MathsCBSE Class 10 · Surface Areas and Volumes

Cube, Cuboid and Cylinder Combination Problems for Class 10

Updated 14 October 2026 · Fact-checked

These problems join a cylinder or hemisphere to a cube or cuboid, or cut one out of it. For surface area you count only the faces you can see: lose the circle where the solids meet and gain the curved surface. For volume you add or subtract. Board papers set them as short answers, long answers and case studies.

Each joined or cut part hides one circle but exposes new surface, so the area gain is only 2πrh or πr²; volumes simply add or subtract.

Illustration drawn for QuestionBuddy.

Understand Cube, Cuboid and Cylinder Combination Problems

Picture a wooden block with a dome glued on top, a pen stand with a hole drilled in it, or a trophy with a pillar on a square base. Each one is a cube or cuboid combined with a cylinder or a hemisphere. This page practises exactly these shapes. That means two kinds of solid, which is the syllabus limit.

Joining on top. Suppose a cylinder of radius $r$ stands on the top face of a cuboid. The cuboid loses a circle of area $π r^{2}$ , because it is now covered. The cylinder's base is also hidden. But its top circle is still exposed, and that circle exactly replaces the lost patch. So the net gain is only the curved wall $2 π r h$ . A hemisphere on a face works the same way: lose $π r^{2}$ , gain $2 π r^{2}$ , so the net gain is $+ π r^{2}$ .

Cutting out. Now suppose a hemisphere is scooped out of a face. You lose the flat circle $π r^{2}$ where the hole opens, but you gain the inner bowl $2 π r^{2}$ . The net change is again $+ π r^{2}$ . Scooping actually adds surface. A cylindrical hole drilled right through a block opens on two faces, so you lose $2 π r^{2}$ and gain the inner wall $2 π r h$ . A hole that stops partway (a blind hole) opens only once, and it also has a circular floor. That floor replaces the lost circle, so the net gain is just $2 π r h$ .

Volume is simpler. No space is created or destroyed when you glue two solids together. So the volume of the joined solid is the sum of the two volumes. When you cut something out, subtract the volume removed. There are no hidden faces to worry about.

Conditions. The cut or attached piece must fit. Its diameter cannot be more than the shorter side of the face, and the depth of a hemisphere ( $r$ ) or a hole cannot be more than the block's height. That is why the largest hemisphere on a cube face has diameter equal to the edge. These rules come from the chapter's main idea in sections 12.2 and 12.3: surface area counts exposed faces, and volume adds or subtracts.

Cube, Cuboid and Cylinder Combination Problems: formulas to remember

Cuboid TSA and volume: $2 (l b + bh + h l), V = l bh$; Starting point for every block problem.
Cube TSA and volume: $6 a^{2}, a^{3}$; Used for cube-based combinations.
Cylinder: CSA, TSA and volume: $CSA = 2 π r h, TSA = 2 π r (r + h), V = π r^{2} h$; In combinations you usually need only the CSA $2 π r h$ , because the circular ends are hidden or replaced.
Hemisphere: CSA, TSA and volume: $CSA = 2 π r^{2}, TSA = 3 π r^{2}, V = \frac{2}{3} π r^{3}$; The curved bowl is $2 π r^{2}$ ; the TSA adds the flat circle $π r^{2}$ .
Cylinder standing on a cube or cuboid face: $TSA = TSA of block + 2 π r h$; The cylinder's top circle replaces the patch it covers, so only the curved wall is added. $V = V_{b l oc k} + π r^{2} h$ .
Hemisphere on, or scooped from, a face: $TSA = TSA of block + π r^{2}$; The same either way: lose $π r^{2}$ , gain $2 π r^{2}$ . Volume is $V_{b l oc k} \pm \frac{2}{3} π r^{3}$ .
Cylindrical hole drilled right through: $TSA = TSA of block - 2 π r^{2} + 2 π r h$; Here $h$ is the length of the hole. $V = V_{b l oc k} - π r^{2} h$ .
Blind cylindrical hole (not right through): $TSA = TSA of block + 2 π r h$; The circular floor of the hole replaces the opening. $V = V_{b l oc k} - π r^{2} h$ .
Visible-faces checklist: $TSA = TSA of block - hidden or opened circles + new exposed surfaces$; - Added solid: lose the covered patch, gain the added solid's outer surface (cylinder: wall and top; hemisphere: bowl).
- Scooped hemisphere: lose 1 circle, gain the bowl $2 π r^{2}$ .
- Through-hole: lose 2 circles, gain the wall $2 π r h$ .
- Blind hole: lose 1 circle, gain the wall $2 π r h$ and the floor $π r^{2}$ .
Fitting condition: $2 r \leq shorter side of face, depth \leq height$; The largest hemisphere on a cube face of edge $a$ has $r = \frac{a}{2}$ .

How to solve Cube, Cuboid and Cylinder Combination Problems questions

Use this method on any cube, cuboid and cylinder or hemisphere combination, whether the extra solid is added or cut out.

1Read carefully and note whether the extra solid is placed on the block, scooped out of it, or drilled right through it.
2Write down $l, b, h$ or $a$ for the block and $r, h$ for the cylinder or hemisphere. Convert a diameter to a radius.
3Check that the extra solid fits: its diameter is not more than the face, and its depth is not more than the height.
4If a length is unknown, set up an equation from the given volume or area and solve for it first.
5For surface area, write a word line such as 'TSA = TSA of cuboid − area of opening + inner curved surface', then substitute.
6For volume, add for a joined solid and subtract for a removed one, taking $π r^{2}$ out as a common factor where you can.
7Finish with units (cm² or cm³) and convert to cost or litres if asked.

Cube, Cuboid and Cylinder Combination Problems 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.

Hemisphere on, or scooped from, a face

Suppose you make a wooden diya stand for a festival from a cuboidal block 12 cm long, 8 cm wide and 5 cm high. Two hemispherical hollows, each of radius 2.1 cm, are scooped out of the top face to hold the diyas. The whole block, including the hollows and the bottom, is to be varnished.

Find the area to be varnished and the volume of wood left in the stand. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Fit check: the two hollows need $2 \times 4.2 = 8.4$ cm of length, which is less than 12 cm, and each depth of 2.1 cm is less than 5 cm.
TSA of block $= 2 (12 \times 8 + 8 \times 5 + 5 \times 12) = 2 (96 + 40 + 60) = 392$ cm²
Each hollow adds a net $π r^{2} = \frac{22}{7} \times 4.41 = 13.86$ cm²
Area to varnish $= 392 + 2 \times 13.86 = 419.72$ cm²
Volume of the block $= 12 \times 8 \times 5 = 480$ cm³
Volume of one hollow $= \frac{2}{3} \times \frac{22}{7} \times 9.261 = 19.404$ cm³
Wood left $= 480 - 2 \times 19.404 = 441.192$ cm³

Answer: Area to varnish = 419.72 cm²; wood left = 441.192 cm³

Memory hook: Every diya hollow loses a lid but gains a bowl: a net $+ π r^{2}$ .

Cylinder standing on a cube or cuboid face

Imagine your school builds a flag-pole base: a concrete cuboid platform 1.4 m by 1.4 m by 0.5 m, with a solid concrete cylinder of radius 0.35 m and height 2 m standing at its centre. Everything that can be seen is painted, but not the bottom face that rests on the ground.

Find the area to be painted, the cost of painting at ₹30 per m², and the volume of concrete used. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Cuboid without its bottom face $= 2 (1.96 + 0.7 + 0.7) - 1.96 = 6.72 - 1.96 = 4.76$ m²
The cylinder's top replaces the patch it covers, so add only its curved surface $= 2 \times \frac{22}{7} \times 0.35 \times 2 = 4.4$ m²
Area to paint $= 4.76 + 4.4 = 9.16$ m²
Cost of painting $= 9.16 \times 30 =$ ₹274.80
Volume of concrete $= 1.4 \times 1.4 \times 0.5 + \frac{22}{7} \times 0.1225 \times 2 = 0.98 + 0.77 = 1.75$ m³

Answer: Area painted = 9.16 m², cost = ₹274.80, concrete used = 1.75 m³

Memory hook: A pillar on a platform only adds its wall: its top replaces the patch it covers.

Tricks to remember Cube, Cuboid and Cylinder Combination Problems

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

Lid lost, bowl gained
A hemisphere on or in a face always gives net $+ π r^{2}$ , because $2 π r^{2} - π r^{2} = π r^{2}$ .
Use it when: Write the TSA in one line without listing every face.
Through-hole: two lids lost
A hole right through opens twice: subtract $2 π r^{2}$ , then add the wall $2 π r h$ .
Use it when: Drilled cubes and cuboids.
Glue adds, knife subtracts
For volume, joining a solid means add and cutting one out means subtract. There is nothing else to remember.
Use it when: Every volume part of these questions.

Cube, Cuboid and Cylinder Combination Problems: solved examples

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

Example 1: Cylinder standing on a cube · 2 marks · Very short answerBoard level

A solid cylinder of radius 3.5 cm and height 10 cm stands at the centre of the top face of a solid cube of edge 14 cm. Find the total surface area of the combined solid. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

TSA of cube $= 6 \times 1 4^{2} = 1176$ cm² (½)
The cylinder's top circle replaces the circle it covers on the cube, so add only its curved surface. (½)
CSA of cylinder $= 2 \times \frac{22}{7} \times 3.5 \times 10 = 220$ cm² (½)
TSA $= 1176 + 220 = 1396$ cm² (½)

Answer: 1396 cm²

Example 2: Hemisphere scooped from a cuboid · 3 marks · Short answerBoard level

A solid cuboid measures 25 cm by 21 cm by 14 cm. The largest possible hemisphere is scooped out of its 25 cm by 21 cm face. Find the total surface area and the volume of the remaining solid. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Largest diameter = shorter side of the face = 21 cm, so $r = 10.5$ cm. The depth of 10.5 cm is less than 14 cm, so it fits. (½)
TSA of cuboid $= 2 (525 + 294 + 350) = 2338$ cm² (½)
Net change in area $= + π r^{2} = \frac{22}{7} \times 110.25 = 346.5$ cm², so TSA $= 2684.5$ cm² (1)
Volume of hemisphere $= \frac{2}{3} \times \frac{22}{7} \times 1157.625 = 2425.5$ cm³ (½)
Remaining volume $= 25 \times 21 \times 14 - 2425.5 = 7350 - 2425.5 = 4924.5$ cm³ (½)

Answer: TSA = 2684.5 cm²; volume = 4924.5 cm³

Example 3: Trophy: cylinder on a cuboid base · 3 marks · Short answerBoard level

A trophy is a solid cylinder of radius 3.5 cm and height 14 cm fixed at the centre of the top of a cuboid base 10 cm by 10 cm by 4 cm. Every surface except the bottom of the base is to be polished. Find the area polished and the volume of the trophy. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Cuboid without its bottom face $= 2 (100 + 40 + 40) - 100 = 260$ cm² (1)
Add the cylinder's CSA only $= 2 \times \frac{22}{7} \times 3.5 \times 14 = 308$ cm², so the area polished $= 568$ cm² (1)
Volume $= 10 \times 10 \times 4 + \frac{22}{7} \times 12.25 \times 14 = 400 + 539 = 939$ cm³ (1)

Answer: Area polished = 568 cm²; volume = 939 cm³

Example 4: Find the radius of a drilled hole first · 5 marks · Long answerHarder board level

A cylindrical hole is drilled right through a solid metal cuboid 14 cm by 12 cm by 10 cm. The hole runs along the 10 cm edge and goes through the centres of the two 14 cm by 12 cm faces. The volume of metal left is 1295 cm³. Find (i) the radius of the hole, (ii) the total surface area of the remaining solid, and (iii) the increase in surface area compared with the original cuboid. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Volume of cuboid $= 14 \times 12 \times 10 = 1680$ cm³, so the hole's volume $= 1680 - 1295 = 385$ cm³ (1)
$\frac{22}{7} r^{2} \times 10 = 385$ , so $r^{2} = 12.25$ and $r = 3.5$ cm (1)
TSA of cuboid $= 2 (168 + 120 + 140) = 856$ cm² (½)
Two openings lost $= 2 \times 38.5 = 77$ cm²; inner wall gained $= 2 \times \frac{22}{7} \times 3.5 \times 10 = 220$ cm² (1)
TSA $= 856 - 77 + 220 = 999$ cm² (1)
Increase $= 999 - 856 = 143$ cm² (½)

Answer: (i) 3.5 cm (ii) 999 cm² (iii) 143 cm²

Example 5: Case study: added dome or scooped bowl · 4 marks · Case-basedHOTS / competency

Imagine a craft workshop makes paperweights from cubes of edge 21 cm. Design P has the largest possible solid hemisphere glued flat-side down on the top face. Design Q has the largest possible hemisphere scooped out of the top face. Take $π = \frac{22}{7}$ .
(i) Find the radius of the hemisphere. (1 mark)
(ii) Find the total surface area of design P. (1 mark)
(iii) Show that P and Q have equal total surface area, and find how much more volume P has than Q. (2 marks)

Try it first, then show the solutionSolution

(i) The diameter equals the edge, so $r = \frac{21}{2} = 10.5$ cm (1)
(ii) TSA of P $= 6 \times 441 + π r^{2} = 2646 + 346.5 = 2992.5$ cm² (1)
(iii) In Q you lose $π r^{2}$ and gain the bowl $2 π r^{2}$ , a net $+ π r^{2}$ , the same as P. So both have TSA 2992.5 cm². (1)
P's volume is $a^{3} + V_{h}$ and Q's is $a^{3} - V_{h}$ , so the difference $= 2 V_{h} = 2 \times 2425.5 = 4851$ cm³ (1)

Answer: (i) 10.5 cm (ii) 2992.5 cm² (iii) The TSAs are equal; P has 4851 cm³ more volume.

Example 6: Blind hole: find the depth from the gain in area · 5 marks · Long answerHarder board level

A solid wooden cuboid measures 20 cm by 15 cm by 10 cm. A cylindrical hole of radius 3.5 cm is drilled straight down into the centre of the 20 cm by 15 cm top face, but it does not go right through. After drilling, the total surface area of the block has increased by 132 cm². Find (i) the depth of the hole, (ii) the total surface area and the volume of the remaining solid, and (iii) the cost of varnishing the whole solid, inside and out, at ₹10 per 100 cm². Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Fit check: the diameter 7 cm is less than 15 cm, so the hole fits in the face. (½)
A blind hole loses the opening $π r^{2}$ but gains its floor $π r^{2}$ and the wall $2 π r h$ , so the net gain is $2 π r h$ . (1)
$2 \times \frac{22}{7} \times 3.5 \times h = 132$ , so $22 h = 132$ and $h = 6$ cm. Since 6 cm is less than 10 cm, the hole is indeed blind. (1)
TSA of cuboid $= 2 (300 + 150 + 200) = 1300$ cm², so new TSA $= 1300 + 132 = 1432$ cm² (1)
Volume left $= 20 \times 15 \times 10 - \frac{22}{7} \times 12.25 \times 6 = 3000 - 231 = 2769$ cm³ (1)
Cost $= \frac{1432}{100} \times 10 =$ ₹143.20 (½)

Answer: (i) 6 cm (ii) TSA = 1432 cm², volume = 2769 cm³ (iii) ₹143.20

Example 7: Hemisphere glued on a cuboid: painting cost and volume · 5 marks · Long answerBoard level

A memento is made of a solid cuboid 28 cm by 21 cm by 10 cm, with the largest possible solid hemisphere glued flat-side down at the centre of its 28 cm by 21 cm top face. The whole memento, except the bottom face of the cuboid, is painted at ₹20 per 100 cm². Find the cost of painting and the volume of the memento. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

Largest diameter = shorter side of the top face = 21 cm, so $r = 10.5$ cm. (½)
TSA of cuboid $= 2 (588 + 210 + 280) = 2156$ cm² (1)
The hemisphere covers $π r^{2}$ and adds the bowl $2 π r^{2}$ : net $+ π r^{2} = \frac{22}{7} \times 110.25 = 346.5$ cm², so TSA $= 2502.5$ cm² (1)
Leave out the bottom face: area painted $= 2502.5 - 588 = 1914.5$ cm² (½)
Cost $= \frac{1914.5}{100} \times 20 =$ ₹382.90 (1)
Volume $= 28 \times 21 \times 10 + \frac{2}{3} \times \frac{22}{7} \times 1157.625 = 5880 + 2425.5 = 8305.5$ cm³ (1)

Answer: Cost of painting = ₹382.90; volume = 8305.5 cm³

Example 8: Find the block's height from a given TSA · 3 marks · Short answerHarder board level

A solid metal stand is a cuboid with a square base of side 14 cm and height $H$ cm. A solid cylinder of radius 3.5 cm and height 10 cm is fixed at the centre of its top face. The total surface area of the stand is 1060 cm². Find $H$ and the volume of the stand. Take $π = \frac{22}{7}$ .

Try it first, then show the solutionSolution

TSA = TSA of cuboid + CSA of cylinder, since the cylinder's top replaces the patch it covers. (½)
CSA of cylinder $= 2 \times \frac{22}{7} \times 3.5 \times 10 = 220$ cm² (½)
$2 (196 + 14 H + 14 H) + 220 = 1060$ , so $612 + 56 H = 1060$ , $56 H = 448$ and $H = 8$ cm (1)
Volume $= 14 \times 14 \times 8 + \frac{22}{7} \times 12.25 \times 10 = 1568 + 385 = 1953$ cm³ (1)

Answer: $H = 8$ cm; volume = 1953 cm³

Where students lose marks in Cube, Cuboid and Cylinder Combination Problems

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

Mistake: Subtracting the opening but not adding the bowl
Students forget that the inside of a scooped hollow is an exposed surface.
Wrong working
$6 a^{2} - π r^{2}$
Right working
$6 a^{2} - π r^{2} + 2 π r^{2} = 6 a^{2} + π r^{2}$
Do this: Always ask what new surface the cut reveals.
Usually costs: 1 to 2 marks
Mistake: Adding the full TSA of a cylinder placed on a block
The cylinder's base and the patch it covers are hidden.
Wrong working
$TSA_{b l oc k} + 2 π r (r + h)$
Right working
$TSA_{b l oc k} + 2 π r h$
Do this: List the visible faces in words before using any formula.
Usually costs: 1 mark
Mistake: Subtracting only one circle for a through-hole
A hole that goes right through opens on two faces.
Wrong working
$TSA - π r^{2} + 2 π r h$
Right working
$TSA - 2 π r^{2} + 2 π r h$
Do this: Count the openings: through means two, blind means one (and a blind hole's floor replaces it).
Usually costs: 1 mark
Mistake: Using the diameter as the radius
'Largest hemisphere' gives the diameter equal to the edge.
Wrong working
$r = 21$ for a cube of edge 21 cm
Right working
$r = 10.5$ cm
Do this: Write $r = \frac{a}{2}$ on its own line.
Usually costs: Most of the question

Exam tips

Write the word line (for example 'TSA = TSA of cuboid + CSA of cylinder') first; it shows your method even if the arithmetic slips.
State the fit check for the 'largest' hemisphere or cylinder before using its radius.
Write radii like 10.5 as $\frac{21}{2}$ and cancel 7s before you multiply.
In case studies, reuse the radius and volumes you found in earlier parts and give units in every part.

Practice questions on Cube, Cuboid and Cylinder Combination Problems

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 hemisphere of radius 3.5 cm is scooped out of one face of a cube of edge 10 cm. What is the total surface area of the remaining solid? Take $π = \frac{22}{7}$ .
(a) 561.5 cm²
(b) 638.5 cm²
(c) 677 cm²
(d) 600 cm²
Hint
You lose a lid but gain a bowl.
Answer
(b) 638.5 cm², since $600 + π r^{2} = 600 + 38.5 = 638.5$ cm²
Q2 · 5 marks · Long answerHarder board level
A solid cuboid measures 15 cm by 12 cm by 14 cm. A cylindrical hole is drilled right through it, parallel to the 14 cm edges, joining the centres of the two 15 cm by 12 cm faces. Because of the hole, the total surface area increases by 231 cm². Find the radius of the hole, the cost of painting the remaining solid inside and out at ₹10 per 100 cm², and the volume of the remaining solid. Take $π = \frac{22}{7}$ .
Hint
Increase = wall gained minus two openings lost = $2 π r (14 - r)$ . You get a quadratic; reject the root that does not fit.
Answer
$2 π r h - 2 π r^{2} = \frac{44}{7} r (14 - r) = 231$ , so $r (14 - r) = 36.75$ , i.e. $r^{2} - 14 r + 36.75 = 0$ , giving $r = 3.5$ or $10.5$ . A diameter of 21 cm does not fit in a 15 cm by 12 cm face, so $r = 3.5$ cm. TSA $= 2 (180 + 168 + 210) + 231 = 1116 + 231 = 1347$ cm². Cost $= \frac{1347}{100} \times 10 =$ ₹134.70. Volume left $= 2520 - 38.5 \times 14 = 2520 - 539 = 1981$ cm³.
Q3 · 5 marks · Long answerHarder board level
A solid wooden cube has edge $a$ cm. A blind cylindrical hole of radius 2.1 cm and depth 5 cm is drilled into the centre of the top face, and another identical hole into the centre of the bottom face. The total surface area of the remaining solid is 996 cm². (i) Find $a$ and check that the two holes do not meet. (ii) Find the volume of wood left. (iii) Find the cost of varnishing the whole solid, inside and out, at ₹5 per 100 cm². Take $π = \frac{22}{7}$ .
Hint
Each blind hole loses its opening but gains a floor of the same size and a curved wall, so the net gain is $2 π r h$ per hole.
Answer
(i) Each hole adds $2 π r h = 2 \times \frac{22}{7} \times 2.1 \times 5 = 66$ cm², so $6 a^{2} + 132 = 996$ , $6 a^{2} = 864$ and $a = 12$ cm. The holes reach a total depth of $5 + 5 = 10$ cm, less than 12 cm, so they do not meet. (ii) Each hole has volume $\frac{22}{7} \times 4.41 \times 5 = 69.3$ cm³, so wood left $= 1728 - 2 \times 69.3 = 1589.4$ cm³. (iii) Cost $= \frac{996}{100} \times 5 = 49.80$ , that is ₹49.80.
Q4 · 5 marks · Long answerHarder board level
A solid cuboid has a square face of side 21 cm and height $h$ cm. The largest hemisphere is scooped out of the square top face, and the remaining volume is 3748.5 cm³. Find $h$ , the total surface area of the remaining solid, and the cost of polishing it at ₹2 per 10 cm². Take $π = \frac{22}{7}$ .
Hint
The hemisphere's volume is 2425.5 cm³; add it back to get the cuboid's volume.
Answer
$441 h = 3748.5 + 2425.5 = 6174$ , so $h = 14$ cm. TSA $= 2 (441 + 294 + 294) + 346.5 = 2058 + 346.5 = 2404.5$ cm². Cost $= \frac{2404.5}{10} \times 2 =$ ₹480.90.
Q5 · 3 marks · Short answerHOTS / competency
A cylinder of radius 3.5 cm and height 6 cm is fixed on a cube of edge 10 cm. A student writes: TSA $= 6 a^{2} + 2 π r (r + h) = 600 + 209 = 809$ cm². Find the error and give the correct total surface area. Take $π = \frac{22}{7}$ .
Hint
Which circles are hidden, and which circle is still on the outside?
Answer
The student added both circles of the cylinder and kept the covered patch of the cube. The cylinder's base and that patch are hidden, and its top replaces the patch, so add only $2 π r h = 132$ cm². Correct TSA $= 732$ cm².
Q6 · 1 mark · MCQHarder board level
A cylindrical hole of radius 3.5 cm is drilled right through a solid cube of edge 14 cm, between the centres of two opposite faces. How does the total surface area change? Take $π = \frac{22}{7}$ .
(a) It increases by 231 cm²
(b) It increases by 308 cm²
(c) It decreases by 77 cm²
(d) It increases by 269.5 cm²
Hint
Wall gained minus two openings lost.
Answer
(a) It increases by 231 cm², since $2 π r h - 2 π r^{2} = 308 - 77 = 231$ cm²
Q7 · 1 mark · MCQHarder board level
From two identical cubes of edge $a$ , solid P is made by gluing the largest possible hemisphere on top of one cube, and solid Q by scooping the largest possible hemisphere out of the top of the other. What is the ratio of the volume of P to the volume of Q? Take $π = \frac{22}{7}$ .
(a) 31 : 53
(b) 53 : 31
(c) 16 : 5
(d) 2 : 1
Hint
With $r = \frac{a}{2}$ , the hemisphere's volume is $\frac{π a ^{3}}{12}$ .
Answer
(b) 53 : 31, since $V_{h} = \frac{2}{3} π \frac{a ^{3}}{8} = \frac{11}{42} a^{3}$ , so P : Q $= (1 + \frac{11}{42}) : (1 - \frac{11}{42}) = 53 : 31$
Q8 · 1 mark · Assertion-reasonBoard level
Assertion (A): Scooping a hemisphere of radius $r$ out of one face of a solid cube increases its total surface area by $π r^{2}$ .
Reason (R): The curved bowl exposed, of area $2 π r^{2}$ , is larger than the flat circle removed, of area $π r^{2}$ .
(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
Work out the net change: what is lost and what is gained?
Answer
(a) Both A and R are true and R is the correct explanation of A. The net change is $2 π r^{2} - π r^{2} = π r^{2}$ , which is exactly why the area goes up.
Q9 · 2 marks · Very short answerBoard level
The largest possible right circular cylinder is to be cut from a solid cuboidal block 14 cm by 10 cm by 7 cm. Find the volume of this cylinder. Take $π = \frac{22}{7}$ .
Hint
Try the axis along each of the three edges. The diameter is limited by the shorter side of the end face.
Answer
Axis along 14 cm: $r = 3.5$ , $V = 38.5 \times 14 = 539$ cm³. Axis along 10 cm: $r = 3.5$ , $V = 385$ cm³. Axis along 7 cm: end face 14 by 10, so $r = 5$ , $V = \frac{22}{7} \times 25 \times 7 = 550$ cm³. The largest cylinder has volume 550 cm³.
Q10 · 4 marks · Case-basedBoard level
Suppose a stationery shop's display unit is a solid wooden cuboid 30 cm long and 20 cm wide. Before any work, the block has a volume of 6000 cm³. A blind cylindrical hole of radius 3.5 cm and depth 8 cm is drilled straight down at the centre of the top face to hold pens. Take $π = \frac{22}{7}$ .
(i) Find the height of the block. (1 mark)
(ii) Find the volume of wood left after drilling. (1 mark)
(iii) The whole remaining solid, including the inside of the hole, is varnished at ₹5 per 100 cm². Find the cost. (2 marks)
Hint
A blind hole's floor replaces its opening, so the area gained is just the wall.
Answer
(i) $30 \times 20 \times h = 6000$ , so $h = 10$ cm (the 8 cm hole fits). (ii) $6000 - 38.5 \times 8 = 6000 - 308 = 5692$ cm³. (iii) TSA $= 2 (600 + 200 + 300) + 2 \times \frac{22}{7} \times 3.5 \times 8 = 2200 + 176 = 2376$ cm²; cost $= \frac{2376}{100} \times 5 =$ ₹118.80.
Q11 · 4 marks · Case-basedHOTS / competency
Case study: Imagine a hostel's storage room is a cube of edge 4 m. A cylindrical water tank of height 2 m stands at the centre of its flat roof. When full, the tank holds 3080 litres (1 m³ = 1000 litres). Take $π = \frac{22}{7}$ .
(i) Find the radius of the tank. (1 mark)
(ii) The outside of the room (except the floor) and the tank are painted at ₹50 per m². Find the cost. (1 mark)
(iii) A second room of the same size has a tank of the same radius but a different height. Painting it in the same way costs ₹4550. Find the height of this tank and its capacity in litres. (2 marks)
Hint
The tank's top replaces the patch of roof it covers, so add only its curved surface. In (iii), work back from the cost to the area.
Answer
(i) $\frac{22}{7} r^{2} \times 2 = 3.08$ , so $r^{2} = 0.49$ and $r = 0.7$ m. (1)
(ii) Room without its floor $= 5 \times 16 = 80$ m²; tank CSA $= 2 \times \frac{22}{7} \times 0.7 \times 2 = 8.8$ m². Area $= 88.8$ m², cost $= 88.8 \times 50 = 4440$ , that is ₹4440. (1)
(iii) Area painted $= \frac{4550}{50} = 91$ m², so the tank CSA $= 91 - 80 = 11$ m². $2 \times \frac{22}{7} \times 0.7 \times h = 4.4 h = 11$ , so $h = 2.5$ m. Capacity $= \frac{22}{7} \times 0.49 \times 2.5 = 3.85$ m³ $= 3850$ litres. (2)

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.Net change in TSA when a hemisphere is scooped from a face?Show answer
$+ π r^{2}$
2.Net change in TSA when a cylinder stands on a face?Show answer
$+ 2 π r h$
3.Net change in TSA for a hole drilled right through?Show answer
$- 2 π r^{2} + 2 π r h$
4.Radius of the largest hemisphere on a cube face of edge $a$ ?Show answer
$\frac{a}{2}$
5.Volume when a solid is cut out of a block?Show answer
Block volume minus the volume removed

Cube, Cuboid and Cylinder Combination Problems: frequently asked questions

Why does scooping out a hemisphere increase the surface area?

You remove a flat circle but expose a curved bowl twice its area. So the net gain is one circle's area, and the total surface area goes up even though the volume goes down.

Do I count the bottom face of the block?

Count it for total surface area unless the question says the base rests on the ground or is not painted. Read words like 'except the base' carefully.

Is melting and recasting part of this topic?

No. Conversion of a solid from one shape to another is not in the 2026-27 syllabus.

Can I take pi as 3.14?

Use 22/7 unless the question says otherwise. Board numbers are usually chosen so that 22/7 cancels neatly.