NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes

The faster ``subtract the hidden faces'' route. You can start from the two separate cubes and just remove what gets hidden on joining.

• Start whole: two cubes have a combined surface of 2× 96 = 192 cm² before joining.
• Hide two faces: joining end to end hides one 16 cm² face on each cube, so 32 cm² is lost and the visible surface is 192-32=160.
• Show the cuboid first: the examiner accepts either route, but writing the dimensions 8× 4× 4 before you substitute reads cleaner.
• Mind the unit: always state the answer in square centimetres, as marks are deducted for a missing or wrong unit.

160 cm², whether found directly or by removing the two 16 cm² hidden faces.

Read ``inner'' and ``hollow'' before you pick a formula. Those two words decide which surfaces you count, and missing them is the top mark-loser here.

• Curved only: the vessel is hollow and you want the inner surface, so include just the two curved parts. No base and no top ring are counted.
• Factor first: pull out 2π r so the sum 2π r h + 2π r² becomes one product 2π r(h+r), and you substitute once instead of twice.
• Clean cancel: with r=7 the 227 cancels the seven straight away, leaving the tidy 2× 22× 13 = 572.
• Plan ahead: a cost-of-plating version would multiply this 572 cm² by a rate, so always finish the area cleanly.

572 cm², found as 2π r(h+r) with r=7, h=6.

Keep the curved surfaces separate, never add a base. The toy's outline has just two skins: the slanted cone wall and the rounded hemisphere bowl.

• No base: many students add the cone base area here, but that flat circle is glued against the hemisphere and stays hidden, so it never shows on the outside and must be left out of the total.
• Easy cancel: with r=3.5=72 the seven in the radius cancels the seven under the pi, so the cone term works out to 137.5 and the hemisphere term to 77, both of which you can finish in your head without a calculator.
• Words first: write the formula in words as cone curved surface plus hemisphere curved surface before you put in any numbers. That sentence alone often carries a method mark even if a later arithmetic slip costs you the final figure.
• Check the slant: the cone height of twelve came from subtracting the hemisphere radius, not its diameter, from the total height. The triangle then matches the familiar seven, twenty-four, twenty-five set halved, so a slant height that is not a neat decimal means you subtracted the wrong length.
• Keep the half: a cost-of-painting version multiplies this area by a rate, so finish the area cleanly and do not round the answer up to the nearest whole number, since the half a square centimetre still matters.

214.5 cm², the sum of cone CSA (137.5) and hemisphere CSA (77).

Two traps hide in this short question. Spot them and the rest is one clean line of arithmetic.

• Diameter, not radius: the side of the cube equals the diameter of the hemisphere, so the radius is half of that. Treating the side as the radius would double your hemisphere term and ruin the answer.
• Remove the patch: the circle of contact under the hemisphere leaves the top of the cube, otherwise that patch of area would be counted twice, once as cube top and once under the dome.
• One expression: cancel the removed flat circle against one of the added curved terms and write the surface as cube area plus one circle term from the start. That keeps three messy terms down to two clean ones and reads well in a board answer.
• Link to scooping: in the next question a hemisphere is carved out of a face instead of stuck on, yet the net change to the area is again the same single circle term, since a depression also opens a curved bowl while removing the same flat circle.

7 cm greatest diameter, surface area 332.5 cm² from 6a²+π r².

A depression behaves exactly like a bump for surface area. This is the surprising idea students should carry away from this pair of questions.

• Same net change: whether you add a hemisphere on a face or scoop one out of it, the flat circle leaves the count and a curved bowl of twice that area joins it, so the change to the surface is exactly the same in both cases. This is the surprising idea the question is built around.
• Stay symbolic: only the letter for the edge is given, with no number to plug in, so keep everything in symbols and factor at the very end. A single tidy bracketed expression is far easier for an examiner to read and for you to substitute into later than a loose sum of two terms.
• Halve carefully: the problem names the diameter, not the radius, so the radius is half of it and its square is a quarter of the edge squared. Squaring the wrong length is the single most common error students make on this question, so write the radius step out plainly.

l²4(24+π) square units, with r=l/2.

Subtract the rounded ends from the length first. Getting the cylinder height right is the whole question.

• Trim the length: each rounded end is a hemisphere whose depth is its radius, so the two of them eat into the length from both sides. After taking those off, the straight cylinder in the middle is shorter than the full capsule, and getting this height right is really the whole question.
• Caps make a sphere: the two hemispherical ends fit together into one whole sphere of the same radius, so the cleanest mental picture is one cylinder wall plus a single sphere skin rather than two separate caps.
• Factor for whole numbers: pulling the common factor out front turns the sum into one neat product, and the bracket then collapses to a whole number, so the final multiplication is painless and easy to check.
• Hold the unit: the capsule is measured in millimetres, so keep that unit through every line. Switching to centimetres halfway is a frequent and costly slip that quietly changes the answer.

220 mm², using 2π r(h+2r) with r=2.5, h=9.

Leave out the floor, and never recompute the slant height. Read the clauses carefully and the work is short.

• No base: ``the base will not be covered'' is a direct instruction to drop the π r² floor term from the canvas area.
• Use the given slant: the slant height is handed to you as 2.8 m, so do not recompute it with √r²+h², which wastes time and may disagree if figures are rounded.
• Factor and cancel: pull out π r; the bracket 2h+l=4.2+2.8=7 is whole, and with r=2 the 227 cancels to give 44 m².
• Finish the cost: the cost line 44× 500 =  22000 carries its own mark, so never stop at the area.

44 m² of canvas costing  22000 at  500 per m².

The cavity opens a cone wall, not a flat circle. Count the three surfaces correctly and the rest is light arithmetic.

• Slant, not base: a hollowed cone exposes its inside wall π r l while the cylinder's top circle disappears, so the top gives a π r l term and the bottom keeps its π r².
• Easy factor: the radius r=0.7=710 pairs with 227 to give 2.2 in one step, and the bracket r+2h+l=0.7+4.8+2.5=8.0 is round.
• Show the exact value: 2.2× 8 = 17.6, so write 17.6 first and only then round to the asked 18 cm².
• Know the version: this is a scoop-out figure where the surface keeps the walls, while a volume question would subtract the cone's volume instead. Mixing the two is a common slip.

≈ 18 cm² (exact 17.6 cm²), from π r(r+2h+l).

Scooping does not change the surface formula, only the volume. This symmetry with the capsule is worth memorising for board papers.

• Same skin: the flat circle that leaves each end and the curved bowl that arrives in its place share the same footprint, so the surface formula here is identical to the capsule with caps added. This symmetry between scooping out and sticking on is well worth memorising for board papers.
• Cancel first: the twice the radius cancels the seven sitting under the pi, so do that cancellation before you touch the bracket. Carrying small whole numbers through the rest keeps the arithmetic clean and quick.
• Then multiply: once the front collapses to a single whole number, the bracket also comes out whole, so the final product is a one-step multiplication you can check at a glance.
• Volume differs: a follow-up asking for the volume of wood would instead subtract the two hemispheres from the cylinder, so keep that contrast clear and never subtract anything on a surface-area question.

374 cm², from 2π r(h+2r) with r=3.5, h=10.

Add volumes; keep π in symbol form. The whole question turns on reading two instructions correctly.

• Volumes add: joined solids lose no volume, so the answer is the cone plus the hemisphere with nothing subtracted.
• Stop at π: ``in terms of π'' means leave the answer as π cm³, never as a decimal like 3.14.
• Light algebra: with r=1 you have r²=r³=1, so the cone gives 13π, the hemisphere gives 23π, and the thirds add to a whole π almost by inspection.
• Quick check: both parts share the radius and the cone height equals it, so the squat solid landing on a tidy π cm³ signals the numbers behaved.

π cm³, the sum of 13π (cone) and 23π (hemisphere).

Keep everything over a denominator of 7. Working in sevenths keeps the arithmetic exact from start to finish.

• Common denominator: because the value of pi used here is twenty-two over seven, every volume term naturally carries a seven in the denominator. If you write the cylinder and the two cones over that same seven, you can add the numerators in one stroke and divide once at the end, with no decimals to round midway and no small errors creeping in.
• No wall thickness: the phrase about the outer and inner dimensions being nearly the same is telling you to treat the aluminium sheet as having no thickness at all, so the air inside fills exactly the full solid volume and no subtraction for the wall is needed.
• Double one cone: the two cones share the same radius and the same height, so the safest book-keeping is to work out one cone and simply double it, rather than inventing a single tall cone of combined height.
• Find the cylinder height: the given length runs from one cone tip to the other, so you must subtract both cone lengths to get the straight cylinder in the middle. Forgetting to remove both cones is the single most common error on this problem, and it makes the cylinder volume far too large.
• Ratio check: a cone and a cylinder of equal radius and height stand in volumes of one to three. The cones here are shorter than the cylinder, so together they add only a small fraction of the total, and a cone share larger than the cylinder would mean a height had been swapped somewhere.
• Whole-number finish: the final fraction divides exactly to a clean whole number, so a leftover decimal would flag a slip, most likely in squaring the radius or in handling the denominator.

66 cm³, from 3967 (cylinder) plus 667 (two cones).

Find one piece, scale up, then take the percentage. Work in that order so the numbers stay manageable.

• One real sum: the volume of a single gulab jamun, about 25.05 cm³, is the only calculation. The rest is multiplying by 45 and then by 0.30.
• Caps make a sphere: treat the two ends as one sphere of radius 1.4 cm, giving the 43π r³ term in one shot instead of 2×23π r³.
• Trim the cylinder: the cylinder length is 5-2.8=2.2 cm after removing the two rounded ends, and missing that subtraction is the usual error.
• Round late: ``about'' allows 338 cm³, but keep three decimals in the working and round only at the last line so the figure stays reliable.

≈ 338 cm³ of syrup in 45 pieces (about 25.05 cm³ each).

Compute one cone, scale to four, subtract once. The cuboid is easy, so the care goes into the four small cones.

• One cone: with r=0.5 the square r²=0.25 is small, and 227× 0.25× 1.4=1.1, so one cone is about 0.3667 cm³ and four remove only about 1.47 cm³.
• Near-equality check: the depressions are tiny next to the block, so 523.53 cm³ sits just below 525. A number far below 525 means a cone was overstated, perhaps by using the radius where the depth belongs.
• Round late: carry four decimals in the cone volume and round only the final answer, since trimming 0.3667 to 0.37 early drifts the total once you multiply by four.

523.53 cm³ of wood (525 cuboid minus 1.47 in four cones).

Cancel π before you reach for a calculator. Keeping π in symbol form to the end is faster and avoids rounding error.

• Pi cancels: both the overflow and one shot carry a factor of π, so it drops out in the division and the count is a pure number.
• Clean divide: the overflow 50π3 divided by one shot π6 has its threes and sixes simplify to exactly 100.
• Whole number expected: you cannot drop part of a lead shot, so a decimal answer means a volume was set up wrongly.
• Exact cube: 0.5³ = 0.125 = 18, so 43π·18 = π6. Spotting this keeps the fractions exact instead of slipping into decimals.

100 lead shots, found by dividing the overflow 50π3 by one shot π6.

Factor π once, convert grams to kilograms at the end. A single multiply and a clean unit are all this question rewards.

• One multiply: keep π outside the bracket while you add 31680 and 3840 to get 35520, then multiply by 3.14 just once rather than scaling each cylinder.
• Grams to kilograms: the density is per cm³, so the mass arrives in grams. Dividing 892262.4 g by 1000 gives the natural 892.26 kg.
• Always label: a bare ``892262.4'' with no gram or kilogram tag loses a mark, so carry the unit through.
• Dominance check: the tall lower cylinder gives 31680π of the 35520π, about nine-tenths. An upper cylinder larger than the lower one means a radius was swapped.

≈ 892.26 kg (i.e. 892262.4 g) for a volume of 35520π cm³.

Keep π in symbol form through the subtraction. Symbol-form π keeps every step clean until the very last line.

• Equal parts: the cone and the hemisphere here come to exactly the same volume, because the cone height happens to be twice the radius, which makes its formula collapse to the hemisphere formula. Spotting that coincidence lets you write both volumes down at once instead of grinding through each.
• Clean subtraction: keeping pi as a symbol, you add the cone and hemisphere together and take the sum away from the cylinder, which leaves a single tidy multiple of pi before you have done any decimal work at all.
• Convert last: only at the very end do you substitute the numerical value of pi to reach the figure in cubic centimetres. A thousand cubic centimetres make one litre and a million make one cubic metre, so the same answer reads as roughly eleven hundred litres or a little over one cubic metre.
• Proportion check: the solid fills almost half of the cylinder, so a little over half the water stays behind. A rough mental check of just under half is a quick way to catch a gross error before you reach the final figure.
• Read the wording: the phrase about touching the bottom is confirming that the solid is fully submerged, so all of its volume displaces water. A solid only partly under the surface would displace just the part below the water line, and the working would change.

360000π cm³ ≈ 1131428.57 cm³ of water remains.

Compute the true value, then judge the claim. A verification question is answered only when you find the real volume and state a verdict.

• Cube decides it: the sphere dominates, so the step (4.25)³ = 76.765625 settles the answer. Rush this cube and the comparison goes wrong.
• Add the parts: the small neck 25.12 cm³ plus the sphere 321.39 cm³ gives about 346.51 cm³.
• State the verdict: since 346.51 differs from 345, the child is wrong, though only slightly, the gap coming partly from her rounding while measuring water.
• Hold two decimals: keys quote 346.51 cm³, so rounding the sphere term early to 321 could mislead you into calling 345 correct. Precision is the whole point here.

True volume ≈ 346.51 cm³; the child's 345 cm³ is not correct.