is sold only in whole litres: how many litres must she buy?

Try it first, then show the solutionSolution

Radius $r = \frac{21}{2} = 10.5$ cm; total length $L = 84$ cm.
Cylinder length $h = L - 2 r = 84 - 21 = 63$ cm.
Cloth area $= 2 π r h + 4 π r^{2} = 2 π r (h + 2 r) = 2 π r L$ .
$= 2 \times \frac{22}{7} \times \frac{21}{2} \times 84 = 66 \times 84 = 5544$ cm².
Volume $= π r^{2} (h + \frac{4 r}{3})$ ; $π r^{2} = \frac{22}{7} \times \frac{441}{4} = 346.5$ cm².
$h + \frac{4 r}{3} = 63 + 14 = 77$ cm.
Volume $= 346.5 \times 77 = 26680.5$ cm³ $= 26.6805$ litres.
Cost of cloth $= \frac{5544}{100} \times 2 = 110.88$ , that is ₹110.88.
She needs 26.68 litres, so she must buy 27 whole litres.

Answer: Cloth needed = 5544 cm², costing ₹110.88; filling = 26680.5 cm³ (about 26.68 litres), so she buys 27 litres.

Memory hook: The bolster trick: a capsule's skin is just $2 π r$ times its full length.

Cone mounted on a hemisphere

Imagine a mason building a wall at your school uses a solid metal plumb bob to check that the wall is vertical. The bob is a cone mounted on a hemisphere, both of radius 2.1 cm, and the total height of the bob is 4.9 cm. Suppose 1 cm³ of the metal weighs 8 g, and the whole outer surface is to be chrome-plated at ₹0.50 per cm².

Find the area to be chrome-plated and the cost of plating. Also find the volume of the plumb bob and its mass in grams.

Try it first, then show the solutionSolution

$r = 2.1$ cm, total height $H = 4.9$ cm.
Cone height $h = H - r = 4.9 - 2.1 = 2.8$ cm.
Slant height $l = 2. 1^{2} + 2. 8^{2} = 4.41 + 7.84 = 12.25 = 3.5$ cm.
Area to plate = CSA of cone + CSA of hemisphere $= π r (l + 2 r)$ .
$= \frac{22}{7} \times 2.1 \times (3.5 + 4.2) = 6.6 \times 7.7 = 50.82$ cm².
Cost of plating $= 50.82 \times 0.50 =$ ₹25.41.
$π r^{2} = \frac{22}{7} \times 4.41 = 13.86$ cm².
Volume $= \frac{1}{3} π r^{2} (h + 2 r) = \frac{1}{3} \times 13.86 \times (2.8 + 4.2) = 4.62 \times 7 = 32.34$ cm³.
Mass $= 32.34 \times 8 = 258.72$ g.

Answer: Area to plate = 50.82 cm², cost = ₹25.41; volume = 32.34 cm³, mass = 258.72 g.

Memory hook: The plumb bob: a top-shaped weight, cone height is total height minus one radius.

Tricks to remember Cone on a Hemisphere and Similar Shapes

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

Height minus one, length minus two
A toy loses one radius from its total height ( $h = H - r$ ). A capsule loses two radii from its total length ( $h = L - 2 r$ ), one for each end.
Use it when: Use this before any formula whenever a total height or total length is given.
Glue hides two circles
Where two solids are glued, picture two coins pressed together: both disappear from the outside. So never add $π r^{2}$ for a joined face.
Use it when: Use when writing the TSA word line for any combination.
Capsule = open tube of full length
Capsule TSA $= 2 π r (h + 2 r) = 2 π r L$ . Its skin equals the side of a cylinder as long as the whole capsule.
Use it when: A quick check or shortcut for capsule surface areas.
Cone to hemisphere is h to 2r
$\frac{1}{3} π r^{2} h : \frac{2}{3} π r^{3} = h : 2 r$ . When $h = 2 r$ the two parts hold equal volume.
Use it when: Use for MCQs and ratio questions on toys.

Cone on a Hemisphere and Similar Shapes: solved examples

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

Example 1: Equal volumes in a toy · 1 mark · MCQBoard level

A toy is a cone mounted on a hemisphere of the same radius $r$ . The volume of the conical part equals the volume of the hemispherical part. The height of the cone is:
(a) $r$
(b) $2 r$
(c) $3 r$
(d) $\frac{r}{2}$

Try it first, then show the solutionSolution

Volume of cone $= \frac{1}{3} π r^{2} h$ ; volume of hemisphere $= \frac{2}{3} π r^{3}$ .
Set them equal: $\frac{1}{3} π r^{2} h = \frac{2}{3} π r^{3}$ , so $h = 2 r$ .

Answer: (b) $2 r$

Example 2: Painting a toy: find the cone height first · 3 marks · Short answerBoard level

A wooden toy is a right circular cone mounted on a hemisphere of the same radius 3.5 cm. The total height of the toy is 15.5 cm. Find the total surface area of the toy and the cost of painting it at ₹2 per 10 cm². Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Cone height $h = 15.5 - 3.5 = 12$ cm (total height includes one radius of the hemisphere). [½]
Slant height $l = 3. 5^{2} + 1 2^{2} = 12.25 + 144 = 156.25 = 12.5$ cm. [1]
TSA = CSA of cone + CSA of hemisphere $= π r (l + 2 r)$ (joined circles are hidden).
$= \frac{22}{7} \times 3.5 \times (12.5 + 7) = 11 \times 19.5 = 214.5$ cm². [1]
Cost $= \frac{214.5}{10} \times 2 =$ ₹42.90. [½]

Answer: TSA = 214.5 cm²; cost of painting = ₹42.90.

Example 3: Capsule with an unknown radius: form an equation · 5 marks · Long answerHarder board level

A solid capsule-shaped roller is a right circular cylinder with a hemisphere of the same radius at each end. The length of the cylindrical part is 3 times the radius, and the total surface area of the roller is 3465 cm². Find the radius, the total length and the volume of the roller. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Let the radius be $r$ ; then cylinder length $h = 3 r$ . TSA $= 2 π r h + 4 π r^{2} = 6 π r^{2} + 4 π r^{2} = 10 π r^{2}$ (both flat ends hidden). [1]
$10 \times \frac{22}{7} \times r^{2} = 3465$ , so $r^{2} = \frac{3465 \times 7}{220} = 110.25$ . [1]
$r = 10.5$ cm, $h = 3 \times 10.5 = 31.5$ cm, total length $= h + 2 r = 31.5 + 21 = 52.5$ cm. [1]
Volume $= π r^{2} h + \frac{4}{3} π r^{3} = π r^{2} (h + \frac{4 r}{3})$ . [½]
$π r^{2} = \frac{22}{7} \times 110.25 = 346.5$ cm²; $h + \frac{4 r}{3} = 31.5 + 14 = 45.5$ cm. [½]
$V = 346.5 \times 45.5 = 15765.75$ cm³. (Check TSA: $10 \times 346.5 = 3465$ cm².) [1]

Answer: Radius = 10.5 cm; total length = 52.5 cm; volume = 15765.75 cm³.

Example 4: Case study: a wooden spinning top · 4 marks · Case-basedHarder board level

At a village fair, suppose a craftsman turns a solid wooden spinning top shaped like a right circular cone mounted on a hemisphere. Both have radius 10.5 cm and the total height of the top is 24.5 cm. Take $π = \frac{22}{7}$ .
(i) Find the height of the conical part. (1 mark)
(ii) Find the slant height of the cone. (1 mark)
(iii) The craftsman claims that if he made the cone 7 cm taller (same radius), the cone and the hemisphere would contain equal amounts of wood. Check his claim with reasons, and find by how many cm³ the volume of the top would increase. (2 marks)

Try it first, then show the solutionSolution

(i) $h = 24.5 - 10.5 = 14$ cm. [1]
(ii) $l = 10. 5^{2} + 1 4^{2} = 110.25 + 196 = 306.25 = 17.5$ cm. [1]
(iii) Equal volumes need $\frac{1}{3} π r^{2} h = \frac{2}{3} π r^{3}$ , i.e. $h = 2 r = 21$ cm. New height $= 14 + 7 = 21$ cm, so the claim is correct. (Check: $π r^{2} = 346.5$ ; cone $= \frac{1}{3} \times 346.5 \times 21 = 2425.5$ cm³ and hemisphere $= \frac{2}{3} \times 346.5 \times 10.5 = 2425.5$ cm³.) [1]
Increase in volume $= \frac{1}{3} π r^{2} \times 7 = \frac{1}{3} \times 346.5 \times 7 = 808.5$ cm³. [1]

Answer: (i) 14 cm (ii) 17.5 cm (iii) The claim is correct, since the new cone height 21 cm equals $2 r$ ; the volume increases by 808.5 cm³.

Example 5: Cylinder with a hemisphere on one end · 3 marks · Short answerBoard level

A solid is a cylinder with a hemisphere of the same radius on one end. The radius is 7 cm and the total height is 19 cm. Find its total surface area and volume. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Cylinder height $h = 19 - 7 = 12$ cm. [½]
Surfaces seen: CSA of cylinder, CSA of hemisphere and the flat base of the cylinder. TSA $= 2 π r h + 2 π r^{2} + π r^{2} = π r (2 h + 3 r)$ . [1]
TSA $= \frac{22}{7} \times 7 \times (24 + 21) = 22 \times 45 = 990$ cm². [½]
Volume $= π r^{2} h + \frac{2}{3} π r^{3} = π r^{2} (h + \frac{2 r}{3}) = 154 \times (12 + \frac{14}{3}) = 154 \times \frac{50}{3} \approx 2566.67$ cm³. [1]

Answer: TSA = 990 cm²; volume $\approx 2566.67$ cm³.

Example 6: An open vessel: inner area and cost · 3 marks · Short answerBoard level

A vessel is a hollow hemisphere with a hollow cylinder of the same radius on top, open at the top. Its diameter is 14 cm and its total height is 13 cm. Find the inner surface area of the vessel and the cost of tin-plating the inside at ₹10 per 100 cm². Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$r = 7$ cm. The hemisphere takes up 7 cm of the height, so the cylinder height $H = 13 - 7 = 6$ cm. [½]
The top is open, so the inner surface is the CSA of the cylinder plus the CSA of the hemisphere: $2 π r H + 2 π r^{2} = 2 π r (H + r)$ . [1]
$= 2 \times \frac{22}{7} \times 7 \times 13 = 44 \times 13 = 572$ cm². [1]
Cost $= \frac{572}{100} \times 10 = 57.20$ , that is ₹57.20. [½]

Answer: Inner surface area = 572 cm²; cost = ₹57.20.

Example 7: Cone on a cylinder: a toy rocket · 3 marks · Short answerBoard level

A solid toy rocket is a cylinder of radius 3.5 cm and height 10 cm with a cone of the same radius and height 12 cm on top. Find its total surface area and volume. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Slant height $l = 3. 5^{2} + 1 2^{2} = 12.25 + 144 = 156.25 = 12.5$ cm. [½]
The circle where cone meets cylinder is hidden; the bottom circle of the cylinder is outside. TSA $= 2 π r h + π r^{2} + π r l = π r (2 h + r + l)$ . [1]
TSA $= \frac{22}{7} \times 3.5 \times (20 + 3.5 + 12.5) = 11 \times 36 = 396$ cm². [½]
Volume $= π r^{2} (h + \frac{h _{c}}{3}) = 38.5 \times (10 + 4) = 539$ cm³. [1]

Answer: TSA = 396 cm²; volume = 539 cm³.

Example 8: A cone hollowed out of a cylinder · 3 marks · Short answerHarder board level

From a solid cylinder of radius 6 cm and height 8 cm, a cone of the same radius and height is hollowed out from the top. Find the total surface area and the volume of the remaining solid. Take $π = 3.14$ .

Try it first, then show the solutionSolution

Slant height of the cavity $l = 6^{2} + 8^{2} = 10$ cm. [½]
The top circle is removed and the inside of the cone is exposed. Surfaces: CSA of cylinder, bottom circle and CSA of cone: TSA $= 2 π r h + π r^{2} + π r l = π r (2 h + r + l) = π \times 6 \times (16 + 6 + 10) = 192 π$ cm². [1]
TSA $= 192 \times 3.14 = 602.88$ cm². [½]
Volume $= π r^{2} h - \frac{1}{3} π r^{2} h = \frac{2}{3} π r^{2} h = \frac{2}{3} \times π \times 36 \times 8 = 192 π \approx 602.88$ cm³. (The two numbers match only for these measurements; the units differ.) [1]

Answer: TSA $= 192 π \approx 602.88$ cm²; volume $= 192 π \approx 602.88$ cm³.

Where students lose marks in Cone on a Hemisphere and Similar Shapes

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

Mistake: Using the total height as the cone's height
The total height of a toy includes the hemisphere's radius below the cone.
Wrong working
Total height 15.5 cm, $r = 3.5$ : $l = 3. 5^{2} + 15. 5^{2}$
Right working
$h = 15.5 - 3.5 = 12$ cm, so $l = 3. 5^{2} + 1 2^{2} = 12.5$ cm
Do this: Always write 'cone height = total height − r' as your first line.
Usually costs: Every later step is wrong; you may keep only the formula mark.
Mistake: Using vertical height in the cone's curved surface
The curved surface unrolls into a sector whose radius is the slant height.
Wrong working
CSA of cone $= π r h = \frac{22}{7} \times 3.5 \times 12 = 132$ cm²
Right working
CSA of cone $= π r l = \frac{22}{7} \times 3.5 \times 12.5 = 137.5$ cm²
Do this: Find $l$ by Pythagoras before any cone surface area.
Usually costs: 1 to 2 marks in a 3-mark question.
Mistake: Adding the hidden flat circles
Using the full TSA of each solid counts the glued faces, which cannot be seen.
Wrong working
TSA $= π r (l + r) + 3 π r^{2}$
Right working
TSA $= π r l + 2 π r^{2} = π r (l + 2 r)$
Do this: Write the word line 'CSA of cone + CSA of hemisphere' before any numbers.
Usually costs: The answer mark and often the method mark.
Mistake: Taking the capsule's total length as the cylinder length
Each hemisphere adds one radius to the length.
Wrong working
$L = 63$ , $r = 10.5$ : $V = π r^{2} (63) + \frac{4}{3} π r^{3}$
Right working
$h = 63 - 21 = 42$ : $V = π r^{2} (42) + \frac{4}{3} π r^{3} = 19404$ cm³
Do this: Write $h = L - 2 r$ on its own line.
Usually costs: 2 to 3 marks in a 5-mark question.

Exam tips

Write the surface area as a sum of named parts in words before substituting; this line can earn a method mark even if your arithmetic slips.
Show each found length (cone height, slant height, cylinder length) on its own line with units, as examiners give step marks for these.
Take common factors like $π r$ or $\frac{1}{3} π r^{2}$ out first, and keep radii like $\frac{21}{2}$ as fractions so $π = \frac{22}{7}$ cancels.
In case studies, later parts often use answers from earlier parts; carry them forward and give each sub-part its own final line with units.
Read words like 'open', 'solid', 'inner' and 'total height' carefully: they decide which faces and lengths you use.

Practice questions on Cone on a Hemisphere and Similar Shapes

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

Q1 · 3 marks · Short answerHarder board level
A solid capsule-shaped roller is a cylinder with a hemisphere of the same radius at each end, and the length of the cylindrical part equals the radius. Its volume is 8489.25 cm³. Find the radius, the total length and the total surface area. Take $π = \frac{22}{7}$ .
Hint
Write the volume in terms of r alone: $π r^{2} (r) + \frac{4}{3} π r^{3}$ .
Answer
$V = π r^{3} + \frac{4}{3} π r^{3} = \frac{7}{3} π r^{3} = \frac{22}{3} r^{3}$ . So $r^{3} = \frac{8489.25 \times 3}{22} = 1157.625$ and $r = 10.5$ cm. Total length $= r + 2 r = 31.5$ cm. TSA $= 2 π r (h + 2 r) = 2 π r (3 r) = 6 π r^{2} = 6 \times 346.5 = 2079$ cm².
Q2 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a carpenter makes pen stands from solid wooden cylinders of radius 3.5 cm and height 10 cm. In design A he scoops out a hemisphere of radius 3.5 cm from the top. In design B he scoops out a cone of radius 3.5 cm and depth 8.4 cm from the top. Take $π = \frac{22}{7}$ .
(i) Find the volume of one uncut cylinder. (1 mark)
(ii) Using (i), find the volume of wood left in design A. (1 mark)
(iii) Polishing covers every surface, inside and out, at ₹1 per 10 cm². Which design costs less to polish, and by how much? (2 marks)
Hint
Scooping removes the top circle but adds the inside surface of the cavity.
Answer
(i) $π r^{2} h = 38.5 \times 10 = 385$ cm³. (1)
(ii) Hemisphere volume $= \frac{2}{3} \times 38.5 \times 3.5 \approx 89.83$ cm³, so wood left $\approx 385 - 89.83 = 295.17$ cm³. (1)
(iii) A: $2 π r h + π r^{2} + 2 π r^{2} = π r (2 h + 3 r) = 11 \times 30.5 = 335.5$ cm². B: cone slant $l = 3. 5^{2} + 8. 4^{2} = 82.81 = 9.1$ cm, so $2 π r h + π r^{2} + π r l = π r (2 h + r + l) = 11 \times 32.6 = 358.6$ cm². Costs: ₹33.55 and ₹35.86, so design A is cheaper by ₹2.31. (2)
Q3 · 5 marks · Long answerHarder board level
A solid top is a cone mounted on a hemisphere of the same radius. The height of the cone and its radius are in the ratio $12 : 5$ , and the total surface area of the top is 177.1 cm². Find the radius and the height of the cone. The cone is painted at ₹10 per 100 cm² and the hemisphere at ₹5 per 100 cm². Find the total cost. Take $π = \frac{22}{7}$ .
Hint
With $h = \frac{12 r}{5}$ , the slant height is $\frac{13 r}{5}$ . Write the TSA in terms of $r$ .
Answer
$l = r^{2} + \frac{144 r ^{2}}{25} = \frac{13 r}{5}$ . TSA $= π r (\frac{13 r}{5} + 2 r) = \frac{23}{5} π r^{2} = 177.1$ , so $π r^{2} = 38.5$ and $r^{2} = 12.25$ , $r = 3.5$ cm. Then $h = 8.4$ cm and $l = 9.1$ cm. Cone CSA $= π r l = 11 \times 9.1 = 100.1$ cm²; hemisphere CSA $= 2 \times 38.5 = 77$ cm² (check: sum 177.1). Cost $= 10.01 + 3.85 = 13.86$ , that is ₹13.86.
Q4 · 3 marks · Short answerBoard level
A solid toy is a cone mounted on a hemisphere, both of radius 7 cm. The volume of the toy is 1078 cm³. Find the total height of the toy and the ratio of the volume of the cone to that of the hemisphere. Take $π = \frac{22}{7}$ .
Hint
Set $\frac{1}{3} π r^{2} (h + 2 r) = 1078$ and solve for $h$ .
Answer
$\frac{1}{3} \times 154 \times (h + 14) = 1078$ , so $h + 14 = 21$ , $h = 7$ cm. Total height $= 7 + 7 = 14$ cm. Ratio $= h : 2 r = 7 : 14 = 1 : 2$ .
Q5 · 5 marks · Long answerHarder board level
A solid decorative top is a right circular cone mounted on a hemisphere of the same radius. The height of the cone and its radius are in the ratio $4 : 3$ , and the total height of the top is 49 cm. Find (i) the radius, (ii) the ratio of the curved surface area of the conical part to that of the hemispherical part, and (iii) the total cost of painting the top if the cone is painted at ₹10 per 100 cm² and the hemisphere at ₹5 per 100 cm². Take $π = \frac{22}{7}$ .
Hint
Write $h = \frac{4 r}{3}$ , so total height $= \frac{4 r}{3} + r = \frac{7 r}{3}$ . The area ratio is $π r l : 2 π r^{2} = l : 2 r$ .
Answer
$\frac{7 r}{3} = 49$ , so $r = 21$ cm and $h = 28$ cm. $l = 441 + 784 = 1225 = 35$ cm. Cone CSA $= π r l = 66 \times 35 = 2310$ cm²; hemisphere CSA $= 2 π r^{2} = 132 \times 21 = 2772$ cm². Ratio $= l : 2 r = 35 : 42 = 5 : 6$ . Cost $= \frac{2310}{100} \times 10 + \frac{2772}{100} \times 5 = 231 + 138.60 = 369.60$ , that is ₹369.60.
Q6 · 3 marks · Short answerHOTS / competency
A solid toy is a cone mounted on a hemisphere of the same radius $r$ . The curved surface area of the cone equals that of the hemisphere. Show that the height of the cone is $3 r$ . Hence, for $r = 7$ cm, find the total height and the volume of the toy. Take $π = \frac{22}{7}$ and $3 = 1.732$ .
Hint
Set $π r l = 2 π r^{2}$ to find $l$ , then use Pythagoras.
Answer
$π r l = 2 π r^{2}$ gives $l = 2 r$ , so $h = l^{2} - r^{2} = 3 r^{2} = 3 r$ . For $r = 7$ : $h = 73 \approx 12.12$ cm, total height $\approx 19.12$ cm. Volume $= \frac{1}{3} π r^{2} (h + 2 r) = \frac{1}{3} \times 154 \times (12.124 + 14) \approx 1341$ cm³.

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 cone mounted on a hemisphere of the same radius?Show answer
$π r l + 2 π r^{2} = π r (l + 2 r)$
2.Volume of a cone mounted on a hemisphere?Show answer
$\frac{1}{3} π r^{2} (h + 2 r)$
3.Cone height in a toy of total height $H$ ?Show answer
$h = H - r$
4.Cylinder length in a capsule of total length $L$ ?Show answer
$h = L - 2 r$
5.Capsule TSA in terms of total length $L$ ?Show answer
$2 π r L$
6.Volume of a capsule?Show answer
$π r^{2} h + \frac{4}{3} π r^{3}$

Cone on a Hemisphere and Similar Shapes: frequently asked questions

Why do we not add the base of the cone in a toy?

The base of the cone is glued to the flat face of the hemisphere. Both circles are inside the toy and cannot be seen or painted, so neither is part of the surface area. Only the two curved surfaces count.

Is the volume of a combination also reduced by the hidden faces?

No. A face has area but no volume, so joining two solids loses no space. The volume of the combination is simply the sum of the two volumes.

What if the cone and the hemisphere have different radii?

Then the flat circles do not match exactly, and a ring of flat surface stays exposed, which you must add. In these combination problems the radii are the same, so always check the question to confirm this.

Is the frustum of a cone in this chapter for 2026-27?

No. The frustum of a cone is not in the 2026-27 syllabus, so you do not need to study it.