PrepShorts · Study sheet · Class 10 Mathematics · Chapter 12, Surface Areas and Volumes
Chapter 12 · Surface Areas and Volumes
Which faces vanish at the join, and why you cannot simply add
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Paint a cone. Paint a dome. Add the two amounts up, then put the dome on the cone and paint that instead - and the number goes down. Two circles of paint, bought and never used. Working out exactly which paint gets wasted is the whole of this topic, and it is not always two circles.
The idea
Stick two solids together and their surface areas do not add, because surface area measures the boundary — the set of points you could put paint on — and the moment the two pieces touch, the region where they touch stops being on the outside of anything. That loss is not a fixed penalty of two circles. What leaves the boundary is exactly the overlap of the two flat faces, so the correction depends on how the faces meet: matched circle to equal circle, both discs go and the answer is just the curved parts; a small circle on a large flat face, only that circle goes and the large face keeps the rest; a wide circle on a narrow one, a ring survives and has to be counted. Once you stop reciting a rule and start asking what can still be painted, all three cases and the hollowed ones besides come out of the same single question — and the surprise falls out with them, that sticking a dome onto a face and scooping an identical bowl into it change the area by exactly the same amount.
What you should be able to do
- Explain why a joined solid's total surface area falls short of what the two pieces' total surface areas would give
- Identify, for a given join, exactly which parts of which faces leave the outside of the solid
- Compute the surface area of a cylinder capped by hemispheres, where every flat face disappears
- Compute the surface area of a solid where a circular face sits inside a larger flat face, and explain why the total goes up rather than down
- Compute the surface area where two circular faces meet but differ in radius, and account for the ring that survives
- Treat a hollowed-out depression by the same reasoning as an added piece, and show that a dome added and a bowl of the same radius scooped both raise the area by πr²
- Decide, in a physical question, which faces are to be counted at all — the base of a tent, the underside of a bird-bath
- Set out a surface-area calculation as an audit of faces rather than as a remembered formula
Words to know
| Term | Definition in one line | First introduced |
|---|---|---|
| total surface area | the area of everything on the outside of the finished solid, abbreviated TSA | printed in §12.2, p. 162, with the abbreviation expanded |
| curved surface area | the area of a solid's curved skin alone, its flat ends excluded, abbreviated CSA | printed in §12.2, p. 162, with the abbreviation expanded |
| base area | the area of the flat circular end of a cone or cylinder, πr² | printed in Example 2, p. 164, and again in Example 3, p. 165 |
| hemisphere | half a sphere; its curved dome measures 2πr² and its flat face πr² | printed in §12.1, p. 161 |
| depression | a hollow shaped into a solid rather than a piece added to it | printed in Example 4, p. 166, and in Exercise 12.1 q. 5, p. 166 |
| cavity | a piece hollowed out from inside a solid | printed in Exercise 12.1 q. 8, p. 167 |
| slant height | the apex-to-rim distance of a cone, the length a cone's curved area is built on | printed in Example 1's solution, p. 164 |
| surmounted | sitting on top of, the chapter's word for the upper piece of a join | printed in Example 1, p. 163 |
| hidden face | a flat face that stops being on the outside once the join is made | an added term; the chapter describes the disappearance in §12.3, p. 167, without labelling it |
| face audit | listing every candidate face and marking it visible or hidden before adding | an added shorthand for the procedure the chapter carries out example by example |
Where people slip up
- "Total surface area of a composite = TSA of one piece + TSA of the other." This is the error the chapter interrupts itself to warn against on p. 164, and the playing top shows its size: 58.85 cm² claimed against 39.6 cm² actual, an overcount of exactly twice the contact circle.
- "So the rule is: add the two, then subtract two circles." Only when the two faces are equal circles laid on one another. For the block only one small circle goes; for the rocket the overlap is the smaller circle and a ring of the larger one survives. Memorising the subtraction is how a student gets the rocket wrong.
- "Covering part of a surface must reduce the area." Putting a dome on a cube increases it, because the dome's curved skin is twice the flat circle it covers. Ask the class to vote before revealing Example 2's 163.86 cm².
- "Hollowing something out must reduce the surface area." It increases it, and by the same πr² as adding a dome would. Solid material was removed but boundary was created.
- "The cone sits on the cylinder, so the cone's base is hidden." Only the part of it that is actually in contact. Fig. 12.8's plan view shows the ring left over and the ring has to be painted.
- "CSA and TSA are interchangeable if you are careful." They differ by exactly the flat faces, which are the objects this whole topic is about. Write which one you mean at every line.
- "Every flat face of the finished object gets counted." The tent's floor is excluded because the question says so; the bird-bath's underside is excluded because it stands on legs. Deciding what counts is part of the problem, not a detail before it.
- "The hemisphere in the bird-bath adds a flat ring at the rim." It does not — the hollow and the cylinder share a radius, so the rim has no width. If the radii differed, that ring would be real and would have to be added.
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Worked answers: Exercise 12.1 · Exercise 12.2 · this video explains Exercise 12.1 Q2, Exercise 12.1 Q4, Exercise 12.1 Q5, Exercise 12.1 Q6, Exercise 12.1 Q7, Exercise 12.1 Q8, Exercise 12.1 Q9
Transcript2,068 words
Here are two solids, and here is a tin of paint. Paint the cone. Every part of it: the sloping side, and the flat circle it stands on. Now paint the dome. The curved skin, and the flat circle underneath it. Add the two amounts up, and you have a number. Now put the dome on the cone, and paint that. The number is smaller. It has to be. Because the circle on the bottom of the dome and the circle on the top of the cone are pressed together, and you cannot get a brush in there.
Two circles of paint, bought and never used. That is the whole of this lesson, and everything else is working out exactly which paint gets wasted, because it is not always two circles. So it is worth being precise about what surface area is. It is not a property of the material. It is a property of the boundary. Surface area measures the set of points you could put paint on: the points that are on the solid and have nothing but air just beyond them.
Everything inside is not surface. It never was. And that gives you a test you can apply to any point of any object, one point at a time. Stand at a point. Is it on the solid? Step a hair outwards. Are you still on the solid? If you are on and then off, that point is outside, and it counts. If you are on and then still on, you are in the middle of the material, and it does not.
Every question in this topic is that one test, asked at the right places. Watch what the test says as two pieces come together. Here is a cylinder with a flat circular end, and a dome with a flat circular face. Apart, both of those flat faces are outside. Stand on either one, step outwards, and you are in the air. Now slide them together. The instant they touch, every point of those two circles fails the test. Step outwards from the cylinder's end and you are inside the dome. Step outwards from the dome's face and you are inside the cylinder.
The two faces have not gone anywhere. They are still there, made of the same material. But they have stopped being boundary, and boundary is the only thing surface area was ever measuring. On the finished object there is no seam to see at all, and that is the difficulty: the picture hides exactly the information you need. Start with the easiest case, where the two faces are the same circle laid exactly on top of each other.
A capsule: a straight tube with a rounded end at each end. Fourteen millimetres long, five across. Five across means a radius of two and a half. The two rounded ends take up two and a half each, so the straight part is nine. There are four flat circles inside that object, and every one of them is pressed against another one, so not one survives. What is left is the curved wall and the two curved caps.
That comes to two hundred and twenty square millimetres. And here is something worth noticing. Two hundred and twenty is exactly what you would get from the curved wall of a plain tube fourteen millimetres long and five across. The two domes together give back precisely what rounding the ends took away. Now the object that shows you the cost of getting this wrong. A spinning top: a cone standing on its point, with a dome on top of it. One point seven five from the axis, three and a quarter of cone, and a slant of three point seven.
Its surface is thirty nine point six square centimetres. Add the two pieces' surfaces instead, the way the first instinct says, and you get fifty eight point eight five. The gap is nineteen point two five, and that number is not mysterious. The circle where they touch measures nine point six two five. You counted it once as the cone's base and once as the dome's flat face, and neither one is on the outside of anything.
Nineteen point two five is exactly twice nine point six two five. Not roughly. Exactly, and that is the shape of the correction. So write the rule down properly. The outside of a joined solid is the two outsides added together, less twice the region where the pieces are in contact. Twice, because each piece was carrying its own copy of that region and both copies have to go. And read that again, because there is a trap in it.
It says twice the CONTACT. It does not say two circles. The contact is whatever the two faces actually have in common, and that depends entirely on how they meet. If they match, the contact is the whole circle and the whole of both faces goes. If a small circle sits on a big flat face, only that small circle goes and the rest of the big face is still out in the air.
And if two circles of different size are pressed together, the contact is the smaller one, and a ring of the bigger one survives. Three cases. One rule. Take the second case, and predict the answer before you compute it. A cube five centimetres on a side, with a dome sitting on its top face. The dome is four point two across, so its radius is two point one. The bare cube has six faces of twenty five, which is a hundred and fifty.
Now put the dome on. Most people expect the total to go down: you have covered something up. It goes up. The circle you covered measures thirteen point eight six, so the top face keeps twenty five less thirteen point eight six. But the dome's curved skin measures twenty seven point seven two, which is exactly twice the circle it stands on. A hemisphere always is. Lose one circle, gain two. The block comes to a hundred and sixty three point eight six, which is thirteen point eight six more than the cube it was made from.
Now the third case, and the one that punishes a memorised rule. A wooden rocket: a cone standing on a cylinder. The cone's base is five across; the cylinder is only three across. Look at it from directly above and you see two circles, one inside the other. The inner one, radius one and a half, is where the two pieces are actually in contact. That is the part that stops being outside.
The ring between them is not in contact with anything. It is the underside of the cone's overhang, and it is out in the open air, and it needs painting like everything else. That ring measures pi times two point five squared less one point five squared, which is four pi, twelve point five six. Subtract two circles here and you have thrown away a piece of the object that is plainly visible from underneath.
The rocket is usually asked in two colours, which makes the audit explicit. The cone gets one colour. Its curved side, using a slant of six point five, and the ring underneath it. Sixteen point two five plus four, times pi: sixty three point five eight five. The cylinder gets the other. Its curved wall, and the circle it stands on at the very bottom, which nothing covers. Its own top face gets nothing, because the cone is sitting on all of it.
Sixty plus two point two five, times pi: one hundred and ninety five point four six five. Two faces of the same solid, handled two different ways in one question. The top is hidden; the bottom is not. And the two amounts add to the whole outside of the rocket, which is how you know the audit missed nothing. Everything so far has been about sticking pieces on. Now take one away.
A cylinder two point four tall and one point four across, with a cone of exactly the same height and width drilled right out of it. Which faces are outside now? The curved wall of the cylinder, still there. The circle it stands on, still there. Its top face is gone completely, because the cone's mouth is the same width as the cylinder. And in place of it there is a new surface that did not exist before: the inside of the cone, curved, with a slant of two point five.
Seventeen point six square centimetres, which is eighteen to the nearest. Drilling a hole out of something removed material and made more surface, not less. Boundary was created. The same on a cylinder with a bowl hollowed into each end: three hundred and seventy four. Which sets up the fact that makes the whole topic click. Take a cube seven centimetres on a side, and the largest dome its face will take, radius three and a half.
Stick that dome on. You lose the circle it covers and you gain twice that circle in curved skin, so the surface goes up by one circle: two hundred and ninety four plus thirty eight point five, three hundred and thirty two point five. Now start again, and scoop the identical bowl into the face instead. You lose the same circle from the flat face, and you gain the same curved skin, now facing inwards.
Three hundred and thirty two point five. The same number. Adding a dome and hollowing an identical bowl change the surface by exactly the same amount, because the bookkeeping is identical and only the sign of the material changed. One more thing decides the answer, and it is not geometry. A bird bath: a cylinder a hundred and forty five centimetres tall, with a bowl of radius thirty hollowed into the top.
Its whole boundary includes the flat underside. The published answer does not: three point three square metres, with the underside left out, because the bath stands on legs. A tent behaves the same way. Cylindrical wall two point one metres high, four metres across, conical top of slant two point eight. Forty four square metres of canvas, and at five hundred a square metre that is twenty two thousand. The floor is not canvas, and the question says so.
And a vessel has an inside as well as an outside. Asked for the inner surface of one thirteen deep and fourteen across, you want five hundred and seventy two, leaving out the mouth. Deciding which faces the question wants is part of the problem, not a detail before it. A word about the numbers themselves, because two of them are not what they look like. The top's slant is not three point seven. It is the square root of thirteen point six two five, which is three point six nine one two and does not stop.
Three point seven squared is thirteen point six nine, not the number we started from. Everything built on it is a rounding. The value used for pi changes the last digits too. Twenty two over seven and three point one four are not the same number. So a good answer says which one it used. But notice what does not depend on any of that. The overcount on the top was exactly twice the contact circle. The block gained exactly one circle. The dome and the bowl came to exactly the same. Those are the facts of the topic, and no rounding touches them.
Which leaves a procedure that works on anything, including something you have never seen. List every face and every curved skin the pieces have. Against each one, ask the paint question: is there air just beyond it? Mark it seen or hidden. Then add up the seen ones. That is it. There is no formula for a rocket or a bird bath or a capsule, and there is never going to be one.
Two cubes make the point in miniature. Sixty four cubic centimetres each, so each edge is four, so each shows ninety six square centimetres. Apart, a hundred and ninety two. Pushed together, a hundred and sixty. Thirty two gone, which is twice the sixteen of the face they touch on. Not a formula. A count of what is still on the outside.
Where this fits
Taken from the notes each video was made from, not from the reading order — these are the ideas this one rests on and the ones that later rest on it.
Builds on
- Decomposing an everyday object into the basic solidsClass 10 · Ch 12, Surface Areas and Volumes
Comes up again in
- Why volumes do add even though surface areas do notClass 10 · Ch 12, Surface Areas and Volumes