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Solids built from solids

A STORY

One scoop on a cone

Nila and Aarav stand by the canteen window, each holding an ice cream cone with one round scoop on top.

“Suppose the ice cream fills the cone right to the rim,” Aarav says, “and the part above the rim is half a ball.”

“Then the ice cream is two solids,” Nila says. “A cone and a hemisphere. Add their volumes and you have all of it.”

“So adding works,” Aarav says. “What if I wanted to wrap the whole thing in foil?”

“Then you cannot just add,” Nila says. “The flat circle where the scoop sits on the cone is hidden inside. No foil goes there.”

“So the foil covers the side of the cone and the round top,” Aarav says. “Nothing else.”

“Volumes add outright,” Nila says. “Surfaces count only what stays outside.”

Aarav looks at his scoop. “Then I need the volume and the surface of each basic solid first.”

You will learn them for the cube, cuboid, cylinder, cone, sphere and hemisphere, and then put the solids together.

Many everyday objects are made of two basic solids joined at one face. A test tube is a cylinder closed by a hemisphere. A tent is a cone sitting on a cylinder. A pen stand is a cylinder with a conical hole bored into it. Look around your own desk, and you can probably spot a fourth.

Six basic solids build every combination here: cube and cuboid, sphere and hemisphere, cylinder and cone. Each one’s own surface area and volume comes first, then two of them join together.

A combination’s volume always adds or subtracts outright, but its surface area counts only the faces still exposed after the join. We never count a hidden face again once a join covers it, on either side.

Two solids drawn apart still show both circles where they will meet, but drawn joined as one solid, neither circle is drawn or counted.
One join counted twice: both solids keep their whole volume, but the outside gives up two circles, one from each solid.
Kanada
A SCHOLAR INDIA REMEMBERS

Kanada is named as the founder of the Vaisheshika school. The name means the school of distinction. It sorted all the things that exist into six kinds. We need that care with solids. A toy top is a cone joined to a hemisphere. Some of its surface shows, and some is hidden at the join. Those are two different kinds of surface.

Kanada · the sparrow
Check yourself
  1. A toy is made by fixing a cone on top of a hemisphere of the same radius. To find its volume and total surface area, this chapter’s rule is to
    1. add the two volumes outright, but count only the faces still exposed for the surface area
    2. add both the volumes and the total surface areas of the two solids outright
    3. average the two volumes and the two surface areas
    4. measure the whole combination as one new solid from scratch, ignoring the two original solids
    Check your answer
    1. ✓ add the two volumes outright, but count only the faces still exposed for the surface area — (A) Volume is added outright for any combination, while surface area counts only the faces still exposed once the two solids are joined.
    2. add both the volumes and the total surface areas of the two solids outright — Volume adds outright because nothing is hidden inside a solid; surface area does not, because the join hides a face on each side.
    3. average the two volumes and the two surface areas — Neither quantity is ever averaged; volume is summed, and surface area is summed after removing the faces hidden at the join.
    4. measure the whole combination as one new solid from scratch, ignoring the two original solids — The chapter’s whole method is built on the two known solids’ own formulas, not on measuring the new shape directly.
  2. When two solids are joined, why does the combination’s volume simply add the two volumes, while its surface area is not simply the sum of the two surface areas?
    1. volume is always a larger number than surface area, so only volume survives a join
    2. volume counts space enclosed; surface area counts faces, and the join hides one
    3. surface area only applies to solids with a curved face, and volume applies to every solid
    4. surface area depends on which solid is named first in the problem, so order matters
    Check your answer
    1. volume is always a larger number than surface area, so only volume survives a join — Volume and surface area measure different things and are never compared by size; the real reason is what a join hides, not which number is bigger.
    2. ✓ volume counts space enclosed; surface area counts faces, and the join hides one — (B) Volume is space enclosed, none of which the join wastes; surface area is a count of faces, and the join hides one face on each solid.
    3. surface area only applies to solids with a curved face, and volume applies to every solid — Every solid, flat-faced or curved, has a surface area; the join hides a face regardless of the solid’s shape.
    4. surface area depends on which solid is named first in the problem, so order matters — Neither total volume nor total surface area depends on which solid is described first; both follow from the geometry alone.
  3. A student lists this chapter’s basic solids as: cube, cuboid, sphere, hemisphere, cylinder, cone, and pyramid. What is wrong with this list?
    1. cube and cuboid cannot both be basic solids, since a cube is a special cuboid
    2. hemisphere is not a basic solid; only the sphere is
    3. pyramid is not one of the six basic solids this chapter combines
    4. cylinder and cone cannot both appear in the same combination
    Check your answer
    1. cube and cuboid cannot both be basic solids, since a cube is a special cuboid — The chapter treats the cube and the cuboid as two separate basic solids, even though a cube is a special case of a cuboid.
    2. hemisphere is not a basic solid; only the sphere is — The hemisphere is named as one of the six basic solids in its own right, not only as half a sphere.
    3. ✓ pyramid is not one of the six basic solids this chapter combines — (C) The six basic solids are the cube, cuboid, sphere, hemisphere, cylinder and cone; a pyramid is not among them.
    4. cylinder and cone cannot both appear in the same combination — Any two of the six basic solids may be combined; there is no restriction against a cylinder and a cone appearing together.

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Before you start

You already know how to measure flat shapes. Now measure solids built from them. Try each check below.

If any of these felt new, read the page named before going on.

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Two ways to measure a surface

●KEY-TERM

A solid’s curved surface area counts only its curved part. It leaves out every flat face, if the solid has one.

The total surface area counts every face instead, flat and curved together. A solid with at least one flat face always has a total surface area bigger than its curved surface area.

A sphere has no flat face, so its curved and total surface areas are the same number. We check whether a solid has a flat face before deciding which surface area a question wants.

Your turn: which is bigger for a cylinder, its curved surface area or its total surface area? (Answer: the total surface area, since it also counts the two flat circular ends the curved surface area leaves out.)

A cylinder’s flat ends are its top and bottom circles, and the curved band between them is its curved surface.
Check yourself
  1. For a solid cylinder, what is the difference between its curved surface area and its total surface area?
    1. the curved surface area adds the two flat circular ends to the total surface area
    2. the total surface area adds the two flat circular ends to the curved surface area
    3. they are two different names for exactly the same measurement
    4. the curved surface area includes the volume enclosed, and the total does not
    Check your answer
    1. the curved surface area adds the two flat circular ends to the total surface area — It is the total surface area that includes the flat ends in addition to the curve, not the other way round.
    2. ✓ the total surface area adds the two flat circular ends to the curved surface area — (B) The total surface area is the curved surface area plus the two flat circular ends; the curved measure leaves those ends out.
    3. they are two different names for exactly the same measurement — A cylinder’s flat ends are real area that only the total surface area counts; the two measures are not the same number.
    4. the curved surface area includes the volume enclosed, and the total does not — Neither surface-area measure includes volume at all; both measure the surface alone, never the space enclosed.
  2. A sphere has no flat faces anywhere on it. What does this mean for its curved surface area and its total surface area?
    1. the total surface area is still larger, because every solid’s total exceeds its curved measure
    2. a sphere has no total surface area, only a curved one
    3. they are the same number, since every part of a sphere’s surface is curved
    4. the curved surface area is larger, since it is measured along the curve
    Check your answer
    1. the total surface area is still larger, because every solid’s total exceeds its curved measure — Total surface area exceeds curved surface area only when a solid has flat faces to add; a sphere has none, so the two measures coincide.
    2. a sphere has no total surface area, only a curved one — Every solid has a total surface area; for a sphere it simply equals the curved surface area, since there are no flat faces to add.
    3. ✓ they are the same number, since every part of a sphere’s surface is curved — (C) With no flat face to add, a sphere’s curved and total surface areas coincide exactly.
    4. the curved surface area is larger, since it is measured along the curve — “Measuring along the curve” changes nothing here; a sphere’s curved and total surface areas are numerically equal, neither exceeding the other.
  3. A hemisphere’s curved surface area is $2 \pi r^2$, but its total surface area is $3 \pi r^2$. Why is there an extra $\pi r^2$ in the total?
    1. the extra term corrects an error in the curved surface area formula
    2. a hemisphere has one flat base the curved measure omits
    3. a hemisphere has two flat faces, each contributing half of $\pi r^2$
    4. the extra term accounts for the volume enclosed inside the hemisphere
    Check your answer
    1. the extra term corrects an error in the curved surface area formula — The curved surface area formula is correct as it stands; it deliberately counts only the curved part, leaving the flat base for the total.
    2. ✓ a hemisphere has one flat base the curved measure omits — (B) The extra $\pi r^2$ is the area of the hemisphere’s one flat circular base, which the curved measure never counted.
    3. a hemisphere has two flat faces, each contributing half of $\pi r^2$ — A hemisphere has exactly one flat circular base, not two; that single face accounts for the whole extra $\pi r^2$.
    4. the extra term accounts for the volume enclosed inside the hemisphere — The extra term is the area of a flat circle, a surface measurement; it has nothing to do with the volume enclosed.

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A combination of two solids

●KEY-TERM

A combination of solids is a single object made from two basic solids joined at one matching face. The join can work two ways.

One solid can sit on top of another, the way a hemisphere sits on a cone. Or one solid’s shape can be hollowed out from inside another, the way a conical hole is bored into a cylinder.

A test tube is a cylinder closed by a hemisphere at one end. We name the two solids and their one shared face first for every new combination, and the working gets easier after that.

Your turn: is a cube alone a combination of solids? (Answer: no, since a combination needs two basic solids joined at one shared face, and a cube is only one solid.)

Kanada
A SCHOLAR INDIA REMEMBERS

Name the parts before you measure anything. A capsule is one cylinder and two hemispheres. Write that as a list: cylinder, hemisphere, hemisphere. Then every length you need belongs to one part on the list.

Kanada · the sparrow
Check yourself
  1. What makes an object a ‘combination of solids’ in this chapter’s sense?
    1. it is any object built from more than one material
    2. it is any solid with both a curved and a flat face
    3. joining two basic solids at one matching face
    4. it is any solid with more than one axis of symmetry
    Check your answer
    1. it is any object built from more than one material — A combination is about geometric shape, joining two solids at a face; what material the object is made of plays no part in this.
    2. it is any solid with both a curved and a flat face — Having a curved and a flat face describes solids like a cone or cylinder on their own; a combination specifically joins two such solids together.
    3. ✓ joining two basic solids at one matching face — (C) A combination of solids is a single object formed by joining two basic solids at one matching face.
    4. it is any solid with more than one axis of symmetry — The chapter’s definition is about two solids meeting at a face, not about how many axes of symmetry the result has.
  2. A conical hole is bored into a solid cylinder. Is the result a ‘combination of solids’ in the chapter’s sense?
    1. no, because nothing was joined onto the cylinder from outside
    2. no, because a hole is empty space, not a solid
    3. yes — a solid’s shape is removed from inside another
    4. yes, but only if the hole passes all the way through the cylinder
    Check your answer
    1. no, because nothing was joined onto the cylinder from outside — The definition covers both joining a solid on and hollowing one out; the second counts as a combination just as much as the first.
    2. no, because a hole is empty space, not a solid — The bored region has the exact shape of a cone, even though it is now empty space; that shape is what makes this a combination.
    3. ✓ yes — a solid’s shape is removed from inside another — (C) A combination includes hollowing one solid’s shape out of another, not only fixing one on top of the other.
    4. yes, but only if the hole passes all the way through the cylinder — A cavity that stops partway through the solid is still a combination; there is no requirement that it pass all the way through.
  3. A wooden article is a cuboid block with a hemispherical piece scooped out of its top face. What two basic solids does this chapter say the article combines?
    1. a cuboid and a hemisphere, hollowed out
    2. a cube and a sphere, joined by fixing the sphere onto the cube
    3. a cuboid and a cylinder, joined by hollowing the cylinder’s shape out
    4. a cuboid and a cone, joined by fixing the cone on top
    Check your answer
    1. ✓ a cuboid and a hemisphere, hollowed out — (A) The article is a cuboid with a hemisphere’s shape hollowed out of it.
    2. a cube and a sphere, joined by fixing the sphere onto the cube — A cuboid need not have equal edges to be a cube, and the scoop is a hemisphere removed, not a sphere added.
    3. a cuboid and a cylinder, joined by hollowing the cylinder’s shape out — A scoop of this shape is a hemisphere, curved on one side and flat where it meets the block, not a cylinder.
    4. a cuboid and a cone, joined by fixing the cone on top — The scoop is a hemisphere hollowed out, not a cone added on top of the block.

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The slant height of a cone

●KEY-TERM

A cone has two different lengths that are easy to confuse. Its height runs straight up the axis, from the base to the apex.

Its slant height runs along the curved surface instead, from a point on the rim of the base up to the apex. For a cone of base radius $r$ and height $h$, Pythagoras’ theorem gives the slant height as $l = \sqrt{r^2 + h^2}$.

*Because $l = \sqrt{r^2 + h^2}$, the slant height $l$ is always longer than both $r$ and $h$.* We check which length a question wants before picking a value off the diagram. Reaching for the height when a question needs the slant height is a common slip.

Your turn: a cone has base radius $6$ cm and height $8$ cm. What is its slant height? (Answer: $10$ cm, since $l = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$.)

A right triangle shows a cone’s height, radius and slant height as its three sides, and the slant height is the longest of the three.
Check yourself
  1. For a cone, what is the slant height?
    1. the straight-line distance from the apex straight down to the centre of the base
    2. the distance around the circular rim of the base
    3. the straight-line distance from the apex to a point on the rim of the base
    4. the radius of the circular base
    Check your answer
    1. the straight-line distance from the apex straight down to the centre of the base — The distance straight down to the centre is the cone’s height $h$; the slant height runs to the rim, not the centre.
    2. the distance around the circular rim of the base — The distance around the rim is the base’s circumference; the slant height is a straight line from the apex, not a distance along the rim.
    3. ✓ the straight-line distance from the apex to a point on the rim of the base — (C) The slant height runs in a straight line from the apex down to the rim of the base.
    4. the radius of the circular base — The radius measures across the base; the slant height is measured from the apex down to the rim.
  2. Why is a cone’s slant height $l$ never equal to its height $h$, for any cone with a real base radius?
    1. $l$ and $h$ measure the same line, just named differently by different textbooks
    2. $l$ equals $h$ whenever the cone’s radius is a whole number
    3. $l$ is smaller than $h$ whenever the base radius is small
    4. $l=\sqrt{r^2+h^2}$ always exceeds $h$, since $r^2>0$
    Check your answer
    1. $l$ and $h$ measure the same line, just named differently by different textbooks — The height runs straight down to the centre and the slant height runs to the rim; they are two different lines, not two names for one line.
    2. $l$ equals $h$ whenever the cone’s radius is a whole number — The radius being a whole number changes nothing about the relation; $l$ exceeds $h$ for every cone with $r>0$.
    3. $l$ is smaller than $h$ whenever the base radius is small — A smaller radius makes $l$ closer to $h$, but $l=\sqrt{r^2+h^2}$ is always at least as large as $h$, never smaller.
    4. ✓ $l=\sqrt{r^2+h^2}$ always exceeds $h$, since $r^2>0$ — (D) Since $l=\sqrt{r^2+h^2}$ and $r^2>0$, $l$ is always strictly larger than $h$.
  3. A cone’s height and slant height are related by $l = \sqrt{r^2+h^2}$. What geometric fact makes this the correct relation, rather than some other formula?
    1. they form a right triangle, with $l$ as the hypotenuse
    2. the relation comes from the cone’s volume formula, rearranged for $l$
    3. the relation holds only for cones whose height equals their radius
    4. the relation is an approximation, accurate only for tall, narrow cones
    Check your answer
    1. ✓ they form a right triangle, with $l$ as the hypotenuse — (A) The height, radius and slant height form a right triangle inside the cone, with the slant height as the hypotenuse, so Pythagoras’ relation applies.
    2. the relation comes from the cone’s volume formula, rearranged for $l$ — The relation between $l$, $r$ and $h$ is a right-triangle fact about lengths; it does not come from the volume formula at all.
    3. the relation holds only for cones whose height equals their radius — $l=\sqrt{r^2+h^2}$ holds for every right circular cone, whatever the relation between its own $r$ and $h$.
    4. the relation is an approximation, accurate only for tall, narrow cones — The Pythagorean relation between $l$, $r$ and $h$ is exact for every right circular cone, not an approximation for any particular shape.
  4. A cone has base radius $6$ cm and height $8$ cm. What is its slant height?
    1. $10$ cm
    2. $8$ cm, the same as the height
    3. $14$ cm, from adding the radius and height directly
    4. $100$ cm, from $6^2+8^2$ without taking the square root
    Check your answer
    1. ✓ $10$ cm — (A) $l=\sqrt{6^2+8^2}=\sqrt{100}=10$ cm.
    2. $8$ cm, the same as the height — The slant height depends on both $r$ and $h$ together; ignoring the radius of $6$ cm gives the height, not the slant height.
    3. $14$ cm, from adding the radius and height directly — Slant height comes from $\sqrt{r^2+h^2}$, not from adding $r$ and $h$ directly; $6+8=14$ is not the correct relation.
    4. $100$ cm, from $6^2+8^2$ without taking the square root — $6^2+8^2=100$ is the value under the root; the slant height itself is $\sqrt{100}=10$ cm, not $100$.

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Surface area and volume of boxes

A cuboid is a box-shaped solid with three pairs of matching rectangular faces. For a cuboid of length $l$, breadth $b$ and height $h$, each pair of opposite faces has area $l b$, $b h$ and $h l$.

Adding all three pairs gives the total surface area of a cuboid: $2 l b + 2 b h + 2 h l$. Its volume is the product of the three edges, $l b h$.

A cube is a cuboid whose three edges are all equal, $l = b = h = a$. We measure length, breadth and height for any box, and find both totals the same way.

Your turn: a cube has edge $5$ cm. What are its total surface area and volume? (Answer: $150$ cm² and $125$ cm³, since $6 \cdot 5^2 = 150$ and $5^3 = 125$.)

Five basic solids are drawn, each labelled with only the letters its own formulas need.
Check yourself
  1. For a cuboid of length $l$, breadth $b$ and height $h$, which expression gives its total surface area?
    1. $l b h$ — the cuboid’s volume, not its surface area
    2. $2(l+b+h)$ — twice the sum of the three edge lengths
    3. $2 l b + 2 b h + 2 h l$
    4. $6 l b h$ — six times the cuboid’s own volume
    Check your answer
    1. $l b h$ — the cuboid’s volume, not its surface area — $l b h$ is the cuboid’s volume; the total surface area is $2 l b+2 b h+2 h l$, the sum of all six face areas.
    2. $2(l+b+h)$ — twice the sum of the three edge lengths — $2(l+b+h)$ sums lengths, not areas; each pair of opposite faces contributes a PRODUCT of two edges, not a sum.
    3. ✓ $2 l b + 2 b h + 2 h l$ — (C) The total surface area sums all six face areas: $2 l b+2 b h+2 h l$.
    4. $6 l b h$ — six times the cuboid’s own volume — The factor $6$ belongs to the cube’s formula $6a^2$, where all edges are equal; a general cuboid has three different face-pair areas, not one repeated six times.
  2. A cube is ‘a cuboid with $l=b=h=a$.’ Using this, why does $2 l b+2 b h+2 h l$ simplify to $6a^2$ for a cube?
    1. the formula changes completely for a cube, since a cube is a different solid
    2. each term becomes $2a^2$, and three copies sum to $6a^2$
    3. $6$ comes from multiplying the three edges together
    4. the formula only works for a cube if $a$ is measured in centimetres
    Check your answer
    1. the formula changes completely for a cube, since a cube is a different solid — A cube is not a different solid — it is a cuboid with equal edges, so the very same formula simplifies down to $6a^2$.
    2. ✓ each term becomes $2a^2$, and three copies sum to $6a^2$ — (B) With $l=b=h=a$, each of the three terms equals $2a^2$, and $2a^2+2a^2+2a^2=6a^2$.
    3. $6$ comes from multiplying the three edges together — The $6$ counts the cube’s six identical square faces; it is not a product of the edge lengths, which would instead give the volume.
    4. the formula only works for a cube if $a$ is measured in centimetres — The formula $6a^2$ holds in any consistent unit of length; units never change which formula applies.
  3. A wooden block $8$ cm long, $5$ cm broad and $3$ cm high is to be painted on all its faces. How much surface must the paint cover?
    1. $120$ cm³, from $8 \cdot 5 \cdot 3$
    2. $79$ cm², forgetting to double the three face-pair areas
    3. $32$ cm², from adding the three edge lengths and doubling
    4. $158$ cm²
    Check your answer
    1. $120$ cm³, from $8 \cdot 5 \cdot 3$ — $8 \cdot 5 \cdot 3=120$ cm³ is the block’s volume; painting covers surface area, $158$ cm², not the space it occupies.
    2. $79$ cm², forgetting to double the three face-pair areas — Each of the three areas $40$, $15$ and $24$ belongs to a PAIR of opposite faces, so the sum $79$ must be doubled to $158$.
    3. $32$ cm², from adding the three edge lengths and doubling — $2(8+5+3)=32$ adds lengths; surface area needs the three face-pair PRODUCTS doubled, not the edges themselves.
    4. ✓ $158$ cm² — (D) $2(8 \cdot 5 + 5 \cdot 3 + 3 \cdot 8) = 2(40+15+24) = 158$ cm².

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Surface area and volume, sphere family

Cutting a solid creates a new face, so the surface area of a half is more than half of the whole.

A sphere is a perfectly round solid with every point on its surface the same distance from its centre. For a sphere of radius $r$, the surface area is $4 \cdot \pi r^2$ and the volume is $4/3 \cdot \pi r^3$.

A hemisphere is exactly half a sphere, cut through its centre. Its curved surface area and its volume are both half of the whole sphere’s: curved surface area $2 \cdot \pi r^2$, volume $2/3 \cdot \pi r^3$.

*A hemisphere’s total surface area is not half the sphere’s: the flat circular base has to be added on, giving $3 \cdot \pi r^2$ in total.* We can halve the curved surface and the volume with confidence, but never the total surface area the same way.

Your turn: a hemisphere has radius $14$ cm. What is its curved surface area? (Answer: $1232$ cm², since $2 \cdot 22/7 \cdot 14^2 = 2 \cdot 616 = 1232$.)

Check yourself
  1. For a sphere of radius $r$, which pair correctly gives its surface area and volume?
    1. surface area $2 \pi r^2$, volume $2/3 \pi r^3$
    2. surface area $\pi r^2$, volume $\pi r^3$
    3. surface area $4 \pi r^3$, volume $4/3 \pi r^2$
    4. surface area $4 \pi r^2$, volume $4/3 \pi r^3$
    Check your answer
    1. surface area $2 \pi r^2$, volume $2/3 \pi r^3$ — $2\pi r^2$ and $2/3 \pi r^3$ belong to a hemisphere, exactly half a sphere; the whole sphere’s values are twice these.
    2. surface area $\pi r^2$, volume $\pi r^3$ — Both formulas need their coefficients — $4$ for the surface area and $4/3$ for the volume — dropping them gives numbers far too small.
    3. surface area $4 \pi r^3$, volume $4/3 \pi r^2$ — Surface area, a two-dimensional measure, uses $r^2$; volume, three-dimensional, uses $r^3$ — this option has the powers exchanged.
    4. ✓ surface area $4 \pi r^2$, volume $4/3 \pi r^3$ — (D) A sphere’s surface area is $4\pi r^2$ and its volume is $4/3 \pi r^3$.
  2. A hemisphere’s volume is exactly half a sphere’s, so why is a hemisphere’s TOTAL surface area more than half a sphere’s surface area?
    1. the hemisphere adds a flat circular base that the sphere never had
    2. it is not more than half — $3 \pi r^2$ is exactly half of $4 \pi r^2$
    3. curved surfaces always measure larger once a solid is cut in half
    4. the hemisphere’s radius is larger than the sphere’s, so its surface area is larger
    Check your answer
    1. ✓ the hemisphere adds a flat circular base that the sphere never had — (A) The flat circular base created by the cut adds extra area that a whole sphere never carried.
    2. it is not more than half — $3 \pi r^2$ is exactly half of $4 \pi r^2$ — Half of $4 \pi r^2$ is $2 \pi r^2$, the hemisphere’s CURVED surface area; the TOTAL $3 \pi r^2$ is more than half because of the added flat base.
    3. curved surfaces always measure larger once a solid is cut in half — There is no such general rule; the extra area is specifically the flat circular base the cut creates, not a property of cutting shapes in general.
    4. the hemisphere’s radius is larger than the sphere’s, so its surface area is larger — The comparison uses the SAME radius $r$ for the sphere and the hemisphere; the difference comes only from the added flat base.
  3. A hemisphere’s curved surface area and its volume are both found by simply halving the sphere’s formula. Why does the same halving trick fail for the hemisphere’s total surface area?
    1. it does not fail — $3\pi r^2$ is what you get by halving $4\pi r^2$ and rounding
    2. total surface area cannot be halved because it is measured in different units from curved surface area
    3. the halving trick fails for volume too, not just for total surface area
    4. halving the sphere also creates a brand-new flat face, which has no half to come from
    Check your answer
    1. it does not fail — $3\pi r^2$ is what you get by halving $4\pi r^2$ and rounding — There is no rounding involved; $3\pi r^2$ is an exact value, built from a genuine flat face, not from rounding $2\pi r^2$.
    2. total surface area cannot be halved because it is measured in different units from curved surface area — Curved and total surface area are measured in exactly the same units; the mismatch is not units, but the new flat face the cut creates.
    3. the halving trick fails for volume too, not just for total surface area — The hemisphere’s volume genuinely is exactly half the sphere’s, $2/3 \pi r^3$; it is only the total surface area where a new face breaks the simple halving.
    4. ✓ halving the sphere also creates a brand-new flat face, which has no half to come from — (D) The cut creates an entirely new flat face; there is no ‘half’ of that face in the original sphere to derive it from.
  4. A solid hemisphere has radius $7$ cm. Using $\pi=22/7$, what is its total surface area?
    1. $308$ cm², using only the curved surface
    2. $462$ cm²
    3. $616$ cm², from doubling the curved surface area
    4. $718.67$ cm³, treating the answer as a volume
    Check your answer
    1. $308$ cm², using only the curved surface — $308$ cm² is the CURVED surface area alone; the flat circular base adds another $154$ cm² to reach the total, $462$ cm².
    2. ✓ $462$ cm² — (B) $3 \cdot 22/7 \cdot 49 = 462$ cm².
    3. $616$ cm², from doubling the curved surface area — The flat base is added ONCE, not by doubling the curved surface; $308+154=462$ cm², not $308$ doubled.
    4. $718.67$ cm³, treating the answer as a volume — The question asks for surface area, measured in cm², not a volume in cm³; the correct total surface area is $462$ cm².

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Surface area and volume: cylinder

A right circular cylinder has two flat circular ends and one curved surface joining them. For radius $r$ and height $h$, the curved surface area is $2 \cdot \pi r h$.

Adding the two flat circular ends gives the total surface area, $2 \cdot \pi r h + 2 \cdot \pi r^2$. The volume is $\pi r^2 h$, the area of one circular end times the height.

Curved surface area, by contrast, needs the height and the radius together, not the base area. If a question only asks for volume, we can skip the curved surface area completely.

Your turn: a cylinder has radius $7$ cm and height $15$ cm. What is its curved surface area? (Answer: $660$ cm², since $2 \cdot 22/7 \cdot 7 \cdot 15 = 2 \cdot 22 \cdot 15 = 660$.)

Check yourself
  1. For a cylinder of radius $r$ and height $h$, which expression gives its total surface area?
    1. $\pi r^2 h$ — the cylinder’s volume, not its surface area
    2. $2 \pi r h + 2 \pi r^2$
    3. $2 \pi r h$ — the curved part alone, missing the two flat ends
    4. $4 \pi r h$ — doubling the curve instead of adding the ends
    Check your answer
    1. $\pi r^2 h$ — the cylinder’s volume, not its surface area — $\pi r^2 h$ is the cylinder’s volume; the total surface area adds the curved part $2\pi r h$ to the two flat ends $2\pi r^2$.
    2. ✓ $2 \pi r h + 2 \pi r^2$ — (B) The total surface area is the curved part $2\pi r h$ plus the two flat ends $2\pi r^2$.
    3. $2 \pi r h$ — the curved part alone, missing the two flat ends — $2\pi r h$ is the CURVED surface area alone; the total also needs the two circular ends, $+2\pi r^2$.
    4. $4 \pi r h$ — doubling the curve instead of adding the ends — Doubling $2\pi r h$ has no basis; the two missing pieces are the flat circular ends, each of area $\pi r^2$, not a second copy of the curve.
  2. A cylinder’s total surface area formula adds $2\pi r h$ to two copies of $\pi r^2$. What does each of the two $\pi r^2$ pieces represent?
    1. the two flat circular ends, top and bottom
    2. the curved surface, counted in two different ways
    3. one end of the cylinder, counted twice by mistake
    4. the inside and outside of the curved surface
    Check your answer
    1. ✓ the two flat circular ends, top and bottom — (A) Each $\pi r^2$ is the area of one flat circular end, top and bottom.
    2. the curved surface, counted in two different ways — The curved surface is already fully counted once, in $2\pi r h$; the two $\pi r^2$ pieces are the separate flat ends, not another view of the curve.
    3. one end of the cylinder, counted twice by mistake — A cylinder genuinely has two flat ends, top and bottom; each $\pi r^2$ belongs to one real end, not to counting one end twice.
    4. the inside and outside of the curved surface — A solid cylinder’s curved surface has one outward-facing area, already in $2\pi r h$; the two $\pi r^2$ terms are the flat ends, not an inside surface.
  3. A metal rod is a solid cylinder of radius $3.5$ cm and height $10$ cm. Taking $\pi=22/7$, what is its volume?
    1. $220$ cm³, from the curved surface area formula instead
    2. $297$ cm³, from the total surface area formula instead
    3. $110$ cm³, from $22/7 \cdot 3.5 \cdot 10$ without squaring the radius
    4. $385$ cm³
    Check your answer
    1. $220$ cm³, from the curved surface area formula instead — $220$ cm² is the CURVED SURFACE AREA, from $2 \cdot 22/7 \cdot 3.5 \cdot 10$; the volume uses $\pi r^2 h$, giving $385$ cm³.
    2. $297$ cm³, from the total surface area formula instead — $297$ cm² is the TOTAL SURFACE AREA; the volume is a different formula, $\pi r^2 h=385$ cm³.
    3. $110$ cm³, from $22/7 \cdot 3.5 \cdot 10$ without squaring the radius — The volume formula needs $r^2$, not $r$ alone; using $r$ once gives $110$, but squaring it correctly gives $385$ cm³.
    4. ✓ $385$ cm³ — (D) $22/7 \cdot 3.5 \cdot 3.5 \cdot 10 = 385$ cm³.

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Surface area and volume: cone

A right circular cone has one curved surface that slants from a circular base up to a single apex. Its slant height is $l = \sqrt{r^2 + h^2}$, for base radius $r$ and height $h$.

The curved surface area of a cone is $\pi r l$, built from the slant height, never the vertical height. Adding the flat circular base gives the total surface area, $\pi r l + \pi r^2$.

*The volume of a cone, $1/3 \cdot \pi r^2 h$, uses the vertical height $h$: the slant height plays no part in it at all.* We reach for $h$ when we need the volume, and for $l$ only when we need a curved surface.

Your turn: a cone has radius $7$ cm and slant height $10$ cm. What is its curved surface area? (Answer: $220$ cm², since $\pi r l \approx 22/7 \cdot 7 \cdot 10 = 220$.)

A cone with radius 7 and slant height 10 has curved surface area pi r l, about 220 square centimetres.
Check yourself
  1. For a cone of base radius $r$ and slant height $l$, which expression gives its curved surface area?
    1. $\pi r h$
    2. $1/3 \pi r^2 h$
    3. $\pi r l$
    4. $2 \pi r l$
    Check your answer
    1. $\pi r h$ — The curved surface area formula needs the SLANT height $l$, not the vertical height $h$; the two are different lengths.
    2. $1/3 \pi r^2 h$ — $1/3 \pi r^2 h$ is the cone’s volume; the curved surface area is $\pi r l$, using the slant height.
    3. ✓ $\pi r l$ — (C) A cone’s curved surface area is $\pi r l$, using the slant height.
    4. $2 \pi r l$ — A cone has one curved surface, counted once; there is no second copy to justify doubling $\pi r l$.
  2. A cone’s total surface area formula is $\pi r l + \pi r^2$, using the slant height $l$ for one term but the radius $r$ for the other. Why does the first term need $l$ while the second needs only $r$?
    1. both terms could use either $l$ or $r$; the choice makes no difference to the answer
    2. $l$ belongs to the base and $r$ belongs to the curved surface, the reverse of what the formula shows
    3. the curved part slants and needs $l$; the flat base needs only $r$
    4. the flat base also needs $l$, but the formula simplifies it away
    Check your answer
    1. both terms could use either $l$ or $r$; the choice makes no difference to the answer — $l$ and $r$ are different lengths for any real cone with $h>0$; swapping them changes the answer, so the choice is not free.
    2. $l$ belongs to the base and $r$ belongs to the curved surface, the reverse of what the formula shows — It is the CURVED part that slants and needs $l$; the flat circular base is measured by its radius $r$ alone, not the reverse.
    3. ✓ the curved part slants and needs $l$; the flat base needs only $r$ — (C) The slanting curved part needs $l$; the flat circular base is measured by its radius $r$ alone.
    4. the flat base also needs $l$, but the formula simplifies it away — A flat circle’s area, $\pi r^2$, genuinely needs only its radius; there is no $l$ term hidden or simplified away for it.
  3. A conical tent has base radius $7$ m and slant height $25$ m. Taking $\pi=22/7$, how much canvas is needed for the tent, ignoring the base?
    1. $704$ m², including the base’s floor area
    2. $1232$ m³, treating the answer as a volume
    3. $275$ m², from halving the curved surface area for the base being flat
    4. $550$ m²
    Check your answer
    1. $704$ m², including the base’s floor area — The tent’s floor is not canvas; the question asks to ignore the base, so only the curved part, $550$ m², is needed.
    2. $1232$ m³, treating the answer as a volume — $1232$ m³ is the cone’s VOLUME, computed with the height $24$ m; the question asks for canvas needed, a surface area in m², which is $550$.
    3. $275$ m², from halving the curved surface area for the base being flat — There is no reason to halve the curved surface area; $22/7 \cdot 7 \cdot 25=550$ m² is already the full curved surface, needing no adjustment for the base.
    4. ✓ $550$ m² — (D) Curved surface area $=22/7 \cdot 7 \cdot 25=550$ m²; the base is ignored since it is not made of canvas.

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Only exposed faces count

▲CONCEPT
A cone joined to a cylinder hides the two circles at their join, leaving only the curved and flat faces exposed.

A combination of two solids has more faces than either solid alone, but not every one of them counts. The surface area of a combination adds up only the faces still visible from outside the finished object.

A face where the two solids meet is covered over. It drops out of the total, no matter which of the two solids it originally belonged to.

Checking whether a face is covered at the join settles which faces belong in a combination’s surface area and which do not.

Your turn: does the face where two solids touch belong in the combination’s surface area? (Answer: no, since that face is covered and no longer visible from outside.)

Kanada
A SCHOLAR INDIA REMEMBERS

Look at each face and ask what kind it is. Can you touch it from outside, or is it hidden inside the join? Take a cone standing on a hemisphere. The two flat circles are pressed together and hidden. We count the curved surface of the cone and the curved surface of the hemisphere. The hidden circles do not count.

Kanada · the sparrow
Check yourself
  1. When two solids are joined to form a combination, which faces count toward the combination’s total surface area?
    1. every face of both original solids, whether hidden or visible
    2. only the faces that belonged to the larger of the two solids
    3. only the faces still visible from outside the combined solid
    4. only the curved faces, never the flat ones
    Check your answer
    1. every face of both original solids, whether hidden or visible — The face where the two solids meet is covered by the join and is no longer visible; it must be left out, not counted along with the visible faces.
    2. only the faces that belonged to the larger of the two solids — There is no rule favouring the larger solid; visible faces from BOTH solids count, and the rule is about visibility, not size.
    3. ✓ only the faces still visible from outside the combined solid — (C) Only the faces still visible from outside the combined solid count toward its total surface area.
    4. only the curved faces, never the flat ones — Flat faces that remain exposed count exactly as much as curved ones; the only face excluded is the one hidden at the join.
  2. A hemisphere is fixed onto one flat end of a cylinder. Why does the cylinder’s OTHER flat end still count fully toward the total surface area?
    1. it was never touched by the join, so it remains just as exposed as before
    2. it does not count fully — every face of a joined solid loses some area
    3. it counts fully only if the hemisphere is smaller than the cylinder’s radius
    4. it stops counting once any face of the solid is joined to another solid
    Check your answer
    1. ✓ it was never touched by the join, so it remains just as exposed as before — (A) The other end is untouched by the join and so remains fully exposed.
    2. it does not count fully — every face of a joined solid loses some area — The join only hides the ONE face where the two solids actually touch; every other face, including the cylinder’s untouched end, keeps its full area.
    3. it counts fully only if the hemisphere is smaller than the cylinder’s radius — Whether the untouched end counts fully has nothing to do with the hemisphere’s size; it counts fully simply because the join never reaches it.
    4. it stops counting once any face of the solid is joined to another solid — A join removes exactly the one face it touches; it has no effect on the solid’s other, untouched faces.
  3. Why does the rule ‘count only exposed faces’ apply the same way whether one solid is fixed ONTO another or hollowed OUT of another?
    1. it does not apply the same way — hollowing always adds surface area instead of removing it
    2. it does not apply the same way — a hollowed solid has no faces left to be exposed
    3. the joined face leaves the outer boundary in both cases
    4. it applies the same way only when both solids are the same shape
    Check your answer
    1. it does not apply the same way — hollowing always adds surface area instead of removing it — Hollowing can increase, decrease or leave unchanged the total surface area depending on the shapes involved; the RULE about counting only the outer boundary is what stays the same, not a fixed direction of change.
    2. it does not apply the same way — a hollowed solid has no faces left to be exposed — A hollowed solid still has plenty of exposed faces; only the boundary shifts to include the newly revealed inner surface where the cavity was cut.
    3. ✓ the joined face leaves the outer boundary in both cases — (C) In both cases the face at the join leaves the outer boundary, whether that face is covered by an added solid or opened up by a removed one.
    4. it applies the same way only when both solids are the same shape — The rule about counting only the outer boundary applies to any two solids joined or hollowed, regardless of whether they share a shape.

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A join hides two faces

▲CONCEPT
Count the flat circles before and after: three become one, because a single join swallows a face from each of the two solids.

A join between two solids hides a face on each side of it, not just one. Fixing a hemisphere onto the flat end of a cylinder removes two faces at once.

The cylinder’s own flat end disappears, because the hemisphere now sits exactly where it used to be. The hemisphere’s flat base disappears too, because it sits flush against the cylinder’s end.

Adding the two full total surface areas counts the joined faces twice, before anyone has subtracted anything. One fix adds only the curved surfaces that remain. A second fix subtracts the hidden circle from a total that still includes it. We check our own working the same way before trusting any sum of two totals.

Your turn: when a hemisphere is glued onto a cylinder, how many faces does the join hide? (Answer: two, since both the cylinder’s flat end and the hemisphere’s flat base disappear together.)

Check yourself
  1. When a hemisphere is fixed onto a flat end of a cylinder of the same radius, how many faces does the join hide?
    1. two — the cylinder’s end and the hemisphere’s base, both at once
    2. one — only the cylinder’s flat end disappears
    3. one — only the hemisphere’s flat base disappears
    4. none — both flat faces are still there, simply touching each other
    Check your answer
    1. ✓ two — the cylinder’s end and the hemisphere’s base, both at once — (A) Both the cylinder’s flat end and the hemisphere’s flat base are hidden together at the single join.
    2. one — only the cylinder’s flat end disappears — The hemisphere’s flat base is pressed against the cylinder’s end and is equally hidden; both faces vanish together, not just the cylinder’s.
    3. one — only the hemisphere’s flat base disappears — The cylinder’s flat end is pressed against the hemisphere’s base and is equally hidden; both faces vanish together, not just the hemisphere’s.
    4. none — both flat faces are still there, simply touching each other — Once two faces touch and press together at a join, neither is visible from outside any more; both are hidden, not merely touching while staying exposed.
  2. Why does a join between two solids always hide a face on BOTH sides of it, rather than on just one side?
    1. because the smaller solid’s face always covers the larger solid’s face
    2. it does not always hide both sides — only curved solids hide both faces
    3. one side is hidden by the join, and the other side is hidden by the paint or material added later
    4. the two faces used to sit in the very same place once joined, so both are covered at once
    Check your answer
    1. because the smaller solid’s face always covers the larger solid’s face — Neither solid’s face ‘covers’ the other by size; both faces occupy the identical spot at the join and are hidden together, regardless of which solid is larger.
    2. it does not always hide both sides — only curved solids hide both faces — The both-sides rule holds for any two solids meeting at a matching face, curved or flat; it is not limited to curved solids.
    3. one side is hidden by the join, and the other side is hidden by the paint or material added later — Nothing external is needed; the geometry of the join itself hides both faces, because they occupy the same location the moment the solids meet.
    4. ✓ the two faces used to sit in the very same place once joined, so both are covered at once — (D) Both faces occupy the identical spot at the join, so both are hidden together.
  3. A capsule is a cylinder with a hemisphere fixed on EACH flat end. How many faces does joining hide in total, across both ends?
    1. two — one face per join, not two
    2. four — each of the two joins hides a pair of faces
    3. three — the cylinder only has one flat end to lose
    4. eight — every face of all three solids disappears
    Check your answer
    1. two — one face per join, not two — Each single join hides a face on both sides of it, so two joins hide two faces each — four faces in total, not two.
    2. ✓ four — each of the two joins hides a pair of faces — (B) Each join hides two faces, and there are two joins, so four faces are hidden in total.
    3. three — the cylinder only has one flat end to lose — A cylinder has two flat ends, and BOTH are used in a capsule, one at each hemisphere; three faces undercounts by one.
    4. eight — every face of all three solids disappears — Only the faces actually AT a join disappear; the two hemispheres’ curved surfaces and everywhere else on the cylinder stay fully exposed.
  4. A hemisphere of radius $4$ cm is fixed onto a cylinder of the same radius. Since the join hides a face on both sides, how much flat area is removed from the combination’s total surface area, compared with simply adding the two solids’ total surface areas?
    1. the area of one circle of radius $4$ cm, since only one face is ever really hidden
    2. the curved surface area of the hemisphere, since that is what replaces the missing faces
    3. twice the area of one circle of radius $4$ cm, since two flat faces are hidden
    4. no area is removed; the two total surface areas are simply added unchanged
    Check your answer
    1. the area of one circle of radius $4$ cm, since only one face is ever really hidden — Both the cylinder’s end and the hemisphere’s base are hidden at this join, not just one; the removed area is TWO circles’ worth, not one.
    2. the curved surface area of the hemisphere, since that is what replaces the missing faces — The hemisphere’s curved surface is what the combination GAINS at the join; the amount REMOVED from the simple total is the two hidden flat circles, a separate quantity.
    3. ✓ twice the area of one circle of radius $4$ cm, since two flat faces are hidden — (C) Two flat circles, one from each solid, are hidden at the join, so twice one circle’s area is removed.
    4. no area is removed; the two total surface areas are simply added unchanged — If nothing were removed, the combination’s surface area would equal the sum of the two total surface areas — exactly the mistake this chapter singles out; two hidden circles’ worth of area is removed.

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Only the shared face disappears

▲CONCEPT
Check yourself
  1. A hemisphere sits on one face of a cube. What happens to the cube’s other five faces?
    1. they stay fully exposed and fully counted, untouched by the join
    2. each loses a small circle of area, matching the hemisphere’s base
    3. they are halved in area, since the cube now shares its shape with the hemisphere
    4. they disappear along with the joined face, since the cube is no longer a separate solid
    Check your answer
    1. ✓ they stay fully exposed and fully counted, untouched by the join — (A) The other five faces are never touched by the join and so stay fully exposed and fully counted.
    2. each loses a small circle of area, matching the hemisphere’s base — Only the ONE face the hemisphere actually touches loses area; the other five faces are never in contact with the hemisphere and lose nothing.
    3. they are halved in area, since the cube now shares its shape with the hemisphere — The other five faces are completely unaffected by the join; there is no mechanism by which joining one face would halve the areas of the others.
    4. they disappear along with the joined face, since the cube is no longer a separate solid — The cube remains a real part of the combination with five faces still exposed; only the one face at the join is removed, not the whole solid’s identity.
  2. Why does only the small circle under the hemisphere disappear from the cube’s face, rather than the cube’s whole face disappearing?
    1. the cube’s whole face does disappear, but the hemisphere’s curved surface exactly replaces its full area
    2. the circle disappears because it is inside the cube, hidden from the hemisphere’s own view
    3. the hemisphere’s base only covers a circle matching its own radius
    4. the circle disappears only if the hemisphere’s radius equals the cube’s edge
    Check your answer
    1. the cube’s whole face does disappear, but the hemisphere’s curved surface exactly replaces its full area — Only the small circle matching the hemisphere’s radius is removed, not the whole square face; the rest of that face, outside the circle, stays exposed and counted.
    2. the circle disappears because it is inside the cube, hidden from the hemisphere’s own view — The circle is removed because the hemisphere’s base physically covers it at the join, not because of any notion of what is ‘hidden from view.’
    3. ✓ the hemisphere’s base only covers a circle matching its own radius — (C) The hemisphere only physically covers a circle matching its own radius; the rest of the square face stays exposed around it.
    4. the circle disappears only if the hemisphere’s radius equals the cube’s edge — The circle under the hemisphere is removed whenever the hemisphere sits on that face, whatever its radius is relative to the cube’s edge.
  3. A cube of edge $6$ cm has a hemisphere of radius $2$ cm fixed on its top face. Taking $\pi=22/7$, what area of the top face still counts as exposed, after the hemisphere is fixed?
    1. $23.43$ cm²
    2. $36$ cm², the whole top face, unaffected by the hemisphere
    3. $0$ cm², since the whole top face is now covered by the hemisphere
    4. $12.57$ cm², just the circle the hemisphere covers
    Check your answer
    1. ✓ $23.43$ cm² — (A) $36 - 22/7 \cdot 4 = 36-12.57=23.43$ cm².
    2. $36$ cm², the whole top face, unaffected by the hemisphere — The hemisphere’s base covers a circle of that face, about $12.57$ cm²; the remaining exposed part is $36-12.57=23.43$ cm², not the whole $36$ cm².
    3. $0$ cm², since the whole top face is now covered by the hemisphere — The hemisphere only covers a circle of radius $2$ cm on the face, not the whole $6$ cm square; plenty of the face remains exposed around it.
    4. $12.57$ cm², just the circle the hemisphere covers — $12.57$ cm² is the area the hemisphere COVERS and removes; the area still counted as EXPOSED is the rest of the face, $36-12.57=23.43$ cm².
A hemisphere on a cube’s face removes only the circle beneath it, and the cube’s other five faces stay complete.

A join changes exactly one face of each solid, never the whole solid. Every other face stays fully exposed and fully counted, exactly as it was before the join.

Fix a hemisphere onto one face of a cube. The cube’s other five faces are untouched, and all five still count in full.

Only the small circle directly under the hemisphere is removed from the cube’s total surface area. We name which single face the join touches before doing any arithmetic. Every other face then stays safe from being lost or doubled by mistake.

Your turn: a hemisphere is fixed to one face of a solid. Do the solid’s other faces change? (Answer: no, since a join changes only the one face it touches, and every other face stays untouched.)

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A join needs matching radii

▲CONCEPT
Check yourself
  1. For a cone mounted on a hemisphere to join smoothly, what must be true of the two solids?
    1. they must share the same height
    2. they must share the same total surface area
    3. the same radius at the touching face
    4. they must both be made of the same material
    Check your answer
    1. they must share the same height — The cone and hemisphere can have any heights they like; what must match at the join is the RADIUS of the touching circular faces, not the heights.
    2. they must share the same total surface area — Matching total surface areas has nothing to do with a smooth join; only the RADIUS at the touching face needs to match.
    3. ✓ the same radius at the touching face — (C) A smooth join requires the two solids to share the same radius at the touching face.
    4. they must both be made of the same material — The smoothness of a join is a matter of shape and radius, not of what material the solids are made from.
  2. Why does a mismatched radius at a join leave a step or a gap between the two solids?
    1. because the two solids would then have different volumes
    2. because mismatched radii make the combination’s total surface area negative
    3. a larger circle cannot meet a smaller one edge-to-edge
    4. because the taller solid always pushes the shorter one out of alignment
    Check your answer
    1. because the two solids would then have different volumes — Volumes can differ freely without causing any step; the step comes specifically from the two touching circles having different radii, not different volumes.
    2. because mismatched radii make the combination’s total surface area negative — Surface area is never negative, matched radii or not; a mismatch simply leaves a visible step or gap, not a negative quantity.
    3. ✓ a larger circle cannot meet a smaller one edge-to-edge — (C) A larger circle cannot meet a smaller one edge-to-edge; the mismatch shows up as a visible step.
    4. because the taller solid always pushes the shorter one out of alignment — Height plays no part in whether the join is smooth; the cause of a step is specifically the two circles at the join having different radii.
  3. A test tube is described as a cylinder with a hemisphere at one end. Why is it safe to assume, without being told, that the hemisphere’s radius equals the cylinder’s radius?
    1. it is not safe to assume this; the problem must always state both radii separately
    2. describing it as one smooth solid already implies matching radii
    3. cylinders and hemispheres always have equal radii by definition, whatever the object
    4. the radii match because both shapes are curved solids
    Check your answer
    1. it is not safe to assume this; the problem must always state both radii separately — A real test tube is one continuous smooth shape; describing it that way already tells you the radii match, without needing a separate statement.
    2. ✓ describing it as one smooth solid already implies matching radii — (B) Describing it as one smooth, continuous shape already implies matching radii at the join.
    3. cylinders and hemispheres always have equal radii by definition, whatever the object — Cylinders and hemispheres can have any radius on their own; it is only because the test tube is described as one smooth object that its particular radii must match.
    4. the radii match because both shapes are curved solids — Being curved has nothing to do with matching radii; the reason is that the object is one continuous smooth shape, which forces the join’s radii to agree.
A dome whose base matches the block’s width joins smoothly, but a narrower dome leaves a flat ring exposed at the join.

Two solids join smoothly only when they share the same radius at the face where they meet. A cone mounted on a hemisphere needs the cone’s base radius to match the hemisphere’s radius exactly.

The same rule holds for a hemisphere fixed on a cylinder. It also holds for a hemisphere fixed on a flat face of matching size. A mismatched radius leaves a step or a gap at the join, instead of one smooth surface.

We check that the two radii match before writing down any formula.

Your turn: can a cone of radius $5$ cm sit smoothly on a hemisphere of radius $6$ cm? (Answer: no, since a smooth join needs the two radii to match exactly.)

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Spotting a combination in real life

▲CONCEPT
Check yourself
  1. A capsule of medicine is described in this chapter as a combination of which basic solids?
    1. a cylinder with a cone fixed at each end
    2. a cylinder with a hemisphere fixed at each end
    3. two hemispheres joined directly to each other, with no cylinder
    4. a sphere with a cylinder passing through its centre
    Check your answer
    1. a cylinder with a cone fixed at each end — A capsule’s rounded ends are hemispheres, not cones; a cone would give the object a pointed end, not a rounded one.
    2. ✓ a cylinder with a hemisphere fixed at each end — (B) A capsule is a cylinder with a hemisphere fixed at each end.
    3. two hemispheres joined directly to each other, with no cylinder — A capsule has a straight cylindrical body between its two rounded ends; two hemispheres alone would make a full sphere, with no straight middle.
    4. a sphere with a cylinder passing through its centre — A capsule is a solid cylinder with rounded ends, not a sphere with a hole bored through it.
  2. An ice-cream cone with one scoop of ice cream on top is treated in this chapter as a combination of which two solids?
    1. a cone and a sphere
    2. a cone and a hemisphere
    3. a cylinder and a cone
    4. a cuboid and a hemisphere
    Check your answer
    1. a cone and a sphere — A single rounded scoop resting on the cone’s rim is modelled as a hemisphere, not a full sphere, since only its curved top is exposed above the cone.
    2. ✓ a cone and a hemisphere — (B) The cone-shaped holder and the rounded scoop on top are modelled as a cone and a hemisphere.
    3. a cylinder and a cone — The ice-cream holder narrows to a point, which makes it a cone, not a cylinder, which would have the same width all the way down.
    4. a cuboid and a hemisphere — The holder is a cone, tapering to a point, not a box-shaped cuboid.
  3. A grain silo is a cylindrical tank topped with a conical roof, both of the same radius. A farmer wants its total outer surface area, ignoring the base resting on the ground. Which two surfaces does this combination add?
    1. the cylinder’s total surface area and the cone’s total surface area
    2. the cylinder’s curved surface and the cone’s total surface area
    3. the cylinder’s volume and the cone’s volume
    4. the cylinder’s curved surface and the cone’s curved surface
    Check your answer
    1. the cylinder’s total surface area and the cone’s total surface area — Adding both TOTAL surface areas would double-count the flat circle where the cone meets the cylinder; only the two CURVED surfaces are added, since the base is also excluded here.
    2. the cylinder’s curved surface and the cone’s total surface area — Both solids contribute only their curved parts to this combination; using the cone’s TOTAL surface area would wrongly include its own flat base, which is hidden at the join.
    3. the cylinder’s volume and the cone’s volume — The question asks for outer SURFACE AREA to know how much material covers the silo, not the volume of grain it can hold.
    4. ✓ the cylinder’s curved surface and the cone’s curved surface — (D) With the base ignored and the join hiding a flat circle, only the two curved surfaces are added.

Many everyday objects are combinations of exactly two basic solids, joined at one matching face. Naming the two solids first turns an unfamiliar object into two familiar formulas.

A test tube is a cylinder with a hemisphere at one end. A capsule-shaped tablet is a cylinder with a hemisphere at each end. An ice-cream cone with a scoop on top is a cone with a hemisphere.

Get the naming of the two solids and their shared face wrong, and every next formula is built on the wrong shape.

Your turn: what two solids make up a pencil sharpened to a point at one end? (Answer: a cylinder and a cone, joined at the pencil’s flat circular end.)

A test tube and an ice-cream cone are each pulled apart into the two solids that join to make them.

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Volume simply adds or subtracts

▲CONCEPT
Check yourself
  1. A toy is a cone fixed on top of a hemisphere. How is the toy’s total volume found?
    1. add the volume of the cone to the volume of the hemisphere
    2. add the two volumes, then subtract the volume of the hidden circle at the join
    3. average the two volumes
    4. multiply the two volumes together
    Check your answer
    1. ✓ add the volume of the cone to the volume of the hemisphere — (A) The volumes simply add: cone volume plus hemisphere volume.
    2. add the two volumes, then subtract the volume of the hidden circle at the join — Volume has no hidden region to subtract at a join; unlike surface area, nothing about the enclosed space is lost when two solids are fixed together.
    3. average the two volumes — Volumes of a combination are summed, never averaged; averaging would understate the true enclosed space of the whole toy.
    4. multiply the two volumes together — Multiplying two volumes would give a quantity in a completely different unit (length to the sixth power); combined volumes are added, not multiplied.
  2. A conical cavity is hollowed out of a solid cylinder. How is the volume of the remaining solid found?
    1. add the cone’s volume to the cylinder’s volume
    2. subtract the cylinder’s volume from the cone’s volume
    3. the remaining solid’s volume cannot be found without knowing the surface area first
    4. subtract the cone’s volume from the cylinder’s volume
    Check your answer
    1. add the cone’s volume to the cylinder’s volume — Material is REMOVED to make the cavity, not added; the remaining solid’s volume is the cylinder’s volume minus the cone’s, not the sum.
    2. subtract the cylinder’s volume from the cone’s volume — The cylinder is the larger original solid and the cone is the smaller piece removed from it; the cone’s volume is subtracted FROM the cylinder’s, not the other way round.
    3. the remaining solid’s volume cannot be found without knowing the surface area first — Volume is found directly from the two solids’ own dimensions; it needs no information about surface area at all.
    4. ✓ subtract the cone’s volume from the cylinder’s volume — (D) The remaining volume is the cylinder’s volume minus the cone’s volume.
  3. Surface area drops a hidden face at a join, but volume never drops anything when two solids are joined. What is different about the two quantities that explains this?
    1. volume is always a bigger number than surface area, so a small loss makes no visible difference
    2. surface area is measured in a smaller unit than volume, so it is more sensitive to the join
    3. volume only avoids loss when the two solids are the same shape
    4. surface area counts boundary faces, one erased by the join; volume never loses any space
    Check your answer
    1. volume is always a bigger number than surface area, so a small loss makes no visible difference — The two are never compared by which number is bigger; the real difference is what is being measured — a boundary that can be hidden, versus enclosed space that cannot be lost by joining.
    2. surface area is measured in a smaller unit than volume, so it is more sensitive to the join — Units of measurement do not make one quantity ‘more sensitive’ to a join; the real reason is that surface area counts boundary faces, one of which the join removes, while volume counts space, none of which the join removes.
    3. volume only avoids loss when the two solids are the same shape — Volume adds outright for ANY two solids joined together, whatever their shapes; there is no same-shape requirement.
    4. ✓ surface area counts boundary faces, one erased by the join; volume never loses any space — (D) Surface area counts boundary faces, one of which the join erases; volume counts enclosed space, none of which a join ever wastes.
  4. A solid is a cylinder of radius $3$ cm and height $6$ cm, with a hemisphere of the same radius scooped out of one flat end. Using $\pi=3.14$, what is the volume of the remaining solid?
    1. $226.08$ cm³, the cylinder’s volume plus the hemisphere’s volume
    2. $169.56$ cm³, the cylinder’s volume alone, unaffected by the scoop
    3. $113.04$ cm³
    4. $56.52$ cm³, the hemisphere’s volume alone
    Check your answer
    1. $226.08$ cm³, the cylinder’s volume plus the hemisphere’s volume — The hemisphere is REMOVED from the cylinder, not added onto it; its volume must be subtracted, giving $113.04$ cm³, not added to give $226.08$ cm³.
    2. $169.56$ cm³, the cylinder’s volume alone, unaffected by the scoop — Scooping out the hemisphere genuinely removes $56.52$ cm³ of material; the remaining solid’s volume is less than the full cylinder’s, at $113.04$ cm³.
    3. ✓ $113.04$ cm³ — (C) Cylinder volume $3.14 \cdot 9 \cdot 6=169.56$ cm³ minus hemisphere volume $2/3 \cdot 3.14 \cdot 27=56.52$ cm³ gives $113.04$ cm³.
    4. $56.52$ cm³, the hemisphere’s volume alone — $56.52$ cm³ is the volume TAKEN OUT of the cylinder; the volume that remains in the solid is $169.56-56.52=113.04$ cm³.

The volume of a combination is the plain sum of its two parts’ volumes. Nothing is lost to a hidden face, unlike surface area.

When one solid is hollowed out of another instead of joined onto it, the volume works the other way. It becomes the difference of the two volumes, the larger solid’s volume minus the smaller one’s.

We decide whether a solid is joined on or hollowed out. That single choice tells us at once whether its volume adds or subtracts.

Your turn: does the volume of a scooped-out ice-cream tub add or subtract? (Answer: it subtracts, since the scooped part is hollowed out, not joined on.)

Kanada
A SCHOLAR INDIA REMEMBERS

Surface and volume are different kinds of thing. Keep them apart. When two solids join, some surface is hidden. No volume is hidden. So for a cone on a hemisphere, we simply add the two volumes.

Kanada · the sparrow

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Adding both totals overcounts

✕MISCONCEPTION
Check yourself
  1. A hemisphere of radius $3.5$ cm is fixed onto one flat end of a cylinder of the same radius. A student finds the combination’s total surface area by adding the cylinder’s total surface area to the hemisphere’s total surface area. What is wrong with this method?
    1. the joined face is hidden on both sides and must be dropped from both totals
    2. nothing is wrong — adding the two total surface areas is exactly the right method
    3. the method is wrong because the hemisphere’s total surface area formula is incorrect
    4. the method is wrong because the cylinder’s radius does not match the hemisphere’s
    Check your answer
    1. ✓ the joined face is hidden on both sides and must be dropped from both totals — (A) The joined face is hidden on both sides; it must be dropped from both totals, not counted in twice.
    2. nothing is wrong — adding the two total surface areas is exactly the right method — Adding the two total surface areas double-counts the region at the join; that face is hidden and must be removed from the total, not kept in twice.
    3. the method is wrong because the hemisphere’s total surface area formula is incorrect — The hemisphere’s own formula $3 \pi r^2$ is correct; the error is in adding both TOTALS together, which counts the hidden joined face twice.
    4. the method is wrong because the cylinder’s radius does not match the hemisphere’s — The problem states both solids share the same radius $3.5$ cm; the actual error is adding two total surface areas without removing the face hidden at the join.
  2. For the hemisphere and cylinder of radius $3.5$ cm above, adding the two total surface areas overcounts the true answer. By how much does it overcount, and why?
    1. by $38.5$ cm² — only the cylinder’s hidden face is double-counted
    2. by $115.5$ cm² — the hemisphere’s whole total surface area is the overcount
    3. by $77$ cm² — two hidden flat circles, one on each side
    4. by $0$ cm² — adding the two total surface areas gives the correct answer
    Check your answer
    1. by $38.5$ cm² — only the cylinder’s hidden face is double-counted — Both the cylinder’s end AND the hemisphere’s own base are hidden at this join; the overcount is TWO circles’ worth, $77$ cm², not one.
    2. by $115.5$ cm² — the hemisphere’s whole total surface area is the overcount — The hemisphere’s curved surface, $77$ cm², is genuinely part of the true answer; only its flat base’s share of the double-counting, together with the cylinder’s, totals $77$ cm² overcounted — not the whole $115.5$ cm².
    3. ✓ by $77$ cm² — two hidden flat circles, one on each side — (C) Two hidden circles, each of area $38.5$ cm², account for the full $77$ cm² overcount.
    4. by $0$ cm² — adding the two total surface areas gives the correct answer — Adding the two totals genuinely overcounts by $77$ cm², the two flat circles hidden at the join; it is not the correct method.
  3. Why does ‘just add the two total surface areas’ feel like a safe method to a student, even though it is wrong for a joined combination?
    1. because total surface area formulas are usually wrong on their own
    2. because students confuse curved surface area with total surface area
    3. because the join always makes the combination’s surface area smaller than either solid alone
    4. plain addition works for volume, so it feels safe to assume the same for surface area
    Check your answer
    1. because total surface area formulas are usually wrong on their own — Each solid’s own total surface area formula is correct; the error appears only when two such totals are added across a join, which hides a face.
    2. because students confuse curved surface area with total surface area — This particular trap is about adding two TOTALS across a join, not about confusing curved and total surface area, though that is a separate error this chapter also warns about.
    3. because the join always makes the combination’s surface area smaller than either solid alone — The combination’s surface area is not necessarily smaller than either single solid’s own area; the real issue is specifically that adding two full totals double-counts the hidden join, not a general shrinking effect.
    4. ✓ plain addition works for volume, so it feels safe to assume the same for surface area — (D) Volume genuinely does add outright, so it feels natural, though wrongly, to assume the same addition works for surface area too.
  4. A cone of radius $3.5$ cm is fixed on top of a cylinder of the same radius. A student computes the combination’s total surface area as the cylinder’s total surface area plus the cone’s total surface area. Which single change fixes this method?
    1. replace the cylinder’s total surface area with just its curved surface area, and leave the cone’s total surface area as it is
    2. add a third term for the area of the hidden circle, on top of both totals
    3. replace the cone’s total surface area with just its curved surface area, and subtract one circle from the cylinder’s total
    4. no change is needed — the two total surface areas may simply be added as they are
    Check your answer
    1. replace the cylinder’s total surface area with just its curved surface area, and leave the cone’s total surface area as it is — It is the CONE’s flat base that is hidden at this join, not touched by the cylinder’s other end; the cone’s total must drop to its curved surface, and one circle must leave the cylinder’s total.
    2. add a third term for the area of the hidden circle, on top of both totals — The hidden circle must be REMOVED from the naive sum, not added on top of it; adding a third term in this direction makes the overcount worse, not better.
    3. ✓ replace the cone’s total surface area with just its curved surface area, and subtract one circle from the cylinder’s total — (C) The cone’s flat base is hidden at this join, so its total drops to its curved surface, and the cylinder loses one circle from its own total.
    4. no change is needed — the two total surface areas may simply be added as they are — Adding both total surface areas as they stand double-counts the hidden face at the join; a fix is needed, dropping to curved surface on one side and removing a circle on the other.
  5. Two identical cubes of edge $a$ are glued face to face to form a cuboid. Does the ‘drop the hidden face from both totals’ rule from curved-solid combinations apply here too?
    1. no — the rule only applies when at least one of the two solids is curved
    2. no — cubes glued together do not hide any face, since both faces are flat
    3. yes, but only $a^2$ needs to be dropped, not $2a^2$
    4. yes — gluing hides a face on each cube, dropping $2a^2$
    Check your answer
    1. no — the rule only applies when at least one of the two solids is curved — The rule is about any hidden face at any join, flat or curved; two glued cubes hide a flat square face on each side, exactly the same mechanism.
    2. no — cubes glued together do not hide any face, since both faces are flat — A flat face pressed against another flat face is hidden just as completely as a curved solid’s flat base pressed against another solid; flatness does not exempt a face from the rule.
    3. yes, but only $a^2$ needs to be dropped, not $2a^2$ — Gluing hides a face on BOTH cubes, one square on each side, so $2a^2$ in total must be dropped, not just one cube’s share.
    4. ✓ yes — gluing hides a face on each cube, dropping $2a^2$ — (D) Each cube loses one square face at the glued join, so $2a^2$ in total is dropped from the naive sum of the two total surface areas.

A tempting shortcut: add the total surface areas of the two separate solids to get the combination’s total surface area.

That shortcut is wrong. The face where the two solids meet is hidden inside the join. Adding the two totals counts that face twice, once from each solid, when the combination’s real surface does not carry it at all.

We subtract the hidden faces before trusting any sum of two totals, so this mistake never reaches the final answer.

Your turn: a hemisphere of radius $7$ cm sits on a cylinder of radius $7$ cm, taking $\pi \approx 22/7$. By how much does adding the two totals overcount? (Answer: $308$ cm², since the join circle is $\pi r^2 \approx 22/7 \cdot 7 \cdot 7 = 154$ cm², and adding both totals double-counts it: $2 \cdot 154 = 308$.)

The correct panel counts 77 once, and the wrong panel hides an extra circle of 38.5 inside its 115.5, red-ringed at the join.
Siddharth sits at a school desk, pencil lifted, notebook open with faint ruled lines and no writing, a plain pencil box and an eraser beside it.
First try

Hemisphere of radius $3.5$ cm fixed onto one flat end of a cylinder of the same radius. I added the hemisphere’s whole surface, $3 \cdot \pi r^2 = 115.5$ cm², straight onto the cylinder’s own total.

Second look

The join hides one circle of area $\pi r^2 = 38.5$ cm² inside the cylinder’s flat end. That circle drops out of the cylinder’s total, and only the hemisphere’s curved part, $2 \cdot \pi r^2 = 77$ cm², is added back.

At a join, take away the hidden circle once. Never add a whole surface twice.

What is the total surface area of a cylinder with a hemisphere fixed on one end?

Weaker. A hemisphere of radius $3.5$ cm sits on a cylinder of the same radius and height $6$ cm, taking $\pi \approx 22/7$. The cylinder’s total surface area is $2 \cdot 22/7 \cdot 3.5 \cdot 6 + 2 \cdot 38.5 = 132 + 77 = 209$ cm². The hemisphere’s total surface area is $3 \cdot 38.5 = 115.5$ cm². Adding the two totals gives $209 + 115.5 = 324.5$ cm². But the join hides a circle on the cylinder and a circle on the hemisphere, so the true surface must be smaller than $324.5$ cm² by two circles, $2 \cdot 38.5 = 77$ cm². An answer equal to the sum of the two totals means the join was never taken away.

Stronger. Count only what is outside the joined solid: the cylinder’s curved surface, its one remaining flat end, and the hemisphere’s curved surface. That gives $132 + 38.5 + 77 = 247.5$ cm². And $324.5 - 247.5 = 77$, exactly the two hidden circles.

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Worked: a cylinder’s surface and volume

Check yourself
  1. A cylindrical pipe has radius $7$ cm and height $20$ cm. Taking $\pi=22/7$, what is its curved surface area?
    1. $3080$ cm³, using the volume formula instead
    2. $880$ cm²
    3. $1188$ cm², including the two flat ends
    4. $440$ cm², forgetting to double the curved surface term
    Check your answer
    1. $3080$ cm³, using the volume formula instead — $3080$ cm³ is the pipe’s VOLUME; the curved surface area uses $2 \pi r h$, giving $880$ cm².
    2. ✓ $880$ cm² — (B) $2 \cdot 22/7 \cdot 7 \cdot 20 = 880$ cm².
    3. $1188$ cm², including the two flat ends — $1188$ cm² is the TOTAL surface area, including the two flat ends; the curved surface area alone is $880$ cm².
    4. $440$ cm², forgetting to double the curved surface term — The curved surface area formula is $2 \pi r h$; leaving out the $2$ gives half the correct value, $440$ instead of $880$.
  2. For the same pipe, why is the total surface area, $1188$ cm², greater than the curved surface area, $880$ cm², by exactly $2 \cdot 22/7 \cdot 49$?
    1. because the total surface area formula includes an extra safety margin for real pipes
    2. the total adds two flat ends, each $22/7 \cdot 49$, left out of the curved measure
    3. because the pipe’s volume is added into the total surface area
    4. because the curved surface area formula was applied incorrectly the first time
    Check your answer
    1. because the total surface area formula includes an extra safety margin for real pipes — There is no safety margin in a geometric formula; the extra term is the exact area of the two flat circular ends, nothing more.
    2. ✓ the total adds two flat ends, each $22/7 \cdot 49$, left out of the curved measure — (B) The extra term is exactly the area of the two flat circular ends, which the curved measure never counted.
    3. because the pipe’s volume is added into the total surface area — The extra term, $2 \cdot 22/7 \cdot 49 = 154$, is a surface area (the two ends), not the pipe’s volume, $3080$ cm³.
    4. because the curved surface area formula was applied incorrectly the first time — The curved surface area of $880$ cm² is correct as it stands; it is deliberately partial, leaving the two flat ends for the total to add.
  3. A student needs to know how much sheet metal covers the pipe’s OUTSIDE curved surface only, to wrap it once. Which of the pipe’s three measurements — $880$ cm², $1188$ cm² or $3080$ cm³ — should they use?
    1. $1188$ cm², the total surface area, since it is the larger and safer number
    2. $3080$ cm³, since it is the pipe’s overall size
    3. $880$ cm², the curved surface area
    4. half of $1188$ cm², since the curved part is roughly half the total
    Check your answer
    1. $1188$ cm², the total surface area, since it is the larger and safer number — The total surface area includes the two flat ends, which a curved wrap around the outside never touches; using it would overestimate the sheet metal needed.
    2. $3080$ cm³, since it is the pipe’s overall size — Volume measures the space inside the pipe, not the sheet metal needed to wrap its outside; the wrap needs a surface area, $880$ cm².
    3. ✓ $880$ cm², the curved surface area — (C) Wrapping only the outside curve needs exactly the curved surface area, $880$ cm².
    4. half of $1188$ cm², since the curved part is roughly half the total — The curved surface area is an exact value, $880$ cm², not an approximation of half the total; halving $1188$ gives $594$, a different and incorrect number.
Worked example

A cylindrical pipe’s surface and volume

  1. radius $r = 7$ cm, height $h = 20$ cm, taking $\pi \approx 22/7$
    These are the pipe’s own numbers, and $\pi \approx 22/7$ holds for the whole problem.
  2. curved surface area $= 2 \cdot \pi r h \approx 2 \cdot 22/7 \cdot 7 \cdot 20 = 880$ cm²
    Put $r$ and $h$ into the cylinder’s curved surface area formula.
  3. total surface area $= 880 + 2 \cdot \pi r^2 \approx 880 + 2 \cdot 22/7 \cdot 49 = 880 + 308 = 1188$ cm²
    Add the two flat circular ends, $2 \cdot \pi r^2$, to the curved surface area from the step before.
  4. volume $= \pi r^2 h \approx 22/7 \cdot 49 \cdot 20 = 3080$ cm³
    The volume formula uses only the base area and the height. It never needs the curved surface area.
  5. $2 \cdot 22/7 \cdot 49 = 308$, and $880 + 308 = 1188$
    Check the total by re-adding the curved surface and the two flat ends on their own. The pieces must add back to the $1188$ cm² found above.
A cylinder’s surface and volume
  1. Find the volume first Multiply the base area by the height, $\pi r^2 h$, using the pipe’s radius $7$ cm and height $20$ cm. That gives $3080$ cm³.
  2. Find the curved surface Multiply $2 \cdot \pi r h$ using the same radius and height. That gives $880$ cm².
  3. Add the two flat ends Add $2 \cdot \pi r^2$, here $308$ cm², to the curved surface for the total surface area, $1188$ cm².
  4. Check the curved surface against the volume The curved surface is always $2/r$ times the volume, since $2 \pi r h = (2/r) \cdot \pi r^2 h$. Here $2 \cdot 3080 / 7 = 880$, which matches.

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Worked: a toy’s total surface

Check yourself
  1. A toy is a cone of radius $7$ cm and height $24$ cm mounted on a hemisphere of the same radius. What is the cone’s slant height?
    1. $25$ cm
    2. $24$ cm, the same as the cone’s height
    3. $31$ cm, from adding the radius and height directly
    4. $625$ cm, forgetting to take the square root
    Check your answer
    1. ✓ $25$ cm — (A) $l=\sqrt{49+576}=\sqrt{625}=25$ cm.
    2. $24$ cm, the same as the cone’s height — The height and slant height are different lengths; the slant height uses $l=\sqrt{49+576}=\sqrt{625}=25$ cm, not the height alone.
    3. $31$ cm, from adding the radius and height directly — Slant height comes from $\sqrt{r^2+h^2}$, not from $7+24=31$; the correct value is $25$ cm.
    4. $625$ cm, forgetting to take the square root — $49+576=625$ is the value under the square root; the slant height itself is $\sqrt{625}=25$ cm, not $625$.
  2. For this toy, why does the total surface area add the cone’s CURVED surface area to the hemisphere’s CURVED surface area, rather than either solid’s total surface area?
    1. because a toy’s surface area is always measured using curved surfaces only
    2. because the cone and hemisphere have no flat faces to begin with
    3. both flat bases are pressed together at the join and vanish
    4. because using total surface area would give a smaller, and therefore wrong, answer
    Check your answer
    1. because a toy’s surface area is always measured using curved surfaces only — There is no such general rule for toys; the reason curved surfaces are used here is specifically that both flat bases are hidden at this particular join.
    2. because the cone and hemisphere have no flat faces to begin with — Both the cone and the hemisphere do have a flat circular face; those two faces are the ones hidden at the join, not faces that never existed.
    3. ✓ both flat bases are pressed together at the join and vanish — (C) Both flat bases are hidden together at the join, leaving only the two curved surfaces exposed.
    4. because using total surface area would give a smaller, and therefore wrong, answer — The reason to use curved surfaces is the hidden join, not whether the resulting number seems large or small; using the totals would actually give a LARGER, over-counted answer.
  3. A similar toy has a cone of radius $3$ cm and height $4$ cm mounted on a hemisphere of the same radius. Taking $\pi=3.14$, what is its total surface area to colour?
    1. $131.88$ cm², using the cone’s total surface area instead of just its curved part
    2. $94.2$ cm³, treating the answer as a volume
    3. $103.62$ cm²
    4. $56.52$ cm², the hemisphere’s curved surface alone, forgetting the cone
    Check your answer
    1. $131.88$ cm², using the cone’s total surface area instead of just its curved part — The cone’s flat base is pressed against the hemisphere and hidden; only its curved surface, $47.1$ cm², should be counted, not its full total of $75.36$ cm².
    2. $94.2$ cm³, treating the answer as a volume — $94.2$ cm³ is the toy’s VOLUME; the question asks for surface area to colour, which is $103.62$ cm², not a volume.
    3. ✓ $103.62$ cm² — (C) Cone curved surface $3.14 \cdot 3 \cdot 5=47.1$ cm² plus hemisphere curved surface $2 \cdot 3.14 \cdot 9=56.52$ cm² gives $103.62$ cm².
    4. $56.52$ cm², the hemisphere’s curved surface alone, forgetting the cone — The cone’s curved surface, $47.1$ cm², must also be counted; $56.52$ cm² alone leaves out more than half of the toy’s surface.
Worked example

A toy’s total surface, cone on hemisphere

  1. cone radius $r = 7$ cm, cone height $h = 24$ cm, hemisphere of the same radius, taking $\pi \approx 22/7$
    These are the toy’s own numbers, and the hemisphere’s radius matches the cone’s so the two join smoothly.
  2. slant height $l = \sqrt{r^2 + h^2} = \sqrt{49 + 576} = \sqrt{625} = 25$ cm
    Pythagoras’ theorem gives the slant height from the cone’s radius and height.
  3. curved surface area of the cone $= \pi r l \approx 22/7 \cdot 7 \cdot 25 = 550$ cm²
    Put $r$ and the slant height $l$ into the cone’s curved surface area formula.
  4. curved surface area of the hemisphere $= 2 \cdot \pi r^2 \approx 2 \cdot 22/7 \cdot 49 = 308$ cm²
    Put the same radius into the hemisphere’s curved surface area formula.
  5. total surface area of the toy $= 550 + 308 = 858$ cm²
    Both flat circular faces vanish at the join, so the total is just the two curved surfaces added.
  6. $7^2 + 24^2 = 49 + 576 = 625 = 25^2$
    Check the slant height by confirming that 7, 24 and 25 form a Pythagorean triple. Squaring $25$ must return the same $625$ we started with.
A cone of radius 7 and height 24 sits on a hemisphere of the same radius, with no circle drawn at their join.

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Worked: a hemisphere on a cube

Check yourself
  1. A cube of edge $10$ cm has a hemisphere of radius $3.5$ cm fixed on one face. What is the cube’s own total surface area, before the hemisphere is added?
    1. $600$ cm²
    2. $1000$ cm³, from $10^3$, its volume
    3. $100$ cm², the area of one face only
    4. $60$ cm², from $6 \cdot 10$
    Check your answer
    1. ✓ $600$ cm² — (A) $6 \cdot 10^2 = 600$ cm².
    2. $1000$ cm³, from $10^3$, its volume — $1000$ cm³ is the cube’s VOLUME; its total surface area is $6a^2=600$ cm².
    3. $100$ cm², the area of one face only — $100$ cm² is the area of ONE face; the cube’s TOTAL surface area counts all six faces, giving $600$ cm².
    4. $60$ cm², from $6 \cdot 10$ — Each face’s area is $10^2=100$ cm², not $10$ cm²; six such faces give $600$ cm², not $60$.
  2. Fixing the hemisphere removes a circle of area $38.5$ cm² from the cube’s top face and adds the hemisphere’s curved surface area of $77$ cm². Why does the combination’s total surface area, $638.5$ cm², come from $600-38.5+77$ rather than just $600+77$?
    1. the hemisphere’s base covers, and removes, that circle of the face
    2. because $600+77$ would give too large a number for a toy this size
    3. because the cube’s formula $6a^2$ already overcounts by $38.5$ cm² on its own
    4. because the hemisphere’s curved surface area formula already includes a subtraction
    Check your answer
    1. ✓ the hemisphere’s base covers, and removes, that circle of the face — (A) The hemisphere’s base physically covers a circle on the cube’s top face; that area is removed before the hemisphere’s own curved surface is added.
    2. because $600+77$ would give too large a number for a toy this size — The reason to subtract $38.5$ is the hidden circle at the join, a geometric fact; it has nothing to do with what number ‘feels’ reasonable.
    3. because the cube’s formula $6a^2$ already overcounts by $38.5$ cm² on its own — $6a^2=600$ cm² is exactly correct for a plain cube; the $38.5$ cm² is subtracted only because the hemisphere now covers that circle on the cube’s top face, not because the formula was ever wrong.
    4. because the hemisphere’s curved surface area formula already includes a subtraction — The hemisphere’s curved surface area, $2 \pi r^2=77$ cm², is a plain, un-adjusted formula; the subtraction of $38.5$ cm² happens separately, on the cube’s side of the calculation.
  3. A cube of edge $8$ cm has a hemisphere of radius $2$ cm fixed on one face. Taking $\pi=3.14$, what is the total surface area of the combination?
    1. $409.12$ cm², forgetting to remove the covered circle
    2. $371.44$ cm², forgetting to add the hemisphere’s curved surface
    3. $396.56$ cm²
    4. $421.68$ cm², adding the covered circle instead of removing it
    Check your answer
    1. $409.12$ cm², forgetting to remove the covered circle — The hemisphere’s base covers a circle of $12.56$ cm² on the cube’s face; that area must be removed before adding the curved surface, giving $396.56$ cm², not $409.12$.
    2. $371.44$ cm², forgetting to add the hemisphere’s curved surface — After removing the covered circle, the hemisphere’s own curved surface, $25.12$ cm², must be added back in; without it, the toy is missing its rounded top entirely.
    3. ✓ $396.56$ cm² — (C) $384 - 12.56 + 25.12 = 396.56$ cm².
    4. $421.68$ cm², adding the covered circle instead of removing it — The circle under the hemisphere is COVERED, not gained; it must be subtracted from the cube’s face, not added on top of the total.
The hemisphere covers 38.5 cm² of the cube’s top face and adds a curved 77 cm², so the surface is 600 − 38.5 + 77 = 638.5 cm².
Worked example

A hemisphere fixed on a cube

  1. cube of edge $a = 10$ cm, hemisphere of radius $r = 3.5$ cm fixed on one face, taking $\pi \approx 22/7$
    These are the block’s own numbers, with the hemisphere sitting entirely on one of the cube’s six faces.
  2. total surface area of the cube $= 6 a^2 = 6 \cdot 100 = 600$ cm²
    Find the cube’s total surface area before the hemisphere is added at all.
  3. area covered by the hemisphere’s base $= \pi r^2 \approx 22/7 \cdot 3.5 \cdot 3.5 = 38.5$ cm²
    The circle where the hemisphere meets the cube is hidden, so its area comes off the cube’s total.
  4. curved surface area of the hemisphere $= 2 \cdot \pi r^2 \approx 77$ cm²
    The hemisphere gives back its own curved surface in exchange for the circle it covers.
  5. total surface area of the combination $= 600 - 38.5 + 77 = 638.5$ cm²
    Remove the covered circle, then add the hemisphere’s curved surface, matching only the one face the join touches.
  6. $2 \cdot 38.5 = 77$
    Check that the hemisphere’s curved surface is exactly double the flat circle it replaces. $2 \cdot \pi r^2$ is always twice $\pi r^2$, for any radius.
A hemisphere fixed on a cube’s face
  1. Find the cube’s total surface Use $6 a^2$ with edge $a = 10$ cm. That gives $600$ cm².
  2. Find the circle the hemisphere covers Use $\pi r^2$ with the hemisphere’s radius $r = 3.5$ cm. That gives $38.5$ cm².
  3. Find the hemisphere’s curved surface Use $2 \cdot \pi r^2$ at the same radius. That gives $77$ cm².
  4. Take the circle away, add the curved surface Subtract $38.5$ cm² from the cube’s total, then add $77$ cm². The combination’s total is $638.5$ cm².
  5. Check the curved surface against the circle The hemisphere’s curved surface is always twice the circle it covers. Here $2 \cdot 38.5 = 77$, which matches.

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Worked: a capsule’s surface area

Check yourself
  1. A capsule is a cylinder of radius $3.5$ cm and height $8$ cm with a hemisphere of the same radius fixed on each flat end. How many flat faces does this capsule have left, once both hemispheres are fixed on?
    1. none — all four flat faces are hidden at the two joins
    2. two — the cylinder’s two flat ends are still exposed underneath the hemispheres
    3. two — each hemisphere’s own flat base is still exposed, facing outward
    4. four — both cylinder ends and both hemisphere bases are all still exposed
    Check your answer
    1. ✓ none — all four flat faces are hidden at the two joins — (A) All four flat faces, two on the cylinder and two on the hemispheres, are hidden at the two joins.
    2. two — the cylinder’s two flat ends are still exposed underneath the hemispheres — Each cylinder end is pressed directly against a hemisphere’s base and is hidden by the join, not merely covered while staying ‘exposed underneath.’
    3. two — each hemisphere’s own flat base is still exposed, facing outward — Each hemisphere’s flat base faces INWARD, toward the cylinder, and is hidden at the join; it does not face outward.
    4. four — both cylinder ends and both hemisphere bases are all still exposed — All four of these flat faces are hidden, two at each join; none of them remain part of the capsule’s outer, visible surface.
  2. Why does the capsule’s total surface area use ONLY curved surfaces — the cylinder’s curved surface plus both hemispheres’ curved surfaces — and no flat-face term at all?
    1. because a capsule, by definition, has no flat faces anywhere in its formula
    2. every flat face is hidden at one of the two joins
    3. because flat faces do not count toward surface area unless a solid stands on a table
    4. because curved surfaces always outnumber flat surfaces in any combination
    Check your answer
    1. because a capsule, by definition, has no flat faces anywhere in its formula — It is not a rule about capsules in general; it is a direct consequence of every flat face in THIS shape being hidden at a join, calculated the same way as any other combination.
    2. ✓ every flat face is hidden at one of the two joins — (B) With every flat face hidden at one join or the other, only the curved surfaces remain exposed.
    3. because flat faces do not count toward surface area unless a solid stands on a table — Whether a solid rests on a table has no bearing on which faces count; a face counts only if it is exposed, not resting flat on a surface.
    4. because curved surfaces always outnumber flat surfaces in any combination — There is no such general pattern; this capsule happens to have zero exposed flat faces because of exactly where its two joins fall, not because curved surfaces always dominate.
  3. A capsule has a cylinder of radius $1.5$ cm and height $6$ cm, with a hemisphere of the same radius at each end. Taking $\pi=3.14$, what is its total surface area?
    1. $84.78$ cm²
    2. $56.52$ cm², the cylinder’s curved surface alone, forgetting both hemispheres
    3. $70.65$ cm², counting only one hemisphere’s curved surface, not both
    4. $98.91$ cm², adding the cylinder’s own flat ends as well as both hemispheres’ curved surfaces
    Check your answer
    1. ✓ $84.78$ cm² — (A) $56.52 + 28.26 = 84.78$ cm², the cylinder’s curved surface plus both hemispheres’ curved surfaces.
    2. $56.52$ cm², the cylinder’s curved surface alone, forgetting both hemispheres — Both rounded ends add curved surface of their own, $28.26$ cm² together; leaving them out gives only $56.52$ cm², less than half the true total.
    3. $70.65$ cm², counting only one hemisphere’s curved surface, not both — The capsule has a rounded hemisphere at EACH end; counting only one leaves out $14.13$ cm² that the second hemisphere contributes.
    4. $98.91$ cm², adding the cylinder’s own flat ends as well as both hemispheres’ curved surfaces — The cylinder’s two flat ends are exactly where the hemispheres are fixed and are hidden by them; keeping them adds $14.13$ cm² that no longer belongs to the exposed surface.
Worked example

A capsule’s total surface area

  1. cylinder of radius $r = 3.5$ cm and height $h = 8$ cm, with a hemisphere of the same radius fixed on each flat end, taking $\pi \approx 22/7$
    These are the capsule’s own numbers, with two joins instead of one.
  2. curved surface area of the cylinder $= 2 \cdot \pi r h \approx 2 \cdot 22/7 \cdot 3.5 \cdot 8 = 176$ cm²
    Put $r$ and $h$ into the cylinder’s curved surface area formula.
  3. curved surface area of one hemisphere $= 2 \cdot \pi r^2 \approx 77$ cm², so both hemispheres give $154$ cm²
    Put the same radius into the hemisphere’s curved surface area formula, then double it for the two hemispheres.
  4. total surface area of the capsule $= 176 + 154 = 330$ cm²
    Both flat ends of the cylinder and both flat faces of the hemispheres vanish at the two joins, so only curved surfaces remain.
  5. $4 \cdot 22/7 \cdot 3.5 \cdot 3.5 = 154$
    Check the two curved surfaces against a full sphere of the same radius. Two hemispheres together always carry a sphere’s own surface area, $4 \cdot \pi r^2$.
A capsule’s cylinder and its two rounded ends meet at joins where no circle is drawn, since none survives.
A workbench in the school craft room. Kabir sits on a stool at the left, painting a large plain wooden toy shaped like a medicine capsule that lies on its side. A smooth straight cylinder runs between two rounded half-ball caps, each exactly as wide as the cylinder and meeting it in a smooth ring with no step or ledge. Bare wood with no paint on it yet, no stand, no other objects on the bench.
  • the cylinder
  • one hemisphere
  • the other hemisphere
  • a join
Kabir paints a wooden capsule: a cylinder with a hemisphere of the same width on each end.

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Worked: a toy’s volume

Check yourself
  1. A toy is a hemisphere of radius $3$ cm topped by a cone of the same radius and height $4$ cm. Taking $\pi=3.14$, what is the cone’s volume?
    1. $113.04$ cm³, from $3.14 \cdot 9 \cdot 4$, forgetting the $1/3$
    2. $37.68$ cm³
    3. $56.52$ cm³, using the hemisphere’s volume formula instead
    4. $47.1$ cm², treating the answer as a surface area
    Check your answer
    1. $113.04$ cm³, from $3.14 \cdot 9 \cdot 4$, forgetting the $1/3$ — A cone’s volume formula includes a factor of $1/3$; leaving it out triples the correct value, giving $113.04$ instead of $37.68$.
    2. ✓ $37.68$ cm³ — (B) $1/3 \cdot 3.14 \cdot 9 \cdot 4 = 37.68$ cm³.
    3. $56.52$ cm³, using the hemisphere’s volume formula instead — $56.52$ cm³ is the HEMISPHERE’s volume, from $2/3 \cdot \pi r^3$; the cone’s volume uses a different formula, $1/3 \pi r^2 h$, giving $37.68$ cm³.
    4. $47.1$ cm², treating the answer as a surface area — The question asks for volume, in cm³; $47.1$ would be a surface-area-style number in cm², not the cone’s volume of $37.68$ cm³.
  2. For this toy, the total volume is found as $37.68+56.52=94.2$ cm³, simply adding the cone’s and hemisphere’s volumes. Why is no term ever subtracted here, unlike some surface area calculations for the very same toy?
    1. joining never removes any enclosed space
    2. because volume calculations never involve subtraction, even for a hollowed-out solid
    3. because this particular cone and hemisphere happen to have the same radius
    4. because a subtraction term is needed, but it happens to equal zero for this toy
    Check your answer
    1. ✓ joining never removes any enclosed space — (A) Joining two solids never removes enclosed space, so their volumes simply add.
    2. because volume calculations never involve subtraction, even for a hollowed-out solid — Volume DOES subtract when one solid is hollowed out of another; it is only for a JOINED combination like this toy that nothing needs to be subtracted.
    3. because this particular cone and hemisphere happen to have the same radius — Volumes add outright at a join regardless of whether the radii match; the shared radius here only makes the join smooth, not the volumes additive.
    4. because a subtraction term is needed, but it happens to equal zero for this toy — There is no subtraction term to begin with for a joined combination; volume simply adds, with nothing lost to compute in the first place.
  3. A similar toy has a hemisphere of radius $2$ cm topped by a cone of radius $2$ cm and height $9$ cm. Taking $\pi=22/7$, what is its total volume, approximately?
    1. $37.71$ cm³, the cone’s volume alone, forgetting the hemisphere
    2. $20.95$ cm³, from the cone’s volume minus the hemisphere’s volume
    3. $113.14$ cm³, the cone’s volume with the $1/3$ factor left out, and the hemisphere forgotten
    4. $54.48$ cm³, from the cone’s volume $37.71$ cm³ plus the hemisphere’s volume $16.76$ cm³
    Check your answer
    1. $37.71$ cm³, the cone’s volume alone, forgetting the hemisphere — The hemisphere adds $16.76$ cm³ of its own volume; leaving it out gives only the cone’s share, not the whole toy’s volume.
    2. $20.95$ cm³, from the cone’s volume minus the hemisphere’s volume — The hemisphere is fixed ON the cone, not hollowed out of it; a joined combination’s volume adds, giving $54.48$ cm³, not a subtraction.
    3. $113.14$ cm³, the cone’s volume with the $1/3$ factor left out, and the hemisphere forgotten — The cone’s volume needs its $1/3$ factor, giving $37.71$ cm³, and the hemisphere’s $16.76$ cm³ must be added too; this option does neither correctly.
    4. ✓ $54.48$ cm³, from the cone’s volume $37.71$ cm³ plus the hemisphere’s volume $16.76$ cm³ — (D) $37.71+16.76=54.48$ cm³.
Worked example

A toy’s volume, hemisphere and cone

  1. hemisphere of radius $r = 3$ cm, topped by a cone of the same radius and height $h = 4$ cm, taking $\pi \approx 3.14$
    These are the toy’s own numbers, and $\pi \approx 3.14$ holds for this problem instead of $22/7$.
  2. volume of the cone $= 1/3 \cdot \pi r^2 h \approx 1/3 \cdot 3.14 \cdot 9 \cdot 4 = 37.68$ cm³
    Put the cone’s own radius and height into the cone’s volume formula.
  3. volume of the hemisphere $= 2/3 \cdot \pi r^3 \approx 2/3 \cdot 3.14 \cdot 27 = 56.52$ cm³
    Put the same radius into the hemisphere’s volume formula.
  4. volume of the toy $= 37.68 + 56.52 = 94.2$ cm³
    The two parts are simply joined, so their volumes add with nothing subtracted.
  5. $3.14 \cdot 9 \cdot 4 = 113.04$, and $113.04 / 3 = 37.68$
    Check the cone against the cylinder sharing its base and height. A cone always holds exactly a third of that cylinder, and $113.04$ divided by $3$ returns the $37.68$ found above.
A cone of radius 3 and height 4 sits on a hemisphere of radius 3, joined with no face subtracted from either volume.
A hemisphere-and-cone toy’s volume
  1. Find the cone’s volume Use $1/3 \cdot \pi r^2 h$ with the cone’s own radius $r = 3$ cm and height $h = 4$ cm, taking $\pi = 3.14$. That gives $37.68$ cm³.
  2. Find the hemisphere’s volume Use $2/3 \cdot \pi r^3$ at the same radius, $r = 3$ cm. That gives $56.52$ cm³.
  3. Add them, nothing taken away The two parts are joined, not hollowed out, so simply add the volumes. $37.68 + 56.52 = 94.2$ cm³.
  4. Check the cone against a cylinder A cone always holds a third of the cylinder sharing its base and height. $3.14 \cdot 9 \cdot 4 = 113.04$, and a third of that is $37.68$.

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Worked: volume after hollowing out

Check yourself
  1. A solid cylinder of radius $3.5$ cm and height $10$ cm has a conical cavity of the same radius and depth $6$ cm hollowed from one end. What is the volume of material removed by the cavity?
    1. $231$ cm³, from $22/7 \cdot 3.5 \cdot 3.5 \cdot 6$, forgetting the $1/3$
    2. $385$ cm³, the same as the cylinder’s own full volume
    3. $0$ cm³, since a cavity is empty space and has no volume of its own
    4. $77$ cm³
    Check your answer
    1. $231$ cm³, from $22/7 \cdot 3.5 \cdot 3.5 \cdot 6$, forgetting the $1/3$ — A conical cavity’s volume needs the $1/3$ factor, just like any cone; leaving it out triples the value, giving $231$ instead of $77$ cm³.
    2. $385$ cm³, the same as the cylinder’s own full volume — $385$ cm³ is the FULL cylinder’s volume, using the full height of $10$ cm; the cavity only reaches a depth of $6$ cm and removes just $77$ cm³.
    3. $0$ cm³, since a cavity is empty space and has no volume of its own — The cavity’s SHAPE still has a definite volume, $77$ cm³, even though it is now empty; that is exactly the amount of material removed.
    4. ✓ $77$ cm³ — (D) $1/3 \cdot 22/7 \cdot 3.5 \cdot 3.5 \cdot 6 = 77$ cm³.
  2. Why is the remaining solid’s volume found as $385-77=308$ cm³, rather than $385+77$?
    1. because $385+77$ would exceed the volume of the original block, which is impossible
    2. because subtraction always applies whenever two different solids appear in one problem
    3. because the cavity’s depth, $6$ cm, is less than the cylinder’s height, $10$ cm
    4. the cavity removes material, so its volume is subtracted
    Check your answer
    1. because $385+77$ would exceed the volume of the original block, which is impossible — Whether a number ‘seems too big’ is not the reason; the actual reason is that the cavity is hollowed OUT, physically removing material, which is why its volume is subtracted, not added.
    2. because subtraction always applies whenever two different solids appear in one problem — Two-solid problems do not always subtract; a solid fixed ON another, like the toys in this chapter, has its volume ADDED. Subtraction applies only when one shape is hollowed out, as here.
    3. because the cavity’s depth, $6$ cm, is less than the cylinder’s height, $10$ cm — The subtraction happens because material is REMOVED to form the cavity, whatever its depth is; even a cavity as deep as the whole cylinder would still be subtracted, not added.
    4. ✓ the cavity removes material, so its volume is subtracted — (D) Hollowing out the cavity removes material, so its volume is subtracted from the cylinder’s.
  3. A solid cylinder of radius $2$ cm and height $9$ cm has a conical cavity of the same radius and depth $3$ cm hollowed from one end. Taking $\pi=3.14$, what is the volume of the remaining solid?
    1. $100.48$ cm³
    2. $125.6$ cm³, adding the cavity instead of subtracting it
    3. $113.04$ cm³, the full cylinder’s volume, ignoring the cavity
    4. $12.56$ cm³, the cavity’s own volume alone
    Check your answer
    1. ✓ $100.48$ cm³ — (A) $113.04 - 12.56 = 100.48$ cm³.
    2. $125.6$ cm³, adding the cavity instead of subtracting it — The cavity removes material from the cylinder; its volume must be subtracted, giving $100.48$ cm³, not added to give $125.6$.
    3. $113.04$ cm³, the full cylinder’s volume, ignoring the cavity — The cavity genuinely removes $12.56$ cm³ of material; the remaining solid’s volume is less than the full cylinder’s, at $100.48$ cm³.
    4. $12.56$ cm³, the cavity’s own volume alone — $12.56$ cm³ is the volume TAKEN OUT of the cylinder; the volume that remains in the solid is $113.04-12.56=100.48$ cm³.
Worked example

Volume after hollowing out a cavity

  1. solid cylinder of radius $r = 3.5$ cm and height $h = 10$ cm, with a conical cavity of the same radius and depth $6$ cm hollowed from one end, taking $\pi \approx 22/7$
    These are the block’s own numbers, with the cavity’s radius matching the cylinder’s own radius.
  2. volume of the cylinder $= \pi r^2 h \approx 22/7 \cdot 3.5 \cdot 3.5 \cdot 10 = 385$ cm³
    Find the full cylinder’s volume before any cavity is taken out.
  3. volume removed by the cavity $= 1/3 \cdot \pi r^2 h \approx 1/3 \cdot 38.5 \cdot 6 = 77$ cm³
    The cavity is a cone, so use the cone’s volume formula with its own depth as the height.
  4. volume of the remaining solid $= 385 - 77 = 308$ cm³
    A hollowed-out cavity takes its own volume away from the whole. It is never added.
  5. $22/7 \cdot 3.5 \cdot 3.5 \cdot 6 = 231$, and $231 / 3 = 77$
    Check the cavity against the cylinder it was cut from. A cone is always a third of a cylinder with the same radius and height, and $231$ divided by $3$ returns the $77$ cm³ removed.
A cone-shaped cavity of depth 6 is hollowed from the top of a cylinder of radius 3.5 and height 10.
The volume left after hollowing a cylinder
  1. Find the whole cylinder’s volume Use $\pi r^2 h$ with radius $r = 3.5$ cm and height $h = 10$ cm. That gives $385$ cm³.
  2. Find the cavity’s volume The cavity is a cone, so use $1/3 \cdot \pi r^2 h$ with the same radius and its own depth, $6$ cm. That gives $77$ cm³.
  3. Subtract the cavity Take the cavity’s volume away from the whole cylinder. $385 - 77 = 308$ cm³.
  4. Check the cavity against the cylinder A cone is always a third of a cylinder sharing its radius and height. $22/7 \cdot 3.5 \cdot 3.5 \cdot 6 = 231$, and a third of that is $77$.

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Recap

■RECAP
Check yourself
  1. This chapter’s whole idea splits into two halves, one for surface area and one for volume. What is the volume half of that idea?
    1. volume counts only the faces still exposed once two solids are joined
    2. volume is always found by multiplying the two solids’ surface areas together
    3. volume adds the two parts, or subtracts one when hollowed out
    4. volume drops the shared face at the join, exactly like surface area does
    Check your answer
    1. volume counts only the faces still exposed once two solids are joined — Counting exposed faces is the SURFACE AREA half of the idea; the volume half is about adding or subtracting the parts’ volumes, not counting faces at all.
    2. volume is always found by multiplying the two solids’ surface areas together — Volume is computed from each solid’s own dimensions using its own formula; it is never found by multiplying two surface areas together.
    3. ✓ volume adds the two parts, or subtracts one when hollowed out — (C) Volume adds the two parts’ volumes outright, or subtracts one when a solid is hollowed out of another.
    4. volume drops the shared face at the join, exactly like surface area does — Volume has no shared face to drop; that rule belongs to surface area alone. Volume simply adds, or subtracts for a hollowed solid.
  2. Why does this chapter organise every combination problem around the SAME one idea — exposed faces for area, plain addition or subtraction for volume — instead of a separate rule for each pair of solids?
    1. because there are too many possible pairs of solids to give each one its own rule
    2. because the six basic solids all have the same formulas for surface area and volume
    3. because CBSE examinations only ever test one specific pair of solids
    4. the same reasoning about hidden faces and enclosed space applies to any pair
    Check your answer
    1. because there are too many possible pairs of solids to give each one its own rule — The single rule is not a shortcut adopted for convenience; it holds because the underlying geometry — a hidden face at any join, enclosed space that is never lost — is genuinely the same for every pair of solids.
    2. because the six basic solids all have the same formulas for surface area and volume — The six basic solids have different formulas for their own surface area and volume; what is shared is the RULE for combining any two of them, not the formulas themselves.
    3. because CBSE examinations only ever test one specific pair of solids — The one-idea structure reflects a genuine mathematical fact about any combination, not a pattern of which solids happen to appear in examinations.
    4. ✓ the same reasoning about hidden faces and enclosed space applies to any pair — (D) The same reasoning about hidden faces and enclosed space holds for any pair of solids, so one rule covers every combination.
  3. A water tank is a cylinder with a hemispherical bottom, and a separate tank is a cuboid with a conical funnel welded underneath for draining. Both are combinations of solids from this chapter. What is the SAME question to ask about surface area for both tanks, whatever solids each one combines?
    1. which tank holds a greater volume of water
    2. which solid in the pair has the larger individual surface area
    3. which face is hidden at the join
    4. how many basic solids are used to build each tank
    Check your answer
    1. which tank holds a greater volume of water — The question asked is about SURFACE AREA, not which tank holds more water; that volume comparison is a different question altogether.
    2. which solid in the pair has the larger individual surface area — Comparing which part is individually larger says nothing about the combination’s own total; the one question that matters for EVERY combination is which face the join hides.
    3. ✓ which face is hidden at the join — (C) For any combination, the one question that determines its surface area is which face the join hides.
    4. how many basic solids are used to build each tank — Both tanks combine exactly two solids; that count is already known and is not the question that determines either tank’s surface area — the hidden join is.
Check yourself: the whole chapter
  1. A toy is made by fixing a hemisphere on top of a cylinder of the same radius. Which statement about finding its volume and total surface area is correct?
    1. both the volume and total surface area are found by adding the two solids’ own totals. No adjustment for the hidden join is needed
    2. the volume is found by adding the two solids’ volumes. The surface area must leave out the faces hidden where the two solids meet
    3. the surface area is found by adding the two solids’ total surface areas. The volume must leave out the space hidden where the two solids meet
    4. neither the volume nor the surface area can be found without measuring the finished solid directly. Joining changes both formulas entirely
    Check your answer
    1. both the volume and total surface area are found by adding the two solids’ own totals. No adjustment for the hidden join is needed — Surface area is not simply added — the face where the hemisphere meets the cylinder is hidden inside the join and must be left out of the total, on both sides.
    2. ✓ the volume is found by adding the two solids’ volumes. The surface area must leave out the faces hidden where the two solids meet — (B) Volume adds outright, since joining wastes no enclosed space; surface area does not, since the join hides a face on each side.
    3. the surface area is found by adding the two solids’ total surface areas. The volume must leave out the space hidden where the two solids meet — It is volume that adds outright, since joining wastes no space; surface area is the one that needs an adjustment for the hidden faces. This option has the two swapped.
    4. neither the volume nor the surface area can be found without measuring the finished solid directly. Joining changes both formulas entirely — Both solids’ own volume and surface area formulas still work — the whole method is built on reusing them, with one adjustment for the faces the join hides.
  2. A cone is mounted on top of a cylinder of the same radius. When finding the total surface area of this combined solid, the circular face where the cone’s base meets the cylinder’s top should be
    1. counted twice, once for the cylinder and once for the cone, since both solids have that face
    2. counted only once, since it appears in both solids’ own surface area totals, treated as if it still belonged to one of the two shapes
    3. left out of the total entirely, since it is hidden inside the join and no longer visible from outside
    4. replaced by the average of the cylinder’s and cone’s own surface areas at that face
    Check your answer
    1. counted twice, once for the cylinder and once for the cone, since both solids have that face — A face hidden inside a join is not part of the solid’s outside at all — counting it even once, let alone twice, overcounts the true surface.
    2. counted only once, since it appears in both solids’ own surface area totals, treated as if it still belonged to one of the two shapes — The joined face is not visible from outside the finished solid, so it does not belong in the total even once.
    3. ✓ left out of the total entirely, since it is hidden inside the join and no longer visible from outside — (C) The joined face sits inside the combination, not on its outside, so it drops out of the total surface area entirely.
    4. replaced by the average of the cylinder’s and cone’s own surface areas at that face — There is no averaging step for a hidden face — it simply drops out of the total, on both sides of the join.
  3. A hemisphere is fixed onto one flat end of a cylinder, both with the same radius. How many faces does this join hide?
    1. one face — only the cylinder’s flat end disappears
    2. one face — only the hemisphere’s flat base disappears
    3. no faces — both flat circles are still part of the total surface area, since they belong to two different solids
    4. two faces — the cylinder’s flat end and the hemisphere’s flat base both disappear
    Check your answer
    1. one face — only the cylinder’s flat end disappears — The hemisphere’s own flat base is hidden too — it used to sit in the very same place as the cylinder’s flat end, so both disappear together.
    2. one face — only the hemisphere’s flat base disappears — The cylinder’s flat end is hidden too, for the same reason — the join removes a face from each side, not just one.
    3. no faces — both flat circles are still part of the total surface area, since they belong to two different solids — The two flat circles used to occupy the same spot before the join — neither one can still be exposed once the solids are fixed together.
    4. ✓ two faces — the cylinder’s flat end and the hemisphere’s flat base both disappear — (D) A join hides a face on both sides of it — the cylinder’s flat end and the hemisphere’s flat base both disappear, since the two used to sit in the very same place.
  4. A cylindrical block has a conical cavity hollowed out from one end. How is the volume of the remaining solid found?
    1. add the cylinder’s volume and the cone’s volume, the same as for two solids joined together
    2. subtract the cone’s volume from the cylinder’s volume, since the cavity removes that space
    3. subtract half the cone’s volume from the cylinder’s volume, since only part of the cavity is hollow
    4. find the average of the cylinder’s and the cone’s volumes
    Check your answer
    1. add the cylinder’s volume and the cone’s volume, the same as for two solids joined together — A cavity is hollowed out, not fixed on — it removes volume from the cylinder instead of adding a new solid’s volume to it.
    2. ✓ subtract the cone’s volume from the cylinder’s volume, since the cavity removes that space — (B) Hollowing out a solid removes volume instead of adding it, so the remaining volume is the cylinder’s volume minus the cone’s.
    3. subtract half the cone’s volume from the cylinder’s volume, since only part of the cavity is hollow — The entire cavity is hollow, not just half of it — the whole cone-shaped volume is removed, not a fraction of it.
    4. find the average of the cylinder’s and the cone’s volumes — Volume is never averaged between two solids — it is added when one is fixed onto another, and subtracted when one is hollowed out of another.
  5. A cone of base radius $3$ cm is to be mounted on a hemisphere to make one smooth combined solid. What must the hemisphere’s radius be?
    1. any radius larger than $3$ cm, so the hemisphere’s base fully covers the cone’s base
    2. any radius smaller than $3$ cm, so the cone’s base fully covers the hemisphere’s base
    3. exactly $3$ cm, so the two faces meet without a gap or a step
    4. it does not matter, since the two solids join at a single point regardless of radius
    Check your answer
    1. any radius larger than $3$ cm, so the hemisphere’s base fully covers the cone’s base — A hemisphere wider than the cone’s base leaves its own edge sticking out past the cone — a visible step, not a smooth join.
    2. any radius smaller than $3$ cm, so the cone’s base fully covers the hemisphere’s base — A hemisphere narrower than the cone’s base leaves the cone’s edge sticking out past the hemisphere — a step in the other direction.
    3. ✓ exactly $3$ cm, so the two faces meet without a gap or a step — (C) Two solids join smoothly only when they share the same radius at the joined face — here, exactly $3$ cm on both sides.
    4. it does not matter, since the two solids join at a single point regardless of radius — Two solids joined at mismatched radii leave a step or a gap right where they meet — the radius at the join matters a great deal.

We can see every combination as one idea, split into two halves.

Surface area counts only the faces still exposed once two solids are joined. The shared face disappears from both sides of the join, not just one.

Volume never loses anything to a hidden face, whether it adds two parts together or subtracts one from the other after hollowing out.

Name the two basic solids and their one shared face first, and the formulas you need are already familiar.

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Where you will meet this

A tank of radius 70 cm and height 100 cm holds 22/7 × 70² × 100 = 1540000 cm³, which is 1540 litres.

You meet cylinders, cones and their combinations in ordinary objects all the time. Here are eight places their surface area or volume decides a real number.

A cylinder with radius 7 cm and height 10 cm has a curved wall of 440, ends of 308, total metal 748 square centimetres.
Pythagoras comes first: turn the radius and the height into the slant height, and only then does the curved-surface formula apply.
Only what paint can reach counts, so the join between two solids hides a face on each side of it.

Your turn. A wooden dowel is a cylinder of radius $2$ cm and length $25$ cm. It has a cone-shaped tip of the same radius and height $3$ cm glued on one end. Using $\pi = 3.14$, what is its total volume? Answer: The cylinder’s volume is $3.14 \cdot 2^2 \cdot 25 = 314$ cm³. The cone’s volume is $(1/3) \cdot 3.14 \cdot 2^2 \cdot 3 = 12.56$ cm³. The total is $314 + 12.56 = 326.56$ cm³.

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Practice set: Exercise 12.1

Exercise 12.1
  1. practice A cylinder of radius $3.5$ cm and height $10$ cm stands on its flat base, with a hemisphere of the same radius fixed on its top end. Find the total surface area, taking $\pi \approx 22/7$. (Worked in full below — read it, then do the next two the same way.)
  2. practice The same cylinder, radius $3.5$ cm and height $10$ cm on its flat base, but with a cone of the same radius and slant height $12$ cm fixed on top instead of the hemisphere. Find the total surface area, taking $\pi \approx 22/7$. (The same three surfaces are exposed — only the formula for the top one changes, to $\pi r l$.)
  3. practice A cube of edge $14$ cm has a hemisphere of radius $3.5$ cm fixed on its top face. Find the total surface area, taking $\pi \approx 22/7$. (Which face loses a circle here, and what does it gain in exchange?)
  4. practice A cylindrical tin of radius $7$ cm and height $10$ cm is topped by a hemisphere of the same radius. Find its total surface area, taking $\pi \approx 22/7$.
  5. practice A cone of radius $3.5$ cm and slant height $10$ cm is fixed onto a hemisphere of the same radius, forming a toy. Find its total surface area, taking $\pi \approx 22/7$.
  6. practice A cube of edge $7$ cm has a hemispherical depression of radius $3.5$ cm carved into its top face. Find the total surface area of the remaining solid, taking $\pi \approx 22/7$.
  7. practice A tent is shaped like a cylinder of radius $2.8$ m and height $2.1$ m, topped by a cone of the same radius and slant height $2.8$ m. Find the area of canvas needed, taking $\pi \approx 22/7$.
  8. practice A wooden toy is a cone of height $4$ cm and radius $3$ cm fixed on a hemisphere of the same radius. Find the total surface area, taking $\pi = 3.14$.
  9. practice Two identical cubes of edge $5$ cm are joined face to face to form a cuboid. Find the total surface area of the resulting cuboid.
Answers
  1. Cylinder $220$ cm², hemisphere $77$ cm², base $38.5$ cm²; total $335.5$ cm².
  2. Cone $= \pi r l = (22/7) \cdot 3.5 \cdot 12 = 132$ cm², so $220 + 132 + 38.5 = 390.5$ cm².
  3. $6 a^2 = 6 \cdot 196 = 1176$ cm², then $1176 - 38.5 + 77 = 1214.5$ cm².
  4. Curved surface area of the cylinder $= 2 \cdot \pi r h = 440$ cm². Curved surface area of the hemisphere $= 2 \cdot \pi r^2 = 308$ cm². The cylinder’s own flat bottom stays exposed: $\pi r^2 = 154$ cm². Total $= 440 + 308 + 154 = 902$ cm².
  5. Curved surface area of the cone $= \pi r l = 110$ cm². Curved surface area of the hemisphere $= 2 \cdot \pi r^2 = 77$ cm². Both flat faces vanish at the join, so the total is $110 + 77 = 187$ cm².
  6. Total surface area of the cube $= 6 a^2 = 294$ cm². The depression removes a flat circle of area $\pi r^2 = 38.5$ cm² and adds a curved bowl surface of $2 \cdot \pi r^2 = 77$ cm². Total $= 294 - 38.5 + 77 = 332.5$ cm².
  7. Curved surface area of the cylinder $= 2 \cdot \pi r h = 36.96$ m². Curved surface area of the cone $= \pi r l = 24.64$ m². No floor and no hidden flat circle to account for in canvas. Total canvas $= 36.96 + 24.64 = 61.6$ m².
  8. Slant height $l = \sqrt{3^2 + 4^2} = 5$ cm. Curved surface area of the cone $= \pi r l = 47.1$ cm². Curved surface area of the hemisphere $= 2 \cdot \pi r^2 = 56.52$ cm². Total $= 47.1 + 56.52 = 103.62$ cm².
  9. The joined cuboid measures $10$ cm by $5$ cm by $5$ cm. Total surface area $= 2(l b + b h + h l) = 2(50 + 25 + 50) = 250$ cm² — equivalently, the two cubes’ combined $300$ cm² minus the two $25$ cm² faces hidden at the join.
Exercise 12.1 — further practice
  1. practice A test tube is a hollow cylinder of radius $3.5$ cm and height $10$ cm, open at the top and closed at the bottom by a hemisphere of the same radius. Taking the glass as having no thickness, find its total surface area, taking $\pi \approx 22/7$.
  2. practice A cube of edge $10$ cm has a smaller cube of edge $4$ cm placed centrally on its top face, entirely within it. Find the total surface area of the combination.
  3. practice A cuboid of length $10$ cm, breadth $8$ cm and height $6$ cm has a hemispherical depression of radius $3$ cm carved into its top face. Find the total surface area of the remaining solid, taking $\pi = 3.14$.
  4. practice An overhead water tank is a cylinder of radius $7$ m and height $10$ m, topped by a conical roof of the same radius and height $24$ m, sealed with a flat base where it rests on its supports. Find the total surface area of sheet metal needed to build it, taking $\pi \approx 22/7$.
  5. practice A cylindrical block of radius $3$ cm and height $10$ cm has a conical depression of the same radius and depth $4$ cm hollowed into one flat end. Find the total surface area of the remaining solid, taking $\pi = 3.14$.
  6. practice A hemisphere of radius $r$ is fixed onto one flat end of a cylinder of the same radius. Rohan adds the cylinder’s own total surface area to the hemisphere’s own total surface area to get the combination’s total surface area. By how much has he over-counted?
    1. $\pi r^2$
    2. $2 \pi r^2$
    3. $3 \pi r^2$
    4. $4 \pi r^2$
  7. practice A cone is fixed onto one flat end of a cylinder of the same radius, forming a single solid. Which surfaces are left out of the combination’s total surface area?
    1. The cylinder’s curved surface
    2. The cone’s curved surface
    3. The cylinder’s top flat circle and the cone’s flat base
    4. The cylinder’s bottom flat circle
  8. practice A toy is a hemisphere of radius $5$ cm topped by a cone of the same radius. The total height of the toy is $17$ cm. Find its total surface area, taking $\pi = 3.14$.
  9. practice A wooden box is a cuboid measuring $20$ cm by $15$ cm by $10$ cm, standing on its $20$ cm by $15$ cm face on a table. A cubical block of edge $5$ cm is glued centrally on the box’s top face. The box is to be painted, except for the face resting on the table. Find the area to be painted.
  10. practice A decorative bead is made by joining two cones of the same radius $8$ cm base to base — one of height $6$ cm and the other of height $15$ cm. Find its total surface area, taking $\pi = 3.14$.
Answers
  1. $297$ cm².
  2. $664$ cm².
  3. $404.26$ cm².
  4. $1144$ m².
  5. $263.76$ cm².
  6. B — $2 \pi r^2$.
  7. C — The cylinder’s top flat circle and the cone’s flat base.
  8. $361.1$ cm².
  9. $1100$ cm².
  10. $678.24$ cm².

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Practice set: Exercise 12.2

Exercise 12.2
  1. practice A cylinder of radius $7$ cm and height $6$ cm is topped by a cone of the same radius and height $9$ cm. Find the volume of the solid, taking $\pi \approx 22/7$. (Worked in full below — read it, then do the next two the same way.)
  2. practice A solid cylinder of radius $7$ cm and height $9$ cm has a conical cavity of the same radius and depth $9$ cm hollowed out from one end. Find the volume of the remaining solid, taking $\pi \approx 22/7$. (The same cone as above, $462$ cm³ — this time taken away instead of added.)
  3. practice A toy is a hemisphere of radius $3$ cm with a cone of the same radius and height $10$ cm fixed on its flat face. Find the volume of the toy, taking $\pi = 3.14$. (A hemisphere needs no height of its own — its volume is $(2/3) \pi r^3$.)
  4. practice A solid is a cylinder of radius $7$ cm and height $10$ cm, topped by a cone of the same radius and height $6$ cm. Find its volume, taking $\pi \approx 22/7$.
  5. practice A toy is a hemisphere of radius $3$ cm topped by a cone of the same radius and height $8$ cm. Find its volume, taking $\pi = 3.14$.
  6. practice A solid cylinder of radius $6$ cm and height $14$ cm has a conical cavity of the same radius and depth $9$ cm hollowed from one end. Find the volume of the remaining solid, taking $\pi = 3.14$.
  7. practice A gulab jamun is shaped like a cylinder of radius $1.5$ cm with a hemisphere of the same radius at each end, giving a total length of $6$ cm. Find its volume, taking $\pi = 3.14$.
  8. practice A wooden pen stand is a cuboid of length $8$ cm, breadth $6$ cm and height $3$ cm, with two conical holes of radius $0.7$ cm and depth $1.5$ cm drilled into it. Find the volume of wood left in the stand, taking $\pi \approx 22/7$.
  9. practice A solid is formed by scooping a hemispherical cavity of radius $3$ cm out of each end of a solid cylinder of radius $3$ cm and height $10$ cm. Find the volume of the remaining solid, taking $\pi = 3.14$.
Answers
  1. Cylinder $924$ cm³, cone $462$ cm³; total $1386$ cm³.
  2. $154 \cdot 9 = 1386$ cm³ for the cylinder, less $462$ cm³ for the cavity; $924$ cm³ remains.
  3. Hemisphere $= (2/3) \cdot 3.14 \cdot 27 = 56.52$ cm³, cone $= (1/3) \cdot 3.14 \cdot 9 \cdot 10 = 94.2$ cm³; total $150.72$ cm³.
  4. Volume of the cylinder $= \pi r^2 h = 1540$ cm³. Volume of the cone $= 1/3 \cdot \pi r^2 h = 308$ cm³. Both parts are joined, so they add: $1540 + 308 = 1848$ cm³.
  5. Volume of the hemisphere $= 2/3 \cdot \pi r^3 = 56.52$ cm³. Volume of the cone $= 1/3 \cdot \pi r^2 h = 75.36$ cm³. Total $= 56.52 + 75.36 = 131.88$ cm³.
  6. Volume of the cylinder $= \pi r^2 h = 1582.56$ cm³. Volume removed by the cavity $= 1/3 \cdot \pi r^2 h = 339.12$ cm³. Remaining $= 1582.56 - 339.12 = 1243.44$ cm³.
  7. The cylinder’s own length is $6 - 2 \cdot 1.5 = 3$ cm. Volume of the cylinder $= \pi r^2 h = 21.195$ cm³. Volume of each hemisphere $= 2/3 \cdot \pi r^3 = 7.065$ cm³, so both give $14.13$ cm³. Total $= 21.195 + 14.13 = 35.325$ cm³.
  8. Volume of the cuboid $= l b h = 144$ cm³. Volume of one conical hole $= 1/3 \cdot \pi r^2 h = 0.77$ cm³, so two holes remove $1.54$ cm³. Wood left $= 144 - 1.54 = 142.46$ cm³.
  9. Volume of the cylinder $= \pi r^2 h = 282.6$ cm³. Volume of each hemispherical cavity $= 2/3 \cdot \pi r^3 = 56.52$ cm³, so both remove $113.04$ cm³. Remaining $= 282.6 - 113.04 = 169.56$ cm³.
Exercise 12.2 — further practice
  1. practice A wooden cube of edge $8$ cm has a hemispherical hole of radius $3$ cm scooped out of its top face, to hold a candle. Find the volume of wood left in the cube, taking $\pi = 3.14$.
  2. practice A municipal water tank is a cylinder of radius $9$ m and height $15$ m, topped by a hemispherical dome of the same radius. Find its volume, taking $\pi = 3.14$.
  3. practice A solid is formed by fixing a cone onto a cylinder of the same radius. Which statement about its volume is always true?
    1. It equals the sum of the two solids’ volumes.
    2. It is less than the sum of the two solids’ volumes, because of the hidden face.
    3. It is greater than the sum of the two solids’ volumes, because of the hidden face.
    4. It equals the volume of the larger solid alone.
  4. practice A cubical block of wood has edge $10$ cm. A conical hole of radius $3$ cm and depth $4$ cm is drilled into one face. Find the volume of wood left, taking $\pi = 3.14$.
  5. practice A solid metal ball of radius $6$ cm fits exactly inside a cylindrical box of the same radius and height $12$ cm. Find the volume of empty space inside the box around the ball, taking $\pi = 3.14$.
  6. practice A paperweight is a cuboid of length $20$ cm, breadth $15$ cm and height $8$ cm, with a hemisphere of radius $6$ cm fixed on its top face. Find the total volume of the paperweight, taking $\pi = 3.14$.
  7. practice A solid metal rod is made of two cylinders joined end to end: a base cylinder of radius $7$ cm and height $40$ cm, topped by a narrower cylinder of radius $3.5$ cm and height $20$ cm. If $1$ cm³ of the metal has a mass of $8$ g, find the mass of the rod in kilograms, taking $\pi \approx 22/7$.
  8. practice A rectangular stone block measures $2$ m by $1.5$ m by $1$ m. A cylindrical hole of radius $14$ cm is drilled all the way through it, along its height of $1$ m. Find the volume of stone left, in cubic metres, taking $\pi \approx 22/7$. Convert the radius to metres first.
  9. practice A wooden bead is made by joining two cones of the same radius $5$ cm base to base — one of height $6$ cm and the other of height $9$ cm. Find its volume, taking $\pi = 3.14$.
  10. practice A solid hemisphere of radius $6$ cm has a conical cavity of the same radius and depth $6$ cm scooped out from its flat face. Find the volume of the remaining solid, taking $\pi = 3.14$.
Answers
  1. $455.48$ cm³.
  2. $5341.14$ m³.
  3. A — It equals the sum of the two solids’ volumes.
  4. $962.32$ cm³.
  5. $452.16$ cm³.
  6. $2852.16$ cm³.
  7. $55.44$ kg.
  8. $2.9384$ m³.
  9. $392.5$ cm³.
  10. $226.08$ cm³.

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