Join Two Solids Together and Two of Their Faces Disappear Forever
Recall the surface area formulas for all six basic solids, add only the exposed curved surfaces when two are joined, handle a solid with a part scooped out of it, and cost a painting or polishing job from a rate per square metre.
Why is the surface area of two joined solids less than the sum of their surface areas?
Take a wooden cone and a wooden hemisphere of the same radius and glue the flat face of the hemisphere to the base of the cone. The finished toy has a smaller surface area than the two pieces had separately — and not by a little.
The reason is simple once you see it. The two flat circular faces that were glued together are now inside the solid. Nobody can paint them, touch them or see them. They have stopped being surface.
So the rule for every combination question in this chapter is:
Which in practice almost always means: add the curved surfaces and leave out the joined flats. For the cone-on-a-hemisphere toy you add the cone's curved surface and the hemisphere's curved surface, and that is the whole answer — the two circles of radius that used to be the cone's base and the hemisphere's flat face contribute nothing.
The opposite case is scooping. Hollow a hemisphere out of a cylinder and the flat circle you removed disappears from the surface, but a new curved surface appears inside the hollow. So you subtract one and add the other, and the total can even go up.
Getting that bookkeeping right is the whole chapter, and it is why the first step of every answer should be a sketch with the exposed surfaces marked. The formulas themselves are from earlier years and only need recalling:
- The curved surface of a cylinder, a cone and a hemisphere — these are what you add
- The slant height of a cone, which must usually be computed from the radius and the vertical height
- A rate per square metre, which turns an area into a cost in one multiplication
**Take throughout this chapter**, since the radii in these questions are chosen as multiples or halves of .
This page covers the first part of the CBSE Class 10 Maths chapter on surface areas and volumes: the basic formulas, combined solids, hollowed solids, and cost problems.
The reason is simple once you see it. The two flat circular faces that were glued together are now inside the solid. Nobody can paint them, touch them or see them. They have stopped being surface.
So the rule for every combination question in this chapter is:
Which in practice almost always means: add the curved surfaces and leave out the joined flats. For the cone-on-a-hemisphere toy you add the cone's curved surface and the hemisphere's curved surface, and that is the whole answer — the two circles of radius that used to be the cone's base and the hemisphere's flat face contribute nothing.
The opposite case is scooping. Hollow a hemisphere out of a cylinder and the flat circle you removed disappears from the surface, but a new curved surface appears inside the hollow. So you subtract one and add the other, and the total can even go up.
Getting that bookkeeping right is the whole chapter, and it is why the first step of every answer should be a sketch with the exposed surfaces marked. The formulas themselves are from earlier years and only need recalling:
- The curved surface of a cylinder, a cone and a hemisphere — these are what you add
- The slant height of a cone, which must usually be computed from the radius and the vertical height
- A rate per square metre, which turns an area into a cost in one multiplication
**Take throughout this chapter**, since the radii in these questions are chosen as multiples or halves of .
This page covers the first part of the CBSE Class 10 Maths chapter on surface areas and volumes: the basic formulas, combined solids, hollowed solids, and cost problems.
Formula
What are the surface area formulas for the six basic solids?
Curved or lateral surface area leaves out the flat ends; total surface area includes them. Questions distinguish the two carefully, so read which one is wanted.
**Cube of edge :**
**Cuboid of length , breadth and height :**
**Cylinder of radius and height :**
**Cone of radius , vertical height and slant height :**
**Sphere of radius :**
**Hemisphere of radius :**
**Why a hemisphere's total is .** Half a sphere's surface is , and the flat circular face adds . Two plus one is three — and in a combination question the flat face is usually glued to something, so it is the that you actually use.
The slant height is the step most often skipped. A cone question almost never gives directly; it gives the vertical height , and
comes from the Pythagoras theorem applied to the right triangle formed by the radius, the axis and the slant side. **Using where belongs is the single commonest error in the whole chapter**, and since always, it gives an answer that is too small.
Worked example — the slant height. A cone has radius cm and vertical height cm. Find its slant height and curved surface area. Take .
**Check that **: . Correct, and if your slant height ever comes out smaller than the vertical height, the Pythagoras step was inverted.
Notice the arithmetic shortcut. exactly, because is half of . **Every radius in this chapter is chosen so that or comes out whole**, which is a useful reassurance mid-calculation: if it does not, check the radius.
**Cube of edge :**
**Cuboid of length , breadth and height :**
**Cylinder of radius and height :**
**Cone of radius , vertical height and slant height :**
**Sphere of radius :**
**Hemisphere of radius :**
**Why a hemisphere's total is .** Half a sphere's surface is , and the flat circular face adds . Two plus one is three — and in a combination question the flat face is usually glued to something, so it is the that you actually use.
The slant height is the step most often skipped. A cone question almost never gives directly; it gives the vertical height , and
comes from the Pythagoras theorem applied to the right triangle formed by the radius, the axis and the slant side. **Using where belongs is the single commonest error in the whole chapter**, and since always, it gives an answer that is too small.
Worked example — the slant height. A cone has radius cm and vertical height cm. Find its slant height and curved surface area. Take .
**Check that **: . Correct, and if your slant height ever comes out smaller than the vertical height, the Pythagoras step was inverted.
Notice the arithmetic shortcut. exactly, because is half of . **Every radius in this chapter is chosen so that or comes out whole**, which is a useful reassurance mid-calculation: if it does not, check the radius.
How do you find the surface area of two solids joined together?
Sketch the solid, mark which surfaces a coat of paint would reach, add only those, and ignore every face that is glued.
Worked example 1 — a cone on a hemisphere. A toy is in the form of a cone of radius cm mounted on a hemisphere of the same radius. The total height of the toy is cm. Find its total surface area. Take .
Step 1 — split the height. The hemisphere's own height is its radius, cm, so the cone's vertical height is
Step 2 — the slant height, as computed above: cm.
Step 3 — the exposed surfaces. A coat of paint reaches the cone's curved surface and the hemisphere's curved surface, and nothing else:
Step 4 — add:
What was deliberately left out. The cone's base circle and the hemisphere's flat face, each of area cm². **Including them would have added cm² of surface that does not exist, and that is precisely the trap the question is set to catch.
Worked example 2 — a capsule.** A medicine capsule is a cylinder with a hemisphere stuck on each end. The whole capsule is mm long and mm in diameter. Find its surface area. Take .
Step 1 — the radius and the cylinder's length. The radius is mm, and each hemisphere adds mm to the length, so the cylindrical part is
Step 2 — the exposed surfaces. The cylinder contributes only its curved surface, since both its flat ends are covered:
Two hemispheres together make a whole sphere:
Step 3 — add:
The two hemispheres making a sphere is the shortcut worth remembering. Two curved halves are one whole, so — and it saves computing them separately.
And notice the length subtraction. The mm is the length of the whole capsule, so both hemispheres must be taken off to leave the cylinder. Subtracting only one radius, or none, is the error this question is built around.
Worked example 3 — a tent. A tent is a cylinder with a conical top. The cylindrical part has height m and diameter m, and the slant height of the cone is m. Find the area of canvas used. Take .
The radius is m. A tent has no floor of canvas and no lid between the two parts, so only two curved surfaces count:
Check the arithmetic a second way. Both terms share , so the total is m². The sevens cancel exactly, which is a strong sign the numbers were meant to work out.
Worked example 4 — a vessel open at the top. A vessel is a hollow hemisphere surmounted by a hollow cylinder. The hemisphere's diameter is cm and the vessel's total height is cm. Find its inner surface area. Take .
The radius is cm, the hemisphere's height is cm, so the cylinder's height is cm.
The inner surface is the inside of the bowl plus the inside of the tube:
**Factorising out first turned two calculations into one.** That is worth looking for whenever a cylinder and a hemisphere share a radius, and here it made the whole answer a single multiplication.
Worked example 1 — a cone on a hemisphere. A toy is in the form of a cone of radius cm mounted on a hemisphere of the same radius. The total height of the toy is cm. Find its total surface area. Take .
Step 1 — split the height. The hemisphere's own height is its radius, cm, so the cone's vertical height is
Step 2 — the slant height, as computed above: cm.
Step 3 — the exposed surfaces. A coat of paint reaches the cone's curved surface and the hemisphere's curved surface, and nothing else:
Step 4 — add:
What was deliberately left out. The cone's base circle and the hemisphere's flat face, each of area cm². **Including them would have added cm² of surface that does not exist, and that is precisely the trap the question is set to catch.
Worked example 2 — a capsule.** A medicine capsule is a cylinder with a hemisphere stuck on each end. The whole capsule is mm long and mm in diameter. Find its surface area. Take .
Step 1 — the radius and the cylinder's length. The radius is mm, and each hemisphere adds mm to the length, so the cylindrical part is
Step 2 — the exposed surfaces. The cylinder contributes only its curved surface, since both its flat ends are covered:
Two hemispheres together make a whole sphere:
Step 3 — add:
The two hemispheres making a sphere is the shortcut worth remembering. Two curved halves are one whole, so — and it saves computing them separately.
And notice the length subtraction. The mm is the length of the whole capsule, so both hemispheres must be taken off to leave the cylinder. Subtracting only one radius, or none, is the error this question is built around.
Worked example 3 — a tent. A tent is a cylinder with a conical top. The cylindrical part has height m and diameter m, and the slant height of the cone is m. Find the area of canvas used. Take .
The radius is m. A tent has no floor of canvas and no lid between the two parts, so only two curved surfaces count:
Check the arithmetic a second way. Both terms share , so the total is m². The sevens cancel exactly, which is a strong sign the numbers were meant to work out.
Worked example 4 — a vessel open at the top. A vessel is a hollow hemisphere surmounted by a hollow cylinder. The hemisphere's diameter is cm and the vessel's total height is cm. Find its inner surface area. Take .
The radius is cm, the hemisphere's height is cm, so the cylinder's height is cm.
The inner surface is the inside of the bowl plus the inside of the tube:
**Factorising out first turned two calculations into one.** That is worth looking for whenever a cylinder and a hemisphere share a radius, and here it made the whole answer a single multiplication.
What changes when a part is hollowed out instead of added on?
Remove the flat face you cut into, add the new curved surface of the hollow, and keep everything else. The total may rise or fall depending on which is bigger.
Worked example 1 — hemispheres scooped from a cylinder. A wooden article is made by scooping out a hemisphere from each end of a solid cylinder. The cylinder's height is cm and its base radius is cm. Find the total surface area of the article. Take .
What the surface consists of.
- The cylinder's curved surface, unchanged by the scooping
- The two hollow hemispherical surfaces, newly exposed inside
- Not the two flat circular ends, because each has been entirely scooped away — the hemisphere's diameter equals the cylinder's
So
Compare that with the unscooped cylinder, whose total surface would have been
Scooping increased the surface area, from cm² to cm². That surprises students, and the reason is clean: each flat circle of cm² was replaced by a curved hemispherical surface of cm², which is twice as much. Hollowing removes material but can add surface.
Worked example 2 — a hemisphere cut into a cube. A hemispherical depression is cut out of one face of a solid cube so that the hemisphere's diameter equals the cube's edge. Find the surface area of the remaining solid if the edge is cm. Take .
The radius is cm.
The surface consists of:
- The cube's six faces: cm²
- minus the flat circle removed from one face: cm²
- plus the curved surface of the depression: cm²
Check with the general expression. Since , the answer is always
and with and :
The two routes agree.
And the general form shows the pattern. Cutting a hemispherical depression always adds to the surface area, because you lose one flat circle and gain two circles' worth of curved surface. Net gain: exactly one circle.
Worked example 3 — a cone scooped from a cylinder. A solid cylinder of radius cm and height cm has a cone of the same radius and the same height scooped out of one end. Find the total surface area of the remaining solid. Take .
The cone's slant height, as before, is cm.
The surface consists of:
- The cylinder's curved surface: cm²
- The one flat end that was not touched: cm²
- The curved surface of the conical hollow: cm²
- Not the scooped end's flat circle, which has been removed entirely
Compare with the solid cylinder: cm². The scooped solid has more surface again, and by the same reasoning — the flat circle of cm² was replaced by a conical surface of cm².
The habit that makes hollowing questions reliable. Write the three lines out as a list before computing anything: what stays, what is removed, what is newly exposed. A student who writes those three headings never double-counts a flat face, and one who does not usually double-counts exactly one.
Worked example 1 — hemispheres scooped from a cylinder. A wooden article is made by scooping out a hemisphere from each end of a solid cylinder. The cylinder's height is cm and its base radius is cm. Find the total surface area of the article. Take .
What the surface consists of.
- The cylinder's curved surface, unchanged by the scooping
- The two hollow hemispherical surfaces, newly exposed inside
- Not the two flat circular ends, because each has been entirely scooped away — the hemisphere's diameter equals the cylinder's
So
Compare that with the unscooped cylinder, whose total surface would have been
Scooping increased the surface area, from cm² to cm². That surprises students, and the reason is clean: each flat circle of cm² was replaced by a curved hemispherical surface of cm², which is twice as much. Hollowing removes material but can add surface.
Worked example 2 — a hemisphere cut into a cube. A hemispherical depression is cut out of one face of a solid cube so that the hemisphere's diameter equals the cube's edge. Find the surface area of the remaining solid if the edge is cm. Take .
The radius is cm.
The surface consists of:
- The cube's six faces: cm²
- minus the flat circle removed from one face: cm²
- plus the curved surface of the depression: cm²
Check with the general expression. Since , the answer is always
and with and :
The two routes agree.
And the general form shows the pattern. Cutting a hemispherical depression always adds to the surface area, because you lose one flat circle and gain two circles' worth of curved surface. Net gain: exactly one circle.
Worked example 3 — a cone scooped from a cylinder. A solid cylinder of radius cm and height cm has a cone of the same radius and the same height scooped out of one end. Find the total surface area of the remaining solid. Take .
The cone's slant height, as before, is cm.
The surface consists of:
- The cylinder's curved surface: cm²
- The one flat end that was not touched: cm²
- The curved surface of the conical hollow: cm²
- Not the scooped end's flat circle, which has been removed entirely
Compare with the solid cylinder: cm². The scooped solid has more surface again, and by the same reasoning — the flat circle of cm² was replaced by a conical surface of cm².
The habit that makes hollowing questions reliable. Write the three lines out as a list before computing anything: what stays, what is removed, what is newly exposed. A student who writes those three headings never double-counts a flat face, and one who does not usually double-counts exactly one.
How do you turn a surface area into the cost of painting or polishing?
Find the area to be covered, convert it to the unit the rate uses, and multiply. The arithmetic is trivial; the marks are in identifying which surfaces are actually covered and in converting the units.
Worked example 1 — canvas for a tent. For the tent of the earlier section, with m² of canvas, find the cost at ₹ per square metre.
Worked example 2 — painting a pillar. A cylindrical pillar has a diameter of cm and a height of m. Find the cost of painting its curved surface at ₹ per square metre. Take .
Convert first. The diameter is cm m, so the radius is m, and the height is already in metres:
The conversion was the whole difficulty. Working in centimetres would have given cm², and multiplying that by a rate quoted per square metre gives a nonsense answer several thousand times too large. Convert lengths to the rate's unit before computing the area, not afterwards — because converting an area needs cm² per m², and that factor is easy to forget.
Worked example 3 — polishing a combined solid. A wooden toy is a cone of radius cm and slant height cm mounted on a hemisphere of the same radius. Find the cost of painting the toy at ₹ per square centimetre. Take .
Check the cone's vertical height for plausibility. From and , the height is cm — a whole number, as these questions are designed to give. That check confirms the slant height was used in the right place, since had been the vertical height the slant would have been , which is not a whole number.
Worked example 4 — a rate per unit of a different size. A hemispherical dome of a building needs to be painted on the outside. Its diameter is m. Find the cost at ₹ per cm². Take .
Now convert to square centimetres, remembering that m² cm²:
**The rate is per cm²**, so the number of units is
Two conversions in one question, and both must be done in the right order. Read the rate before you compute anything — it tells you which unit the area needs to be in, and it may not be a square metre.
The three-step layout that never loses marks. State the surfaces to be covered; compute the area with its unit; then convert and multiply by the rate. Each step is separately markable, so a conversion slip at the end still leaves the earlier marks intact — provided the working shows them.
Worked example 1 — canvas for a tent. For the tent of the earlier section, with m² of canvas, find the cost at ₹ per square metre.
Worked example 2 — painting a pillar. A cylindrical pillar has a diameter of cm and a height of m. Find the cost of painting its curved surface at ₹ per square metre. Take .
Convert first. The diameter is cm m, so the radius is m, and the height is already in metres:
The conversion was the whole difficulty. Working in centimetres would have given cm², and multiplying that by a rate quoted per square metre gives a nonsense answer several thousand times too large. Convert lengths to the rate's unit before computing the area, not afterwards — because converting an area needs cm² per m², and that factor is easy to forget.
Worked example 3 — polishing a combined solid. A wooden toy is a cone of radius cm and slant height cm mounted on a hemisphere of the same radius. Find the cost of painting the toy at ₹ per square centimetre. Take .
Check the cone's vertical height for plausibility. From and , the height is cm — a whole number, as these questions are designed to give. That check confirms the slant height was used in the right place, since had been the vertical height the slant would have been , which is not a whole number.
Worked example 4 — a rate per unit of a different size. A hemispherical dome of a building needs to be painted on the outside. Its diameter is m. Find the cost at ₹ per cm². Take .
Now convert to square centimetres, remembering that m² cm²:
**The rate is per cm²**, so the number of units is
Two conversions in one question, and both must be done in the right order. Read the rate before you compute anything — it tells you which unit the area needs to be in, and it may not be a square metre.
The three-step layout that never loses marks. State the surfaces to be covered; compute the area with its unit; then convert and multiply by the rate. Each step is separately markable, so a conversion slip at the end still leaves the earlier marks intact — provided the working shows them.
Exam tip
Which habits protect a surface-area answer?
Sketch the solid and shade the surfaces a coat of paint would reach. Everything else follows from that picture.
- List what stays, what is removed and what is newly exposed before computing anything
- Leave out every joined flat face — two glued circles contribute nothing
- Two hemispheres make a sphere, so use rather than computing them separately
- Compute the slant height from , and check that
- Split a total height correctly: a hemisphere's own height is , so subtract once for one end and for two
- **Factorise a shared when a cylinder and a hemisphere have the same radius
- Convert lengths to the rate's unit before computing the area, not afterwards
- Remember m² cm² if you must convert an area
- Read the rate first**, since it may be per cm² rather than per unit
- Give areas in square units and costs in rupees, and answer in a sentence
The misconception to name. Scooping material out of a solid does not always reduce its surface area. It usually increases it, because a flat circle of is replaced by a hemispherical surface of or a conical surface of , both larger. The scooped cylinder went from cm² to cm², and the scooped cube gained exactly . Less material, more surface — and the intuition that removal must reduce everything is simply wrong here.
A second trap. Using the vertical height in place of the slant height in . The slant height is always the longer of the two, so the error produces an answer that is too small — and in the ₹ example the plausibility check on the vertical height ( cm, a whole number) is what confirms the two were not swapped.
- List what stays, what is removed and what is newly exposed before computing anything
- Leave out every joined flat face — two glued circles contribute nothing
- Two hemispheres make a sphere, so use rather than computing them separately
- Compute the slant height from , and check that
- Split a total height correctly: a hemisphere's own height is , so subtract once for one end and for two
- **Factorise a shared when a cylinder and a hemisphere have the same radius
- Convert lengths to the rate's unit before computing the area, not afterwards
- Remember m² cm² if you must convert an area
- Read the rate first**, since it may be per cm² rather than per unit
- Give areas in square units and costs in rupees, and answer in a sentence
The misconception to name. Scooping material out of a solid does not always reduce its surface area. It usually increases it, because a flat circle of is replaced by a hemispherical surface of or a conical surface of , both larger. The scooped cylinder went from cm² to cm², and the scooped cube gained exactly . Less material, more surface — and the intuition that removal must reduce everything is simply wrong here.
A second trap. Using the vertical height in place of the slant height in . The slant height is always the longer of the two, so the error produces an answer that is too small — and in the ₹ example the plausibility check on the vertical height ( cm, a whole number) is what confirms the two were not swapped.
Did you know
Why does a cone's curved surface unroll into a slice of a circle?
Cut a paper cone up one slant side and flatten it. What you get is a sector of a circle — and the radius of that sector is the cone's slant height , not its base radius.
That single observation explains the formula , and it connects this chapter directly to the previous one.
Follow the lengths. The arc of the flattened sector was the rim of the cone's base, so its length is the base circumference . And a sector's area, from the previous chapter, is
Which is exactly the cone's curved surface area. The formula is not something to memorise separately — it is the sector area formula with the arc and the radius renamed.
It also tells you the angle of the flattened sector. The full circle of radius would have circumference , and our arc is only of it, so the fraction is and the angle is
For the toy with and that is — a thin slice, which is why a tall narrow cone unrolls into a narrow sector and a wide flat one unrolls into something close to a full circle.
A cylinder unrolls even more simply. Cut it down one side and it flattens into a rectangle whose width is the base circumference and whose height is . **So its curved surface is — the formula is just the area of that rectangle. You can check it with a paper label peeled off a tin.
And a sphere unrolls into nothing at all.** No amount of cutting will flatten a sphere's surface without stretching or tearing it, which is why cannot be derived by unrolling and why every flat map of a globe distorts something. That is the real difference between a sphere and the other solids in this chapter, and it is worth knowing that the difficulty is genuine rather than a gap in your own knowledge.
One consequence for the combination questions. Because a cone's curved surface is a sector of radius , the material needed to make a conical tent is a single flat piece of cloth — which is why tent and funnel problems ask for rather than a total surface. The curved surface is the thing you can actually cut out of a flat sheet, and that is why the syllabus separates curved from total so insistently.
That single observation explains the formula , and it connects this chapter directly to the previous one.
Follow the lengths. The arc of the flattened sector was the rim of the cone's base, so its length is the base circumference . And a sector's area, from the previous chapter, is
Which is exactly the cone's curved surface area. The formula is not something to memorise separately — it is the sector area formula with the arc and the radius renamed.
It also tells you the angle of the flattened sector. The full circle of radius would have circumference , and our arc is only of it, so the fraction is and the angle is
For the toy with and that is — a thin slice, which is why a tall narrow cone unrolls into a narrow sector and a wide flat one unrolls into something close to a full circle.
A cylinder unrolls even more simply. Cut it down one side and it flattens into a rectangle whose width is the base circumference and whose height is . **So its curved surface is — the formula is just the area of that rectangle. You can check it with a paper label peeled off a tin.
And a sphere unrolls into nothing at all.** No amount of cutting will flatten a sphere's surface without stretching or tearing it, which is why cannot be derived by unrolling and why every flat map of a globe distorts something. That is the real difference between a sphere and the other solids in this chapter, and it is worth knowing that the difficulty is genuine rather than a gap in your own knowledge.
One consequence for the combination questions. Because a cone's curved surface is a sector of radius , the material needed to make a conical tent is a single flat piece of cloth — which is why tent and funnel problems ask for rather than a total surface. The curved surface is the thing you can actually cut out of a flat sheet, and that is why the syllabus separates curved from total so insistently.
Exam relevance
How do surface areas of combined solids matter for JEE?
This is foundation work for Class 11 Conic Sections and Class 12 Application of Integrals and Application of Derivatives, and the scaling ideas here are reused in both JEE and NEET.
Where the formulas lead. Class 12 derives surface areas by integration for solids of revolution, and the cone and sphere results you use here are the standard worked cases. **The relation reappears as the slant of a cone in three-dimensional geometry, where the cone becomes a surface defined by an equation.
Where the combination bookkeeping leads. Class 12 Application of Derivatives sets optimisation problems of exactly this shape: minimise the surface area of a closed cylinder of given volume, or of a cylinder with hemispherical ends. Setting up the total area as a sum of the correct exposed pieces is the whole first half of such a problem, and a candidate who includes a face that is not there minimises the wrong function. JEE Main sets these regularly.
Where the unrolling argument leads.** The derivation of from the sector area is the first example of developing a curved surface onto a plane, and the same reasoning is used for the lateral surface of a frustum. It also explains why a sphere has no such development, which is the starting point of a genuine result in higher geometry.
Where the scaling observation leads. Doubling a length multiplies surface area by four and volume by eight, so the surface-area-to-volume ratio falls as a solid gets bigger. That single fact is used in JEE Chemistry when discussing the reactivity of powdered versus lumpy solids, and in NEET Biology to explain why cells stay small and why alveoli are folded. It is one of the few Class 10 Maths results that is quoted directly in two other subjects.
Where the cost problems lead. They train unit conversion, and unit errors are among the most expensive mistakes in JEE Physics. The discipline of converting to the rate's unit before computing is the same discipline as converting to SI before substituting into a formula.
Question types to expect. At this level: formula recall, combined-solid surface areas, hollowed solids, and cost calculations. In competitive papers: optimisation of surface area under a volume constraint, surface areas by integration, and qualitative surface-area-to-volume reasoning in Chemistry and Biology.
The single trap that costs marks. Including a joined flat face. The two circles glued between a cone and a hemisphere are not surface, and adding them inflates the toy's answer from cm² to cm². In a Class 12 optimisation the same error changes which dimensions are optimal, so the entire answer is lost rather than just a mark.
A second trap. Substituting the vertical height into . The slant height is longer, and the quickest guard is to confirm that , and form a Pythagorean set — the questions are built so that they do.
Board versus competitive emphasis. The CBSE paper marks the listed surfaces, each formula, the substitution and the unit; a competitive paper marks the optimal dimension or a ratio. The transferable habit is writing down the exposed surfaces as a list before any formula — because in an optimisation problem that list is the function you are about to differentiate.
Where the formulas lead. Class 12 derives surface areas by integration for solids of revolution, and the cone and sphere results you use here are the standard worked cases. **The relation reappears as the slant of a cone in three-dimensional geometry, where the cone becomes a surface defined by an equation.
Where the combination bookkeeping leads. Class 12 Application of Derivatives sets optimisation problems of exactly this shape: minimise the surface area of a closed cylinder of given volume, or of a cylinder with hemispherical ends. Setting up the total area as a sum of the correct exposed pieces is the whole first half of such a problem, and a candidate who includes a face that is not there minimises the wrong function. JEE Main sets these regularly.
Where the unrolling argument leads.** The derivation of from the sector area is the first example of developing a curved surface onto a plane, and the same reasoning is used for the lateral surface of a frustum. It also explains why a sphere has no such development, which is the starting point of a genuine result in higher geometry.
Where the scaling observation leads. Doubling a length multiplies surface area by four and volume by eight, so the surface-area-to-volume ratio falls as a solid gets bigger. That single fact is used in JEE Chemistry when discussing the reactivity of powdered versus lumpy solids, and in NEET Biology to explain why cells stay small and why alveoli are folded. It is one of the few Class 10 Maths results that is quoted directly in two other subjects.
Where the cost problems lead. They train unit conversion, and unit errors are among the most expensive mistakes in JEE Physics. The discipline of converting to the rate's unit before computing is the same discipline as converting to SI before substituting into a formula.
Question types to expect. At this level: formula recall, combined-solid surface areas, hollowed solids, and cost calculations. In competitive papers: optimisation of surface area under a volume constraint, surface areas by integration, and qualitative surface-area-to-volume reasoning in Chemistry and Biology.
The single trap that costs marks. Including a joined flat face. The two circles glued between a cone and a hemisphere are not surface, and adding them inflates the toy's answer from cm² to cm². In a Class 12 optimisation the same error changes which dimensions are optimal, so the entire answer is lost rather than just a mark.
A second trap. Substituting the vertical height into . The slant height is longer, and the quickest guard is to confirm that , and form a Pythagorean set — the questions are built so that they do.
Board versus competitive emphasis. The CBSE paper marks the listed surfaces, each formula, the substitution and the unit; a competitive paper marks the optimal dimension or a ratio. The transferable habit is writing down the exposed surfaces as a list before any formula — because in an optimisation problem that list is the function you are about to differentiate.
Key takeaways
What must you be able to do from this part?
Six formulas, one rule about joined faces and one conversion habit.
- Add only the exposed surfaces. Faces glued together are no longer surface
- Cube: lateral, total. Cuboid: lateral, total
- Cylinder: curved, total
- Cone: curved, total, with and always
- Sphere: . Hemisphere: curved, total
- Two hemispheres make a sphere, so use for a capsule's ends
- **A hemisphere's own height is **, so subtract once or twice when splitting a total height
- **A cone of and has ** and a curved surface of cm²
- **A toy of that cone on a hemisphere, total height cm**, has surface area cm²
- **A mm capsule of diameter mm** has a mm cylinder and area mm²
- **A tent with a m cylinder of diameter m and cone slant m** needs m², costing ₹ at ₹ per m²
- **A vessel of radius cm and total height cm** has inner surface cm²
- Scooping usually increases surface area. A cm cylinder of radius cm with both ends scooped has cm², against cm² unscooped
- **A hemispherical depression in a cube adds exactly **, so a cm cube gives cm²
- **A cone scooped from a cm cylinder of radius cm** gives cm²
- Convert lengths to the rate's unit first: a pillar of radius m and height m has m² of curved surface, costing ₹ at ₹ per m²
- ** m² cm²**, so a m dome's m² is cm², costing ₹ at ₹ per cm²
- **A cone's curved surface unrolls into a sector of radius **, which is where comes from
The most convincing self-test is a paper cone. Cut a sector out of a sheet, roll it into a cone, measure its base radius and its slant height, and check that the sector's area really equals — then ask yourself which of your two measurements the formula would have been wrong to use.
- Add only the exposed surfaces. Faces glued together are no longer surface
- Cube: lateral, total. Cuboid: lateral, total
- Cylinder: curved, total
- Cone: curved, total, with and always
- Sphere: . Hemisphere: curved, total
- Two hemispheres make a sphere, so use for a capsule's ends
- **A hemisphere's own height is **, so subtract once or twice when splitting a total height
- **A cone of and has ** and a curved surface of cm²
- **A toy of that cone on a hemisphere, total height cm**, has surface area cm²
- **A mm capsule of diameter mm** has a mm cylinder and area mm²
- **A tent with a m cylinder of diameter m and cone slant m** needs m², costing ₹ at ₹ per m²
- **A vessel of radius cm and total height cm** has inner surface cm²
- Scooping usually increases surface area. A cm cylinder of radius cm with both ends scooped has cm², against cm² unscooped
- **A hemispherical depression in a cube adds exactly **, so a cm cube gives cm²
- **A cone scooped from a cm cylinder of radius cm** gives cm²
- Convert lengths to the rate's unit first: a pillar of radius m and height m has m² of curved surface, costing ₹ at ₹ per m²
- ** m² cm²**, so a m dome's m² is cm², costing ₹ at ₹ per cm²
- **A cone's curved surface unrolls into a sector of radius **, which is where comes from
The most convincing self-test is a paper cone. Cut a sector out of a sheet, roll it into a cone, measure its base radius and its slant height, and check that the sector's area really equals — then ask yourself which of your two measurements the formula would have been wrong to use.