MathsCBSE Class 10 · Surface Areas and Volumes

Surface Area of a Combination of Solids for Class 10

Updated 13 October 2026 · Fact-checked

The surface area of a combination of solids is the total area of the surfaces you can see after two solids are joined or one is cut from another. Faces where solids touch are left out. Board questions ask for the total surface area of toys, vessels and blocks, often after finding a missing height, and then a painting cost.

The circle where the cone meets the hemisphere is sealed inside, so add only the two curved surfaces: πrl and 2πr².

Illustration drawn for QuestionBuddy.

Understand Surface Area of a Combination of Solids

A combination of solids is one object made from two simple solids, such as a cone sitting on a hemisphere or a cylinder with a hemispherical end. Its surface area is the area you could paint, touch or wrap from the outside.

Why it works: when two solids are glued together, the faces that press against each other are sealed inside. Paint cannot reach them, so they are not surface any more. That is why the TSA of a combination is usually not the sum of the two TSAs. You add only the exposed parts.

The working habit is to name the parts in words first, for example 'TSA = CSA of cone + CSA of hemisphere'. Then put in the formulas. This line shows your method and earns credit even if arithmetic slips.

Matching faces: if both solids have the same radius and meet at full circles, both circles vanish. If a small hemisphere sits on a big cube face, only a circle of area $π r^{2}$ is covered. The rest of that face still shows. So the cube loses $π r^{2}$ and the hemisphere adds $2 π r^{2}$ .

Cut-outs: when a cavity is scooped or drilled, you lose the patch where the hole opens but gain the inner wall of the cavity. Remaining TSA = outer TSA − opening(s) + inner curved surface.

Cavities in a cylinder: if a cone of the same radius and height is hollowed out of one end of a cylinder, the top circle is lost and the cone's slanting inner wall is gained. Only the bottom circle stays, so the remaining TSA = $2 π r h + π r^{2} + π r l$ . If a hemisphere is scooped out of each end, both circles are lost and two bowls are gained, so TSA = $2 π r h + 4 π r^{2}$ .

Conditions: the smaller face must fit on the bigger one (diameter not more than the edge or base), and every length must be the one the formula expects. A cone's CSA needs the slant height $l = r^{2} + h^{2}$ . In toys and capsules, the given total height or length includes the hemisphere radius. This topic is about area only. Volumes behave differently, because joining solids hides no space, and they are covered on the next page.

Surface Area of a Combination of Solids: formulas to remember

Exposed-surface rule: $TSA = sum of exposed surfaces only$; Leave out every face where the two solids touch.
Building block: cube: $TSA = 6 a^{2}$; Six square faces of edge $a$ .
Building block: cuboid: $TSA = 2 (l b + bh + h l)$; Length $l$ , breadth $b$ , height $h$ .
Building block: cylinder: $CSA = 2 π r h, TSA = 2 π r (r + h)$; TSA adds the two circular ends.
Building block: cone: $CSA = π r l, TSA = π r (r + l)$; Uses slant height $l$ , not vertical height.
Building block: sphere: $SA = 4 π r^{2}$; A sphere has only a curved surface.
Building block: hemisphere: $CSA = 2 π r^{2}, TSA = 3 π r^{2}$; TSA adds the flat circle $π r^{2}$ .
Cone on hemisphere (same r): $TSA = π r l + 2 π r^{2} = π r (l + 2 r)$; Cone height $h$ = total height $- r$ ; then $l = r^{2} + h^{2}$ .
Cylinder with cone on top, closed base (same r): $TSA = π r^{2} + 2 π r H + π r l$; Drop $π r^{2}$ if the base is open, as in a tent.
Cylinder with hemisphere on one end, closed base (same r): $TSA = π r^{2} + 2 π r H + 2 π r^{2} = 3 π r^{2} + 2 π r H$; $H$ is the cylinder's height only.
Capsule: hemisphere at each end (same r): $TSA = 2 π r h + 4 π r^{2} = 2 π r (h + 2 r)$; $h$ = total length $- 2 r$ .
Hemisphere on, or scooped from, a cube face: $TSA = 6 a^{2} - π r^{2} + 2 π r^{2} = 6 a^{2} + π r^{2}$; Needs $2 r \leq a$ .
Two equal cubes joined face to face: $TSA = 10 a^{2}$; Two faces of area $a^{2}$ are hidden, so $12 a^{2} - 2 a^{2}$ .
Cylindrical hole drilled right through: $TSA = outer TSA - 2 π r^{2} + 2 π r h$; $h$ is the length of the hole.
Cylinder with cone scooped from one end (same r and h): $TSA = 2 π r h + π r^{2} + π r l$; Top circle lost, cone wall gained; $l = r^{2} + h^{2}$ .
Cylinder with hemisphere scooped from each end (same r): $TSA = 2 π r h + 4 π r^{2}$; Both circles lost, two bowls gained; needs $h \geq 2 r$ .
Cone slant height: $l = r^{2} + h^{2}$; Find it before any cone CSA.

How to solve Surface Area of a Combination of Solids questions

This method works for any combined or hollowed solid in a board question.

1Draw a rough sketch and mark every radius, height and slant height given.
2Find any hidden length first: cone height = total height − r, capsule cylinder length = total length − 2r, slant height by Pythagoras.
3List in words which surfaces are exposed and which are hidden where the solids meet.
4For a cavity, subtract the opening(s) and add the inner curved wall.
5Write TSA as a sum of named parts, then substitute formulas.
6Take a common factor such as $π r$ and cancel with $π = \frac{22}{7}$ before multiplying.
7Give the area in square units, then convert to cost if asked.

Surface Area of a 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.

Surface area of a combination: count only exposed surfaces

Imagine your class makes 20 wooden dolls for a Navratri display. Each doll is a solid cylinder of radius 3.5 cm and height 10 cm with a solid hemisphere of the same radius fixed on top as the head. Every outer surface, including the flat base, is painted. Suppose the painter charges ₹10 per 100 cm².

Find the area painted on one doll and the total cost of painting all 20 dolls.

Try it first, then show the solutionSolution

Exposed parts: base circle + CSA of cylinder + CSA of hemisphere (the top circle of the cylinder is hidden).
Base $= π r^{2} = \frac{22}{7} \times 3.5 \times 3.5 = 38.5$ cm²
CSA of cylinder $= 2 π r H = 2 \times \frac{22}{7} \times 3.5 \times 10 = 220$ cm²
CSA of hemisphere $= 2 π r^{2} = 77$ cm²
Area per doll $= 38.5 + 220 + 77 = 335.5$ cm²
For 20 dolls: $335.5 \times 20 = 6710$ cm²
Cost $= \frac{6710}{100} \times 10 =$ ₹671

Answer: 335.5 cm² per doll; total cost ₹671

Memory hook: Glue the head on and the neck circle disappears: paint only what shows.

Tricks to remember Surface Area of a Combination of Solids

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

Glue hides, paint shows
Any face touched by glue is gone; any face paint can reach counts.
Use it when: Deciding which circles to drop in joined solids.
Lose a lid, gain a bowl
Scooping a hemisphere loses $π r^{2}$ but gains $2 π r^{2}$ , so net $+ π r^{2}$ .
Use it when: Cube with hemisphere on or scooped out: TSA $= 6 a^{2} + π r^{2}$ .
Total minus the round bits
Total height minus one radius for a toy; total length minus two radii for a capsule.
Use it when: Finding the cone or cylinder height before using any formula.

Surface Area of a Combination of Solids: solved examples

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

Example 1: Two cubes joined end to end · 2 marks · Very short answerBoard level

Two identical solid cubes, each of volume 64 cm³, are joined end to end. Find the total surface area of the resulting cuboid.

Try it first, then show the solutionSolution

Edge: $a^{3} = 64$ , so $a = 4$ cm. (½ mark)
The cuboid is $8 \times 4 \times 4$ cm, as two faces are hidden. (½ mark)
TSA $= 2 (8 \times 4 + 4 \times 4 + 4 \times 8) = 2 (32 + 16 + 32) = 160$ cm². (1 mark)
Check: $12 a^{2} - 2 a^{2} = 10 \times 16 = 160$ cm².

Answer: 160 cm²

Example 2: Toy with a hidden cone height · 3 marks · Short answerBoard level

A 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 its total surface area. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Cone height $h = 15.5 - 3.5 = 12$ cm. (½ mark)
$l = 3. 5^{2} + 1 2^{2} = 12.25 + 144 = 156.25 = 12.5$ cm. (1 mark)
Both flat circles are hidden, so TSA $= π r l + 2 π r^{2} = π r (l + 2 r)$ . (½ mark)
TSA $= \frac{22}{7} \times 3.5 \times (12.5 + 7) = 11 \times 19.5 = 214.5$ cm². (1 mark)

Answer: 214.5 cm²

Example 3: Capsule from its total length · 3 marks · Short answerBoard level

A medicine capsule is a cylinder with a hemisphere at each end. Its diameter is 7 mm and its total length is 21 mm. Find the surface area of the capsule. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$r = 3.5$ mm; cylinder length $h = 21 - 2 \times 3.5 = 14$ mm. (1 mark)
Exposed: CSA of cylinder + two hemisphere CSAs, so TSA $= 2 π r h + 4 π r^{2} = 2 π r (h + 2 r)$ . (1 mark)
TSA $= 2 \times \frac{22}{7} \times 3.5 \times (14 + 7) = 22 \times 21 = 462$ mm². (1 mark)

Answer: 462 mm²

Example 4: Finding the radius first, then cost · 5 marks · Long answerHarder board level

A solid is a cylinder with a hemisphere of the same radius on its top. The curved surface area of the hemisphere is 308 cm² and the total height of the solid is 19 cm. Find the total surface area of the solid (base included) and the cost of polishing it at ₹10 per 100 cm². Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$2 π r^{2} = 308 \Rightarrow r^{2} = \frac{308 \times 7}{44} = 49$ , so $r = 7$ cm. (1 mark)
Cylinder height $H = 19 - 7 = 12$ cm. (1 mark)
TSA = base + CSA of cylinder + CSA of hemisphere $= π r^{2} + 2 π r H + 2 π r^{2}$ . (1 mark)
$= 154 + 2 \times \frac{22}{7} \times 7 \times 12 + 308 = 154 + 528 + 308 = 990$ cm². (1½ marks)
Cost $= \frac{990}{100} \times 10 =$ ₹99. (½ mark)

Answer: TSA = 990 cm²; cost = ₹99

Example 5: Cylinder with a conical cavity · 5 marks · Long answerHarder board level

A solid cylinder has height 12 cm and diameter 7 cm. A right circular cone of the same height and the same diameter is hollowed out from it, so that the base of the cone is the top face of the cylinder and its vertex touches the centre of the bottom face. Find the total surface area of the remaining solid, and find how much greater it is than the total surface area of the original cylinder. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

$r = 3.5$ cm, $h = 12$ cm. Slant height of the cavity: $l = 3. 5^{2} + 1 2^{2} = 156.25 = 12.5$ cm. (1 mark)
Exposed: CSA of cylinder + bottom circle + inner wall of the cone (CSA of cone); the top circle is lost. (1 mark)
TSA $= 2 π r h + π r^{2} + π r l = π r (2 h + r + l) = \frac{22}{7} \times 3.5 \times (24 + 3.5 + 12.5) = 11 \times 40 = 440$ cm². (1½ marks)
Original closed cylinder: TSA $= 2 π r (r + h) = 2 \times \frac{22}{7} \times 3.5 \times 15.5 = 22 \times 15.5 = 341$ cm². (1 mark)
Increase $= 440 - 341 = 99$ cm². Check: $π r l - π r^{2} = 137.5 - 38.5 = 99$ cm². (½ mark)

Answer: Remaining TSA = 440 cm²; it is 99 cm² more than the original cylinder's TSA

Example 6: Capsule radius from the painting cost · 5 marks · Long answerHarder board level

A solid wooden showpiece is a cylinder with a hemisphere of the same radius at each end. The length of the cylindrical part is 3 times its radius. Painting the whole surface of the showpiece at ₹5 per 100 cm² costs ₹77. Find the radius and the total length of the showpiece. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Area painted $= \frac{77}{5} \times 100 = 1540$ cm². (1 mark)
Let the radius be $r$ ; cylinder length $h = 3 r$ . Exposed: CSA of cylinder + two hemisphere CSAs, so TSA $= 2 π r h + 4 π r^{2}$ . (1 mark)
TSA $= 2 π r (3 r) + 4 π r^{2} = 10 π r^{2}$ . (1 mark)
$10 \times \frac{22}{7} \times r^{2} = 1540 \Rightarrow r^{2} = \frac{1540 \times 7}{220} = 49$ , so $r = 7$ cm. (1 mark)
Total length $= 3 r + 2 r = 5 r = 35$ cm. Check: $10 \times \frac{22}{7} \times 49 = 1540$ cm². (1 mark)

Answer: Radius = 7 cm; total length = 35 cm

Example 7: Case study: trophy base · 4 marks · Case-basedHOTS / competency

A trophy has a cuboidal base 10 cm long, 10 cm wide and 5 cm high. A solid cone of base radius 3.5 cm stands on the centre of its top face, and the total height of the trophy is 17 cm. The whole trophy, including its bottom, is to be coated at ₹2 per cm². Take $π = \frac{22}{7}$ .
(i) Find the height and the slant height of the cone. (1 mark)
(ii) Find the area of the top face of the cuboid that remains exposed. (1 mark)
(iii) Find the total surface area of the trophy and the cost of coating it. (2 marks)

Try it first, then show the solutionSolution

(i) Cone height $h = 17 - 5 = 12$ cm. (½ mark)
$l = 3. 5^{2} + 1 2^{2} = 156.25 = 12.5$ cm. (½ mark)
(ii) Exposed top $= 100 - π r^{2} = 100 - 38.5 = 61.5$ cm². (1 mark)
(iii) Cuboid TSA $= 2 (100 + 50 + 50) = 400$ cm². (½ mark)
Cone CSA $= π r l = \frac{22}{7} \times 3.5 \times 12.5 = 137.5$ cm². (½ mark)
TSA $= 400 - 38.5 + 137.5 = 499$ cm². (½ mark)
Cost $= 499 \times 2 =$ ₹998. (½ mark)

Answer: (i) 12 cm and 12.5 cm (ii) 61.5 cm² (iii) 499 cm²; cost ₹998

Where students lose marks in Surface Area of a Combination of Solids

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

Mistake: Adding full TSAs of both solids
The joined faces are hidden but still counted.
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: List exposed surfaces in words before writing formulas.
Usually costs: Usually the method mark and the answer.
Mistake: Using total height as cone height
The total height includes the hemisphere's radius.
Wrong working
$h = 15.5$ , $l = 3. 5^{2} + 15. 5^{2}$
Right working
$h = 15.5 - 3.5 = 12$ , $l = 12.5$
Do this: Always subtract r (or 2r for a capsule) first.
Usually costs: 1 to 2 marks, and every later step goes wrong.
Mistake: Using vertical height in cone CSA
Mixing up h and l.
Wrong working
CSA $= π r h = \frac{22}{7} \times 3.5 \times 12 = 132$
Right working
CSA $= π r l = \frac{22}{7} \times 3.5 \times 12.5 = 137.5$
Do this: Find l by Pythagoras before any cone area.
Usually costs: 1 mark or more.
Mistake: Forgetting the inner wall of a cavity
Only the lost opening is subtracted.
Wrong working
TSA $= 6 a^{2} - π r^{2}$
Right working
TSA $= 6 a^{2} - π r^{2} + 2 π r^{2} = 6 a^{2} + π r^{2}$
Do this: Say 'lose the lid, gain the bowl' for every cut-out.
Usually costs: 1 to 2 marks.

Exam tips

Write a words line like 'TSA = CSA of cone + CSA of hemisphere' before substituting; it carries method marks.
Show each found length (h, l, r) on its own line with units.
Cancel with $π = \frac{22}{7}$ early; radii like 3.5, 7 and 10.5 are chosen for this.
Read for 'open', 'closed', 'base included', 'inside' before deciding which circles count.
In MCQs, check whether the answer choices include the sum of full TSAs; that is the trap option.

Practice questions on Surface Area of a 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 hemisphere of diameter 7 cm is fixed flat-side down on the top face of a solid cube of edge 7 cm. The total surface area of the solid is:
(a) 294 cm²
(b) 332.5 cm²
(c) 371 cm²
(d) 409.5 cm²
Hint
The cube loses one circle; the hemisphere adds its curved surface.
Answer
(b) 332.5 cm². $294 - 38.5 + 77 = 332.5$ .
Q2 · 3 marks · Short answerBoard level
A model rocket is a closed cylinder of radius 3.5 cm and height 10 cm with a cone of the same radius on top. The total height of the model is 22 cm. Find its total surface area. Take $π = \frac{22}{7}$ .
Hint
Cone height is 22 − 10. Find l, then add base, cylinder CSA and cone CSA.
Answer
$h = 12$ , $l = 12.5$ cm. TSA $= 38.5 + 220 + 137.5 = 396$ cm².
Q3 · 3 marks · Short answerBoard level
A cylindrical hole of radius 3.5 cm is drilled right through a solid cube of edge 14 cm, from the centre of one face to the centre of the opposite face. Find the total surface area of the remaining solid. Take $π = \frac{22}{7}$ .
Hint
Two circles are lost; the inner wall of length 14 cm is gained.
Answer
$1176 - 77 + 308 = 1407$ cm².
Q4 · 5 marks · Long answerHarder board level
An open vessel is a cylinder with a hemispherical bottom of the same inner radius 7 cm. Its total inner depth is 20 cm. Find the inner surface area and the cost of tin-plating it inside at ₹4 per 10 cm². Take $π = \frac{22}{7}$ .
Hint
Cylinder height is 20 − 7. No lid, so no circle is added.
Answer
$H = 13$ cm. Area $= 2 π r H + 2 π r^{2} = 572 + 308 = 880$ cm². Cost $= 88 \times 4 =$ ₹352.
Q5 · 1 mark · Assertion-reasonHOTS / competency
Assertion (A): A cone of the same radius and height as a solid cylinder is hollowed out from one end of the cylinder. The total surface area of the remaining solid is greater than that of the original cylinder.
Reason (R): The slant height of a cone is always greater than its base radius.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Hint
Compare the circle lost with the cone wall gained.
Answer
(a). Change in TSA $= π r l - π r^{2} = π r (l - r)$ , which is positive because $l = r^{2} + h^{2} > r$ . So R is true and explains A.
Q6 · 2 marks · Very short answerBoard level
A solid is made of a cylinder of length 10 cm and radius 3.5 cm with a hemisphere of the same radius fixed at each end. Find its curved surface area. Take $π = \frac{22}{7}$ .
Hint
Two hemispheres together have curved area $4 π r^{2}$ .
Answer
Exposed: CSA of cylinder + two hemisphere CSAs. (½ mark) Cylinder: $2 π r h = 2 \times \frac{22}{7} \times 3.5 \times 10 = 220$ cm². (½ mark) Hemispheres: $4 π r^{2} = 154$ cm², so total $= 374$ cm². (1 mark)
Q7 · 2 marks · Very short answerBoard level
Two identical solid cuboids, each 6 cm × 4 cm × 3 cm, are joined face to face along their 6 cm × 4 cm faces. Find the total surface area of the new solid.
Hint
The heights add; the other two dimensions stay the same.
Answer
New cuboid is 6 cm × 4 cm × 6 cm. (½ mark) TSA $= 2 (l b + bh + h l)$ . (½ mark) $= 2 (24 + 24 + 36) = 168$ cm². (1 mark) Check: $2 \times 108 - 2 \times 24 = 168$ cm².
Q8 · 5 marks · Long answerHarder board level
A toy is a right circular cone mounted on a hemisphere of the same radius. The total height of the toy is 31 cm and the slant height of the cone is 25 cm. The height of the cone is greater than its radius. Find the radius and the total surface area of the toy. Take $π = \frac{22}{7}$ .
Hint
Write the cone height as 31 − r and use Pythagoras to form a quadratic in r.
Answer
$h = 31 - r$ . $r^{2} + (31 - r)^{2} = 625 \Rightarrow r^{2} - 31 r + 168 = 0 \Rightarrow (r - 7) (r - 24) = 0$ . Since $h > r$ , $r = 7$ cm and $h = 24$ cm. TSA $= π r (l + 2 r) = \frac{22}{7} \times 7 \times (25 + 14) = 858$ cm².
Q9 · 5 marks · Long answerHarder board level
The radius and height of a solid cylinder are in the ratio 7 : 24, and their sum is 31 cm. A cone of the same radius and height is hollowed out from one end of the cylinder. Find the total surface area of the remaining solid and the cost of polishing it at ₹5 per 100 cm². Take $π = \frac{22}{7}$ .
Hint
Find r and h from the ratio first, then l. Lose the top circle, gain the cone wall.
Answer
$r = 7$ cm, $h = 24$ cm, $l = 49 + 576 = 25$ cm. TSA $= 2 π r h + π r^{2} + π r l = 1056 + 154 + 550 = 1760$ cm². Cost $= \frac{1760}{100} \times 5 =$ ₹88.
Q10 · 5 marks · Long answerHarder board level
From a solid wooden cylinder of height 10 cm, a hemisphere is scooped out of each end. Each hemisphere has the same radius as the cylinder. The cost of polishing the whole surface of the article at ₹50 per 100 cm² is ₹187. Find the radius of the cylinder. Take $π = \frac{22}{7}$ .
Hint
Find the area from the cost, then use TSA = 2πrh + 4πr² to form a quadratic in r.
Answer
Area $= \frac{187}{50} \times 100 = 374$ cm². $2 π r (10 + 2 r) = 374 \Rightarrow r (10 + 2 r) = 59.5 \Rightarrow 4 r^{2} + 20 r - 119 = 0$ . $r = \frac{- 20 + 48}{8} = 3.5$ cm (the negative root is rejected). Check: $220 + 154 = 374$ cm².
Q11 · 4 marks · Case-basedHOTS / competency
Case study: Imagine your school puts up a tent for a science fair. The tent is a cylinder of radius 2.1 m and height 2 m, topped by a cone of the same radius. The total height of the tent is 4.8 m. Canvas is used only for the walls and the roof, because the floor is the ground. Suppose canvas costs ₹120 per m². Take $π = \frac{22}{7}$ .
(i) Find the slant height of the conical roof. (1 mark)
(ii) Find the area of ground covered by the tent. (1 mark)
(iii) Find the area of canvas needed and its cost. (2 marks)
Hint
Cone height is 4.8 − 2. The floor circle is not canvas.
Answer
(i) $h = 2.8$ m, $l = 4.41 + 7.84 = 3.5$ m. (ii) $π r^{2} = 13.86$ m². (iii) Canvas $= 2 π r H + π r l = 26.4 + 23.1 = 49.5$ m²; cost $= 49.5 \times 120 =$ ₹5940.

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.Which faces are left out in a combination's TSA?Show answer
Faces where the two solids touch.
2.TSA of a cone on a hemisphere of same r?Show answer
$π r (l + 2 r)$
3.Capsule cylinder length from total length L?Show answer
$h = L - 2 r$
4.TSA of cube with hemisphere on or scooped from a face?Show answer
$6 a^{2} + π r^{2}$
5.TSA of two equal cubes joined?Show answer
$10 a^{2}$

Surface Area of a Combination of Solids: frequently asked questions

Is the TSA of a combination the sum of the two TSAs?

No. The faces where the solids touch are hidden, so you must leave them out. Only the volume is a straight sum.

Why does scooping a hemisphere increase the surface area?

You lose a flat circle but gain a curved bowl whose area is twice that circle. So the net surface goes up by one circle's area.

Which NCERT exercise covers this topic?

Exercise 12.1 has questions on the surface area of combined solids, such as toys, vessels and blocks. Practise writing each answer as a sum of exposed parts.

Is the frustum of a cone in the 2026-27 syllabus?

No, the frustum of a cone is not in the 2026-27 syllabus.