MathsCBSE Class 9 · Math of Space: Surface Area and Volume

Cones, Spheres, Hemispheres and Recasting Solids: Class 9

Updated 17 October 2026 · Fact-checked

For a cone of radius $r$, height $h$ and slant height $l = \sqrt{r^2 + h^2}$: CSA $= \pi rl$, TSA $= \pi r(r + l)$, volume $= \frac{1}{3}\pi r^2 h$. For a sphere: surface $4\pi r^2$, volume $\frac{4}{3}\pi r^3$. For a solid hemisphere: CSA $2\pi r^2$, TSA $3\pi r^2$, volume $\frac{2}{3}\pi r^3$. When a solid is melted and recast, the volume stays the same, so equate volumes.

With the same radius r and height r, the cone, hemisphere and cylinder hold volumes in the ratio 1 : 2 : 3.

Illustration drawn for QuestionBuddy.

Understand Cones, Spheres, Hemispheres and Recasting Solids

Cone. A right circular cone has a circular base of radius $r$ , height $h$ and slant height $l = r^{2} + h^{2}$ (Pythagoras on the right triangle formed by the height, radius and slant edge). Unrolling the curved surface gives a sector of a circle of radius $l$ and arc length $2 π r$ , whose area is $\frac{1}{2} \times l \times 2 π r = π r l$ . So CSA $= π r l$ and TSA $= π r l + π r^{2} = π r (l + r)$ . The volume is one third of the cylinder on the same base and height: $V = \frac{1}{3} π r^{2} h$ . For $r = 3.5$ , $h = 12$ : $l = 12.5$ , CSA $= 137.5$ , $V = 154$ (using $π = \frac{22}{7}$ ).

Sphere and hemisphere. A sphere of radius $r$ has surface area $4 π r^{2}$ (four times the area of its great circle) and volume $\frac{4}{3} π r^{3}$ . A hemisphere is half a sphere: its curved surface area is $2 π r^{2}$ , its total surface area (including the flat circular face) is $3 π r^{2}$ , and its volume is $\frac{2}{3} π r^{3}$ . For a hollow hemispherical bowl, the CSA of the inside is the area to be coated and the volume is the capacity.

Ratio of volumes. With a cone, a hemisphere and a cylinder all of radius $r$ and (for the cone and cylinder) height $r$ , the volumes are $\frac{1}{3} π r^{3} : \frac{2}{3} π r^{3} : π r^{3} = 1 : 2 : 3$ .

Combined solids. Draw the solid as parts, list the surfaces that remain visible, and add only those. A toy made of a cone $(r = 3, h = 4, l = 5)$ topped by a hemisphere of radius $3$ : surface $= π r l + 2 π r^{2} = 15 π + 18 π = 33 π$ , since the joining circle is not part of the outside; volume $= \frac{1}{3} π r^{2} h + \frac{2}{3} π r^{3} = 12 π + 18 π = 30 π$ .

Melting and recasting. A solid melted and recast keeps its volume. A sphere of radius $6$ cm recast into a cylinder of radius $3$ cm: $\frac{4}{3} π (6)^{3} = π (3)^{2} h$ , so $288 π = 9 π h$ and $h = 32$ cm. Three spheres of radii $3, 4, 5$ melted into one sphere of radius $R$ : $R^{3} = 27 + 64 + 125 = 216$ , so $R = 6$ . Notice that surface areas do not add up: $4 π (9 + 16 + 25) = 200 π$ but the new sphere has only $144 π$ .

Scaling. If all lengths of a solid are multiplied by $k$ , every area is multiplied by $k^{2}$ and every volume by $k^{3}$ . Doubling a toy's dimensions multiplies its paint area by $4$ and its volume by $8$ .

Method. List the given dimensions, find the missing one (often $l$ ), choose between CSA, TSA and the visible surface, substitute with $π = \frac{22}{7}$ or leave $π$ in the answer, and equate volumes for recasting.

Cones, Spheres, Hemispheres and Recasting Solids: formulas to remember

Slant height: $l = r^{2} + h^{2}$; Right triangle of $r$ , $h$ , $l$ .
Cone: $CSA = π r l, TSA = π r (r + l), V = \frac{1}{3} π r^{2} h$; Volume is $\frac{1}{3}$ of the cylinder.
Sphere: $S = 4 π r^{2}, V = \frac{4}{3} π r^{3}$; Surface is four great circles.
Hemisphere: $CSA = 2 π r^{2}, TSA = 3 π r^{2}, V = \frac{2}{3} π r^{3}$; TSA includes the flat face.
Ratio: $V_{cone} : V_{hemisphere} : V_{cylinder} = 1 : 2 : 3$; Same radius $r$ and height $r$ .
Recasting: $V_{before} = V_{after}$; Surface areas need not be equal.
Scaling: $area \times k^{2}, volume \times k^{3}$; All lengths multiplied by $k$ .

How to solve Cones, Spheres, Hemispheres and Recasting Solids questions

Identify each solid, find missing dimensions, and decide whether to add surfaces or volumes.

1Sketch the solid and mark $r$ , $h$ and $l$ .
2Find the slant height by $l = r^{2} + h^{2}$ if it is not given.
3For a combined solid, add volumes of the parts, but add only the visible surfaces.
4For melting and recasting, equate the total volume before and after and solve for the unknown.
5Leave $π$ in the answer when the radii make cancellation easy; otherwise use $\frac{22}{7}$ .
6State the unit and check with a quick estimate.

Cones, Spheres, Hemispheres and Recasting 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.

Recasting a solid

Suppose a goldsmith melts a solid gold ball of radius 3 cm and draws it into a long cylindrical rod of radius 1 cm.

What is the length of the rod?

Try it first, then show the solutionSolution

Volume of the ball $= \frac{4}{3} π (3)^{3} = 36 π$ cm $^{3}$ .
Rod volume $= π (1)^{2} h = π h$ .
Equate: $π h = 36 π$ , so $h = 36$ cm.

Answer: The rod is 36 cm long.

Memory hook: Melting changes shape, not volume.

Tricks to remember Cones, Spheres, Hemispheres and Recasting Solids

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

One third of the cylinder
A cone has one third of the volume of the cylinder with the same base and height.
Use it when: Remembering the cone volume.
Four circles make a sphere
Sphere surface is $4 π r^{2}$ : four times the area of one great circle.
Use it when: Sphere surface formula.
Hemisphere TSA is 3 pi r squared
Curved $2 π r^{2}$ plus flat face $π r^{2}$ .
Use it when: Solid hemisphere total surface.

Cones, Spheres, Hemispheres and Recasting Solids: solved examples

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

Example 1: Conical hat · 3 marks · Short answerBoard level

A conical party hat has base radius 6 cm and height 8 cm. Find its slant height, the paper needed for the curved surface, and the volume of the hat. Use $π = 3.14$ .

Try it first, then show the solutionSolution

$l = 6^{2} + 8^{2} = 100 = 10$ cm. (½)
Curved surface $= π r l = 3.14 \times 6 \times 10 = 188.4$ cm $^{2}$ . (1)
$V = \frac{1}{3} π r^{2} h = \frac{1}{3} \times 3.14 \times 36 \times 8 = 3.14 \times 96 = 301.44$ cm $^{3}$ . (1½)

Answer: $l = 10$ cm, paper $188.4$ cm $^{2}$ , volume $301.44$ cm $^{3}$ .

Example 2: Sphere and hemisphere · 3 marks · Short answerBoard level

A solid sphere has radius 10.5 cm. Find its surface area and volume. Also find the total surface area of a solid hemisphere of the same radius. Use $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Sphere surface $= 4 π r^{2} = 4 \times \frac{22}{7} \times 110.25 = 1386$ cm $^{2}$ . (1)
Volume $= \frac{4}{3} \times \frac{22}{7} \times (10.5)^{3} = \frac{4}{3} \times 3638.25 = 4851$ cm $^{3}$ . (1)
Hemisphere TSA $= 3 π r^{2} = 3 \times \frac{22}{7} \times 110.25 = 1039.5$ cm $^{2}$ . (1)

Answer: Sphere: $1386$ cm $^{2}$ and $4851$ cm $^{3}$ ; hemisphere TSA $1039.5$ cm $^{2}$ .

Example 3: Sphere recast into a cylinder · 5 marks · Long answerHarder board level

A solid metal sphere of radius 6 cm is melted and recast into a solid cylinder of radius 3 cm. Find the height of the cylinder and compare the total surface areas of the sphere and the cylinder. Use $π = \frac{22}{7}$ where needed.

Try it first, then show the solutionSolution

Volume of sphere $= \frac{4}{3} π (6)^{3} = 288 π$ cm $^{3}$ . (1)
Cylinder: $π (3)^{2} h = 9 π h$ . Equating, $9 π h = 288 π$ so $h = 32$ cm. (1)
Sphere surface $= 4 π (6)^{2} = 144 π$ cm $^{2}$ . (1)
Cylinder TSA $= 2 π r (r + h) = 2 π \times 3 \times 35 = 210 π$ cm $^{2}$ , which is $660$ cm $^{2}$ using $π = \frac{22}{7}$ . (1)
The cylinder has the larger surface area ( $210 π > 144 π$ ): volume is conserved, surface area is not. (1)

Answer: $h = 32$ cm; sphere $144 π$ cm $^{2}$ against cylinder $210 π = 660$ cm $^{2}$ .

Example 4: Hemispherical tank · 3 marks · Short answerHarder board level

A hemispherical water tank has an inner radius of 2.1 m. Find its capacity in litres and the area of its inner curved surface to be waterproofed. Use $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Capacity $= \frac{2}{3} π r^{3} = \frac{2}{3} \times \frac{22}{7} \times (2.1)^{3}$ . (1)
$(2.1)^{3} = 9.261$ ; $\frac{22}{7} \times 9.261 = 29.106$ ; times $\frac{2}{3}$ gives $19.404$ m $^{3} = 19404$ litres. (1)
Inner curved surface $= 2 π r^{2} = 2 \times \frac{22}{7} \times 4.41 = 27.72$ m $^{2}$ . (1)

Answer: Capacity $19404$ litres; inner curved surface $27.72$ m $^{2}$ .

Example 5: A toy: cone and hemisphere · 4 marks · Case-basedHOTS / competency

Case: a toy is a cone of base radius 3 cm and height 4 cm with a hemisphere of radius 3 cm fixed on its flat base so that the toy has a smooth spinning top shape. (i) Find the slant height of the cone. (ii) Find the volume in terms of $π$ and numerically with $π = 3.14$ . (iii) Find the paintable outer surface. (iv) A larger toy has all lengths doubled; find its volume and surface area.

Try it first, then show the solutionSolution

(i) $l = 3^{2} + 4^{2} = 5$ cm. (½)
(ii) $V = \frac{1}{3} π (9) (4) + \frac{2}{3} π (27) = 12 π + 18 π = 30 π \approx 94.2$ cm $^{3}$ . (1½)
(iii) Outer surface $= π r l + 2 π r^{2} = 15 π + 18 π = 33 π \approx 103.62$ cm $^{2}$ (the joining circle is hidden). (1)
(iv) Lengths doubled: volume multiplied by $8$ : $240 π \approx 753.6$ cm $^{3}$ ; surface multiplied by $4$ : $132 π \approx 414.48$ cm $^{2}$ . (1)

Answer: (i) 5 cm; (ii) $30 π \approx 94.2$ cm $^{3}$ ; (iii) $33 π \approx 103.62$ cm $^{2}$ ; (iv) $240 π$ cm $^{3}$ and $132 π$ cm $^{2}$ .

Where students lose marks in Cones, Spheres, Hemispheres and Recasting Solids

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

Mistake: Using the height instead of the slant height in the CSA
Writing $π r h$ for a cone.
Wrong working
CSA $= π \times 7 \times 24$ .
Right working
CSA $= π \times 7 \times 25$ with $l = 25$ .
Do this: Compute $l$ first.
Usually costs: 2 of 3 marks.
Mistake: Forgetting the factor 1/3 for the cone
Using the cylinder formula.
Wrong working
$V = π r^{2} h$ for a cone.
Right working
$V = \frac{1}{3} π r^{2} h$ .
Do this: Remember one third of the cylinder.
Usually costs: The answer.
Mistake: Adding the joining circle in combined solids
Using full TSA of each part.
Wrong working
Toy surface $= TSA of cone + TSA of hemisphere$ .
Right working
Outer surface $= CSA of cone + CSA of hemisphere$ .
Do this: Add only the surfaces that you can see.
Usually costs: 2 marks.

Exam tips

Always compute the slant height before any cone area.
For combined solids, list visible surfaces explicitly.
In recasting questions, cancel $π$ and common factors before computing.
Check that units are consistent and give litres if capacity is asked.

Practice questions on Cones, Spheres, Hemispheres and Recasting 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
The surface area of a sphere is $616$ cm $^{2}$ . Its radius is (use $π = \frac{22}{7}$ )
(a) 7 cm
(b) 14 cm
(c) 21 cm
(d) 3.5 cm
Hint
Solve $4 π r^{2} = 616$ .
Answer
(a) 7 cm. $r^{2} = \frac{616 \times 7}{4 \times 22} = 49$ , so $r = 7$ .
Q2 · 3 marks · Short answerBoard level
A cone has slant height 17 cm and height 15 cm. Find its base radius, its curved surface area and its volume in terms of $π$ , and then numerically with $π = 3.14$ .
Hint
$r = l^{2} - h^{2}$ .
Answer
$r = 289 - 225 = 8$ cm. CSA $= π r l = 136 π \approx 427.04$ cm $^{2}$ . $V = \frac{1}{3} π (64) (15) = 320 π \approx 1004.8$ cm $^{3}$ .
Q3 · 3 marks · Short answerHarder board level
A cone, a hemisphere and a cylinder have equal radius $r$ and the cone and the cylinder also have height $r$ . Show that their volumes are in the ratio $1 : 2 : 3$ .
Hint
Write each volume with $π r^{3}$ as a common factor.
Answer
Cone: $\frac{1}{3} π r^{2} \cdot r = \frac{1}{3} π r^{3}$ . Hemisphere: $\frac{2}{3} π r^{3}$ . Cylinder: $π r^{2} \cdot r = π r^{3}$ . Ratio $\frac{1}{3} : \frac{2}{3} : 1 = 1 : 2 : 3$ .
Q4 · 5 marks · Long answerHarder board level
Three solid metal spheres of radii 3 cm, 4 cm and 5 cm are melted to form one solid sphere. Find the radius of the new sphere, and compare the sum of the surface areas of the three spheres with the surface area of the new sphere. Use $π = \frac{22}{7}$ for the numerical difference.
Hint
Equate volumes: the common factor $\frac{4}{3} π$ cancels.
Answer
$R^{3} = 27 + 64 + 125 = 216$ , so $R = 6$ cm. Sum of old surfaces $= 4 π (9 + 16 + 25) = 200 π$ ; new surface $= 4 π \times 36 = 144 π$ . The surface area decreases by $56 π = 56 \times \frac{22}{7} = 176$ cm $^{2}$ , because volume (not area) is conserved when the spheres merge.
Q5 · 4 marks · Case-basedHOTS / competency
Case: a tent has a cylindrical part of radius 14 m and height 3 m, surmounted by a conical roof of the same radius and height 10.5 m. (i) Find the slant height of the cone. (ii) Find the canvas needed for the tent (cylindrical wall and roof, no floor). (iii) Find the cost of the canvas at 80 rupees per square metre. (iv) Find the volume of air inside. Use $π = \frac{22}{7}$ .
Hint
Canvas = CSA of the cylinder + CSA of the cone.
Answer
(i) $l = 1 4^{2} + 10. 5^{2} = 306.25 = 17.5$ m. (ii) Cylinder CSA $= 2 \times \frac{22}{7} \times 14 \times 3 = 264$ m $^{2}$ ; cone CSA $= \frac{22}{7} \times 14 \times 17.5 = 770$ m $^{2}$ ; total $1034$ m $^{2}$ . (iii) $1034 \times 80 = 82720$ rupees. (iv) Cylinder $\frac{22}{7} \times 196 \times 3 = 1848$ m $^{3}$ ; cone $\frac{1}{3} \times \frac{22}{7} \times 196 \times 10.5 = 2156$ m $^{3}$ ; total $4004$ m $^{3}$ .

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.Cone CSA?Show answer
$π r l$ .
2.Sphere volume?Show answer
$\frac{4}{3} π r^{3}$ .
3.Hemisphere TSA?Show answer
$3 π r^{2}$ .
4.Cone volume in terms of cylinder?Show answer
One third.
5.What stays the same in melting and recasting?Show answer
The volume.

Cones, Spheres, Hemispheres and Recasting Solids: frequently asked questions

Why is the cone volume one third of the cylinder?

This is a standard result, shown by filling a cone with sand or water and pouring it into a cylinder of the same base and height three times.

Does the surface area stay the same when a solid is recast?

No. Only the volume stays the same; the surface area usually changes.

When do I use TSA for a hemisphere?

When the flat circular face is also exposed, as in a solid hemisphere or a lid on top.