What is the total surface area of a combination of two solids?

It is NOT simply the sum of each solid's own surface area. Wherever two solids are joined, the flat circle where they meet is buried inside the combined shape and stops being part of the outside, so it must be subtracted from BOTH sides -- unless one solid is a cube or cuboid, in which case only the covered patch on its flat face is subtracted (the hemisphere or cone's own base is glued away entirely, not subtracted twice). The safest method is always to ask, face by face, which surfaces a paintbrush could actually reach.

Does the volume of a combined solid also lose the glued surface?

No -- this is the single most important difference between the two chapters. Volume is simply ADDITIVE: the volume of a combination of solids is always the sum of the volumes of its parts, whatever shape the join is. Nothing about volume depends on which faces happen to be glued together, unlike surface area.

How do you find the surface area of a solid with a hemispherical depression or a conical cavity cut into it?

A depression or cavity works the OPPOSITE way to a bump: the flat circular area where it is cut is REMOVED from the host solid's surface (that patch is now a hole, not a flat face), and the curved wall of the cavity, now facing inward, is ADDED. If the cavity's own mouth is exactly as wide as the face it is cut into (as in a cylinder bored end to end), the whole flat face disappears and only the cavity's curved wall survives there.

How many exercises does Class 10 Maths Chapter 12 Surface Areas and Volumes have?

Two. In the rationalised 2026-27 book, Section 12.2 (Surface Area of a Combination of Solids) carries Examples 1-4 and then Exercise 12.1 with 9 questions; Section 12.3 (Volume of a Combination of Solids) carries Examples 5-7 and then Exercise 12.2 with 8 questions. Exercise 12.2 Q8 is followed directly by Section 12.4, the Summary, with nothing further.

What solids does Class 10 Chapter 12 combine?

Every combination in this chapter is built from the same five basic solids studied since Class 9 -- the cuboid, the cone, the cylinder, the sphere and the hemisphere -- glued, stacked or bored into one another two at a time. No new solid and no new formula is introduced; the whole chapter is about correctly applying the surface-area and volume formulas you already know to a shape made of more than one piece.

Class 10 Maths Ch 12 Surface Areas and Volumes Solutions

7 Examples + 17 Questions Solved

Chapter 12: Surface Areas and Volumes — NCERT Solutions (rationalised syllabus 2026–27), with all 7 worked Examples from the chapter plus every question of Exercise 12.1 and Exercise 12.2. Every shape in this chapter is built from solids you already know — the cuboid, the cone, the cylinder, the sphere and the hemisphere — glued, stacked or bored into one another. Nothing new is being introduced: the whole chapter is about applying formulas you already have correctly to a solid made of more than one piece.

Fig. 12.1 — the four basic solids: cuboid, cone, cylinder, sphere.

The one rule worth carrying through every question in this chapter: surface area and volume behave DIFFERENTLY when two solids are combined. Volume is simply additive — whatever the join looks like, the combined solid's volume is always the sum of its parts' volumes, with nothing lost. Surface area is not: wherever two solids meet, the flat circle glued between them is buried inside the shape and is no longer part of the outside, so it has to be subtracted from BOTH pieces before their curved surfaces are added. A cube or cuboid with a hemisphere or cone stuck on top is the one common exception worth remembering separately — only the small circular patch covered on the flat face is subtracted (the round solid's own flat base is glued away completely, contributing nothing extra), which is why Example 2's answer comes out as '150 + πr²' rather than a plain difference of two areas.

Section 12.2 (Examples 1-4, Exercise 12.1) is entirely about SURFACE AREA; Section 12.3 (Examples 5-7, Exercise 12.2) is entirely about VOLUME. A depression or a cavity cut INTO a solid works the opposite way to a solid stuck ON one — the flat area is removed rather than kept, and the cavity's own curved wall, now facing inward, is added instead. Read every question carefully for which of these three situations — a bump added, a cavity cut, or two solids of different radii simply stacked, leaving an exposed ring (as in Example 3's rocket) — it actually describes before reaching for a formula.

Worked Examples — Surface Area of a Combination of Solids (Examples 1–4)

Example 1

Rasheed got a playing top (lattu) as his birthday present, which surprisingly had no colour on it. He wanted to colour it with his crayons. The top is shaped like a cone surmounted by a hemisphere. The entire top is 5 cm in height and the diameter of the top is 3.5 cm. Find the area he has to colour. (Take π = \(\tfrac{22}{7}\))

Fig. 12.6 — the top (lattu): a cone surmounted by a hemisphere sharing the same flat circle, total height 5 cm, diameter 3.5 cm.

Solution

Step 1 — what is given. The diameter of the top is 3.5 cm, so the radius (shared by both the cone and the hemisphere, since they are joined smoothly) is \(r=\tfrac{3.5}{2}= \tfrac{7}{4}\) cm. The TOTAL height is 5 cm.

Step 2 — which surfaces are actually painted. The cone and the hemisphere share one flat circle, and that circle is glued INSIDE the solid — a crayon can never reach it. So the area to colour is only the two curved surfaces:

\(\text{Area to colour} = \text{CSA of hemisphere} + \text{CSA of cone}\)

Step 3 — find the cone’s own height. The hemisphere sits on top and its highest point is only its OWN RADIUS above the shared circle (not its diameter) — so the cone's height is the total height minus the hemisphere's radius, not minus 2r:

\(h = \text{total height} - r = 5 - \frac{7}{4}\)

\(h = \frac{13}{4} = 3.25\ \text{cm}\)

Step 4 — find the slant height of the cone.

\(l = \sqrt{r^{2}+h^{2}} = \sqrt{\left(\frac{7}{4}\right)^{2}+(3.25)^{2}}\ \text{cm}\)

\(= \sqrt{3.0625+10.5625}\ \text{cm} = \sqrt{13.625}\ \text{cm}\)

\(l \approx 3.7\ \text{cm (approx.)}\)

Step 5 — the curved surface area of the hemisphere.

\(\text{CSA of hemisphere} = 2\pi r^{2}\)

Step 6 — the curved surface area of the cone.

\(\text{CSA of cone} = \pi r l\)

Step 7 — add the two, factoring out the common \(\pi r\).

\(\text{Area} = 2\pi r^{2} + \pi r l = \pi r\,(2r+l)\)

\(= \frac{22}{7}\times \frac{7}{4}\times \left(3.5+3.7\right)\ \text{cm}^{2}\)

\(= \frac{22}{4}\times 7.2\ \text{cm}^{2} = 39.6\ \text{cm}^{2}\ \text{(approx.)}\)

So Rasheed has to colour about 39.6 cm².

Example 2

The decorative block shown in the figure is made of two solids — a cube and a hemisphere. The base of the block is a cube with edge 5 cm, and the hemisphere fixed on the top has a diameter of 4.2 cm. Find the total surface area of the block. (Take π = \(\tfrac{22}{7}\))

Fig. 12.7 — the decorative block: a cube of edge 5 cm with a hemisphere of diameter 4.2 cm fixed on top.

Solution

Step 1 — the trap in this question. The total surface area is not the surface area of the cube plus the surface area of the hemisphere. Where the hemisphere sits, a small circular patch of the cube's own top face is covered up and no longer belongs to the outside — it has to be SUBTRACTED — while the hemisphere's own curved dome is new surface, to be ADDED.

\(\text{Surface area of block} = \text{TSA of cube}\)

\(\phantom{\text{Surface area of block}} {}- \text{base area of hemisphere}\)

\(\phantom{\text{Surface area of block}} {}+ \text{CSA of hemisphere}\)

Step 2 — the cube's total surface area.

\(\text{TSA of cube} = 6\times(\text{edge})^{2} = 6\times 5\times 5 = 150\ \text{cm}^{2}\)

Step 3 — the hemisphere's radius.

\(r = \frac{4.2}{2} = 2.1\ \text{cm}\)

Step 4 — the two circle-radius terms. The circular patch removed from the cube and the flat base of the hemisphere are the SAME circle, of area \(\pi r^{2}\); the hemisphere's curved area is \(2\pi r^{2}\). So

\(-\pi r^{2} + 2\pi r^{2} = \pi r^{2}\)

\(\text{Surface area of block} = 150 + \pi r^{2}\ \text{cm}^{2}\)

Step 5 — substitute and simplify.

\(= 150 + \frac{22}{7}\times 2.1\times 2.1\ \text{cm}^{2}\)

\(= (150 + 13.86)\ \text{cm}^{2} = 163.86\ \text{cm}^{2}\)

So the total surface area of the block is 163.86 cm².

Example 3

A wooden toy rocket is in the shape of a cone mounted on a cylinder, as shown in the figure. The height of the entire rocket is 26 cm, while the height of the conical part is 6 cm. The base of the conical portion has a diameter of 5 cm, while the base diameter of the cylindrical portion is 3 cm. If the conical portion is to be painted orange and the cylindrical portion yellow, find the area of the rocket painted with each of these colours. (Take π = 3.14)

Fig. 12.8 — the rocket: a cone of base radius 2.5 cm mounted on a cylinder of radius 1.5 cm; the exposed ring of the cone’s own base is painted along with its slant surface.

Solution

Step 1 — what is given. Let the cone have radius \(r\), slant height \(l\) and height \(h\), and the cylinder radius \(r'\) and height \(h'\). Then

\(r = 2.5\ \text{cm},\ h = 6\ \text{cm}\)

\(r' = 1.5\ \text{cm}\)

\(h' = 26-6 = 20\ \text{cm}\)

Step 2 — the slant height of the cone.

\(l = \sqrt{r^{2}+h^{2}} = \sqrt{(2.5)^{2}+6^{2}}\ \text{cm}\)

\(= \sqrt{6.25+36}\ \text{cm} = \sqrt{42.25}\ \text{cm} = 6.5\ \text{cm}\)

Step 3 — notice the trap. The cone's base radius (2.5 cm) is BIGGER than the cylinder's radius (1.5 cm), so when the cone sits on the cylinder, a RING of the cone's own base is left uncovered and exposed — and that ring gets painted orange too, along with the cone's slant surface.

\(\text{Ring exposed} = \pi r^{2} - \pi (r')^{2}\)

\(= \pi\left[(2.5)^{2}-(1.5)^{2}\right] = \pi(6.25-2.25) = 4\pi\ \text{cm}^{2}\)

Step 4 — the orange area.

\(\text{Orange area} = \text{CSA of cone} + \text{ring exposed}\)

\(= \pi r l + \pi\left[r^{2}-(r')^{2}\right]\)

\(= 3.14\times\left[(2.5\times 6.5) + (6.25-2.25)\right]\ \text{cm}^{2}\)

\(= 3.14\times 20.25\ \text{cm}^{2} = 63.585\ \text{cm}^{2}\)

Step 5 — the yellow area. The cylinder's curved wall is exposed, and so is its OWN bottom base (the rocket stands on it) — but its TOP is entirely covered by the cone, so that face contributes nothing.

\(\text{Yellow area} = \text{CSA of cylinder} + \text{one base of cylinder}\)

\(= 2\pi r' h' + \pi (r')^{2} = \pi r'\,(2h'+r')\)

\(= 3.14\times 1.5\times(2\times 20+1.5)\ \text{cm}^{2}\)

\(= 4.71\times 41.5\ \text{cm}^{2} = 195.465\ \text{cm}^{2}\)

So the orange area is 63.585 cm² and the yellow area is 195.465 cm².

Example 4

Mayank made a bird-bath for his garden in the shape of a cylinder with a hemispherical depression at one end. The height of the cylinder is 1.45 m and its radius is 30 cm. Find the total surface area of the bird-bath. (Take π = \(\tfrac{22}{7}\))

Fig. 12.9 — the bird-bath: a cylinder of radius 30 cm and height 1.45 m with a hemispherical depression sunk into its open top.

Solution

Step 1 — this trap runs the OPPOSITE way to Example 2. Here the hemisphere is a DEPRESSION, not a bump — a hollow scooped into the cylinder's own flat top, not something sitting on top of it. The depression's radius equals the cylinder's own radius exactly, so the ENTIRE flat top face is replaced by the hemisphere's curved inner wall — nothing of the flat top survives at all.

\(\text{TSA of bird-bath} = \text{CSA of cylinder}\)

\(\phantom{\text{TSA of bird-bath}} {}+ \text{CSA of hemisphere}\)

Step 2 — what is given. \(r = 30\) cm (the common radius) and \(h = 1.45\) m \(= 145\) cm.

Step 3 — substitute and simplify.

\(= 2\pi r h + 2\pi r^{2} = 2\pi r\,(h+r)\)

\(= 2\times \frac{22}{7}\times 30\times (145+30)\ \text{cm}^{2}\)

\(= 2\times \frac{22}{7}\times 30\times 175\ \text{cm}^{2} = 33000\ \text{cm}^{2} = 3.3\ \text{m}^{2}\)

So the total surface area of the bird-bath is 33000 cm² = 3.3 m².

Exercise 12.1

2 cubes each of volume 64 cm³ are joined end to end. Find the surface area of the resulting cuboid.

2 cubes of volume 64 cm³ each (side 4 cm) joined end to end; the shared face between them vanishes from the surface.

Answer

Step 1 — find the side of each cube.

\(\text{side}^{3} = 64\ \text{cm}^{3}\)

\(\text{side} = \sqrt[3]{64} = 4\ \text{cm}\)

Step 2 — the trap. Joining the two cubes face to face makes ONE cuboid of dimensions \(8\times 4\times 4\) cm — and the two square faces glued together are no longer part of the outside at all. The answer is not the sum of the two cubes' own surface areas (that would double-count nothing lost and give \(2\times 96=192\) cm², which is wrong).

Step 3 — the cuboid's dimensions.

\(l = 8\ \text{cm},\ b = 4\ \text{cm},\ h = 4\ \text{cm}\)

Step 4 — apply the cuboid surface-area formula directly. This formula already counts each of the 6 faces of the JOINED shape exactly once, so the two internal faces are automatically excluded — there is no separate subtraction step needed.

\(\text{TSA} = 2\,(lb+bh+hl)\)

\(= 2\,(8\times 4 + 4\times 4 + 4\times 8)\ \text{cm}^{2}\)

\(= 2\times 80 = 160\ \text{cm}^{2}\)

So the surface area of the resulting cuboid is 160 cm².

A vessel is in the form of a hollow hemisphere mounted by a hollow cylinder. The diameter of the hemisphere is 14 cm and the total height of the vessel is 13 cm. Find the inner surface area of the vessel.

A hollow hemisphere of diameter 14 cm mounted by a hollow cylinder, total height 13 cm.

Answer

Step 1 — what is given. The vessel is HOLLOW, so “inner surface area” means exactly the two curved surfaces a liquid poured in would touch — the inside wall of the cylinder and the bowl of the hemisphere.

\(r = \frac{14}{2} = 7\ \text{cm}\)

\(\text{Inner surface area} = \text{CSA of cylinder}\)

\(\phantom{\text{Inner surface area}} {}+ \text{CSA of hemisphere}\)

Step 2 — find the cylinder's height.

\(h = \text{total height} - r = 13-7\)

\(h = 6\ \text{cm}\)

Step 3 — substitute and simplify.

\(= 2\pi r h + 2\pi r^{2} = 2\pi r\,(h+r)\)

\(= 2\times \frac{22}{7}\times 7\times (6+7)\ \text{cm}^{2}\)

\(= 2\times 22\times 13\ \text{cm}^{2}\)

\(= 572\ \text{cm}^{2}\)

So the inner surface area of the vessel is 572 cm².

A toy is in the form of a cone of radius 3.5 cm mounted on a hemisphere of the same radius. The total height of the toy is 15.5 cm. Find the total surface area of the toy.

Answer

Step 1 — same construction as Example 1. Only the numbers differ, so the same figure and formula apply — the shared circle between the cone and the hemisphere is glued away, and only the two curved surfaces are exposed.

\(\text{TSA} = \text{CSA of cone} + \text{CSA of hemisphere}\)

\(\phantom{\text{TSA}} {}= \pi r l + 2\pi r^{2} = \pi r\,(l+2r)\)

Step 2 — find the cone's own height.

\(h = \text{total height} - r = 15.5 - 3.5\)

\(h = 12\ \text{cm}\)

Step 3 — find the slant height.

\(l = \sqrt{r^{2}+h^{2}} = \sqrt{(3.5)^{2}+12^{2}}\ \text{cm}\)

\(= \sqrt{12.25+144}\ \text{cm} = \sqrt{156.25}\ \text{cm} = 12.5\ \text{cm}\)

Step 4 — substitute and simplify.

\(\text{TSA} = \frac{22}{7}\times 3.5\times (12.5+7)\ \text{cm}^{2}\)

\(= 11\times 19.5\ \text{cm}^{2}\)

\(= 214.5\ \text{cm}^{2}\)

So the total surface area of the toy is 214.5 cm².

A cubical block of side 7 cm is surmounted by a hemisphere. What is the greatest diameter the hemisphere can have? Find the surface area of the solid.

Answer

Step 1 — the greatest possible diameter. The hemisphere sits flush on the cube's flat top face, so its base circle can be no bigger than that square face — the biggest circle that fits inside a square of side 7 cm has its diameter equal to the side itself.

\(\text{greatest diameter} = 7\ \text{cm},\ r = 3.5\ \text{cm}\)

Step 2 — the same construction as Example 2.

\(\text{Surface area} = \text{TSA of cube} - \pi r^{2} + 2\pi r^{2} = 6\,(\text{edge})^{2} + \pi r^{2}\)

Step 3 — substitute and simplify.

\(= 6\times 7\times 7 + \frac{22}{7}\times 3.5\times 3.5\ \text{cm}^{2}\)

\(= 294 + 38.5\ \text{cm}^{2}\)

\(= 332.5\ \text{cm}^{2}\)

\(\text{(check: } 6\times49=294,\ \tfrac{22}{7}\times12.25=38.5\text{)}\)

So the greatest possible diameter is 7 cm, and the surface area of the solid is 332.5 cm².

A hemispherical depression is cut out from one face of a cubical wooden block such that the diameter \(l\) of the hemisphere is equal to the edge of the cube. Determine the surface area of the remaining solid.

A hemispherical depression is cut from ONE face of a cubical block, the depression’s diameter equal to the cube’s own edge.

Answer

Step 1 — set up symbols. Let the cube's edge be \(l\), so the depression's diameter is also \(l\) and its radius is \(\tfrac{l}{2}\).

Step 2 — which surfaces are exposed. The flat circular patch where the depression is cut is REMOVED from that face (it is now a hole, not a flat surface), and the hemisphere's own curved wall, now facing INTO the block, is newly exposed.

\(\text{Surface area} = \text{TSA of cube} - \pi r^{2} + 2\pi r^{2} = 6l^{2} + \pi r^{2}\)

Step 3 — substitute \(r=\tfrac{l}{2}\).

\(= 6l^{2} + \pi\left(\frac{l}{2}\right)^{2}\)

\(= 6l^{2} + \frac{\pi l^{2}}{4}\)

\(= \frac{24l^{2}}{4} + \frac{\pi l^{2}}{4}\)

\(= \frac{l^{2}}{4}\,(24+\pi)\)

So the surface area of the remaining solid is \(\dfrac{l^{2}}{4}\,(24+\pi)\) square units.

A medicine capsule is in the shape of a cylinder with two hemispheres stuck to each of its ends (see the figure). The length of the entire capsule is 14 mm and the diameter of the capsule is 5 mm. Find its surface area.

Fig. 12.10 — a medicine capsule: a cylinder with a hemisphere stuck on each end, overall length 14 mm, diameter 5 mm.

Answer

Step 1 — what is given.

\(r = \frac{5}{2} = 2.5\ \text{mm},\ \text{total length} = 14\ \text{mm}\)

Step 2 — find the cylindrical portion's length. Each hemisphere adds its OWN radius to the overall length (not its diameter, since only the curved half sticks out beyond the shared flat circle) — so the two hemispheres together use up \(2r\) of the total 14 mm.

\(h = 14 - 2r = 14-5\)

\(h = 9\ \text{mm}\)

Step 3 — the two flat circles where the hemispheres meet the cylinder are both glued away, so only three curved surfaces remain: the cylinder's wall and the two hemispherical domes.

\(\text{Surface area} = \text{CSA of cylinder} + 2\times \text{CSA of hemisphere}\)

Step 4 — substitute and simplify.

\(= 2\pi r h + 4\pi r^{2} = 2\pi r\,(h+2r)\)

\(= 2\times \frac{22}{7}\times 2.5\times (9+5)\ \text{mm}^{2}\)

\(= 2\times \frac{22}{7}\times 2.5\times 14\ \text{mm}^{2} = 220\ \text{mm}^{2}\)

So the surface area of the capsule is 220 mm².

A tent is in the shape of a cylinder surmounted by a conical top. If the height and diameter of the cylindrical part are 2.1 m and 4 m respectively, and the slant height of the top is 2.8 m, find the area of the canvas used for making the tent. Also, find the cost of the canvas of the tent at the rate of ₹500 per m². (Note that the base of the tent will not be covered with canvas.)

A tent: a cylinder of diameter 4 m and height 2.1 m, surmounted by a conical top of the SAME radius, slant height 2.8 m.

Answer

Step 1 — what is given.

\(r = \frac{4}{2} = 2\ \text{m},\ h = 2.1\ \text{m},\ l = 2.8\ \text{m}\)

Step 2 — which surfaces need canvas. The base is explicitly NOT covered, and this cone and cylinder share the SAME radius (unlike Example 3's rocket), so the join is flush and no ring is exposed there either. The canvas covers only the two curved walls.

\(\text{Canvas area} = \text{CSA of cylinder} + \text{CSA of cone}\)

\(\phantom{\text{Canvas area}} {}= 2\pi r h + \pi r l = \pi r\,(2h+l)\)

Step 3 — substitute and simplify.

\(= \frac{22}{7}\times 2\times (2\times 2.1+2.8)\ \text{m}^{2}\)

\(= \frac{22}{7}\times 2\times (4.2+2.8)\ \text{m}^{2}\)

\(= \frac{22}{7}\times 2\times 7\ \text{m}^{2}\)

\(= 44\ \text{m}^{2}\)

Step 4 — the cost.

\(\text{Cost} = 44\times 500\)

\(= \text{₹}\,22000\)

So the area of canvas used is 44 m², and the cost at ₹500 per m² is ₹22000.

From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest cm².

A solid cylinder of height 2.4 cm, diameter 1.4 cm, with a conical cavity of the same height and diameter hollowed from one flat face.

Answer

Step 1 — what is given.

\(r = \frac{1.4}{2} = 0.7\ \text{cm},\ h = 2.4\ \text{cm}\)

Step 2 — which surfaces survive. The cavity is bored in through ONE flat circular face of the cylinder, and its mouth is EXACTLY as wide as that face — so that entire flat circle disappears (there is no leftover ring, unlike Example 3), replaced by the cavity's own slant wall, which now curves INTO the solid rather than out of it. The OTHER flat face of the cylinder, untouched, still counts.

\(\text{TSA} = \text{CSA of cylinder} + \text{one untouched base}\)

\(\phantom{\text{TSA}} {}+ \text{CSA of the conical cavity}\)

Step 3 — the slant height of the cavity. The cavity is a cone of the SAME radius and height as the cylinder.

\(l = \sqrt{r^{2}+h^{2}} = \sqrt{(0.7)^{2}+(2.4)^{2}}\ \text{cm}\)

\(= \sqrt{0.49+5.76}\ \text{cm} = \sqrt{6.25}\ \text{cm} = 2.5\ \text{cm}\)

Step 4 — substitute and simplify.

\(\text{TSA} = 2\pi r h + \pi r^{2} + \pi r l = \pi r\,(2h+r+l)\)

\(= \frac{22}{7}\times 0.7\times (2\times 2.4+0.7+2.5)\ \text{cm}^{2}\)

\(= 2.2\times 8.0\ \text{cm}^{2}\)

\(= 17.6\ \text{cm}^{2} \approx 18\ \text{cm}^{2}\)

So the total surface area of the remaining solid is about 17.6 cm², which rounds to 18 cm² (nearest cm²).

A wooden article was made by scooping out a hemisphere from each end of a solid cylinder, as shown in the figure. If the height of the cylinder is 10 cm, and its base is of radius 3.5 cm, find the total surface area of the article.

Fig. 12.11 — a wooden article: a hemisphere scooped out from EACH end of a solid cylinder of height 10 cm, base radius 3.5 cm.

Answer

Step 1 — what is given.

\(r = 3.5\ \text{cm},\ h = 10\ \text{cm}\)

Step 2 — which surfaces survive. Each scoop is exactly as wide as its end of the cylinder, so BOTH flat circular faces disappear entirely, each replaced by a hemispherical bowl curving inward. Only the cylinder's own curved wall and the two bowls remain — there is no flat surface left anywhere on this solid.

\(\text{TSA} = \text{CSA of cylinder} + 2\times \text{CSA of hemisphere}\)

\(\phantom{\text{TSA}} {}= 2\pi r h + 4\pi r^{2} = 2\pi r\,(h+2r)\)

Step 3 — substitute and simplify.

\(= 2\times \frac{22}{7}\times 3.5\times (10+7)\ \text{cm}^{2}\)

\(= 22\times 17\ \text{cm}^{2}\)

\(= 374\ \text{cm}^{2}\)

So the total surface area of the article is 374 cm².

Worked Examples — Volume of a Combination of Solids (Examples 5–7)

Example 5

Shanta runs an industry in a shed which is in the shape of a cuboid surmounted by a half cylinder. If the base of the shed is of dimension 7 m × 15 m, and the height of the cuboidal portion is 8 m, find the volume of air that the shed can hold. Further, suppose the machinery in the shed occupies a total space of 300 m³, and there are 20 workers, each of whom occupy about 0.08 m³ space on an average. Then, how much air is in the shed? (Take π = \(\tfrac{22}{7}\))

Fig. 12.12 — the shed: a cuboid 15 m × 7 m × 8 m surmounted by a half-cylinder roof of the same 7 m span.

Solution

Step 1 — volumes always simply ADD, unlike surface areas. No matter how two solids are joined, nothing about their volume disappears at the seam — only surface area loses the glued patch. So the shed's total volume is just the cuboid's volume plus the half-cylinder's.

\(\text{Volume of air} = \text{volume of cuboid} + \text{volume of half-cylinder}\)

Step 2 — what is given. The cuboid is \(15\,\text{m}\times 7\,\text{m}\times 8\,\text{m}\); the half-cylinder has diameter 7 m (so radius 3.5 m) and length 15 m, running the length of the roof.

\(l = 15\ \text{m},\ b = 7\ \text{m},\ h = 8\ \text{m}\)

\(r = 3.5\ \text{m},\ \text{length of roof} = 15\ \text{m}\)

Step 3 — substitute and simplify. (The two terms are shown separately, then added, to keep each line short.)

\(\text{Volume of cuboid} = 15\times 7\times 8 = 840\ \text{m}^{3}\)

\(\text{Volume of half-cylinder} = \frac{1}{2}\times \frac{22}{7}\times 3.5\times 3.5\times 15 = 288.75\ \text{m}^{3}\)

\(\text{Total} = (840 + 288.75)\ \text{m}^{3} = 1128.75\ \text{m}^{3}\)

Step 4 — subtract the machinery and the workers.

\(\text{Space used} = \text{machinery} + \text{workers} = 300 + 20\times 0.08\)

\(= 300 + 1.6 = 301.6\ \text{m}^{3}\)

\(\text{Air with machinery and workers} = 1128.75 - 301.6 = 827.15\ \text{m}^{3}\)

So the volume of air when the shed is empty is 1128.75 m³, and with the machinery and the workers in place it is 827.15 m³.

Example 6

A juice seller was serving his customers using glasses as shown in the figure. The inner diameter of the cylindrical glass was 5 cm, but the bottom of the glass had a hemispherical raised portion which reduced the capacity of the glass. If the height of a glass was 10 cm, find the apparent capacity of the glass and its actual capacity. (Use π = 3.14)

Fig. 12.13 — the juice glass: an inner radius of 2.5 cm and height 10 cm, with a hemispherical raised portion at the bottom reducing its capacity.

Solution

Step 1 — the two capacities are different questions. The “apparent” capacity is what the glass LOOKS like it holds if you only saw its height and diameter — a plain cylinder. The bump at the bottom, solid glass, is not available for juice, so it has to be SUBTRACTED to get the actual capacity.

\(r = \frac{5}{2} = 2.5\ \text{cm},\ h = 10\ \text{cm}\)

Step 2 — find the apparent capacity.

\(\text{Apparent capacity} = \pi r^{2} h\)

\(= 3.14\times 2.5\times 2.5\times 10\ \text{cm}^{3} = 196.25\ \text{cm}^{3}\)

Step 3 — find the volume taken up by the hemispherical bump. The bump is a hemisphere of the same radius as the glass.

\(\text{Volume of hemisphere} = \frac{2}{3}\pi r^{3}\)

\(= \frac{2}{3}\times 3.14\times 2.5\times 2.5\times 2.5\ \text{cm}^{3} \approx 32.71\ \text{cm}^{3}\)

Step 4 — subtract.

\(\text{Actual capacity} = \text{apparent capacity}\)

\(\phantom{\text{Actual capacity}} {}- \text{volume of hemisphere}\)

\(= (196.25 - 32.71)\ \text{cm}^{3} = 163.54\ \text{cm}^{3}\)

So the apparent capacity is 196.25 cm³, and the actual capacity, after the bump is taken away, is 163.54 cm³.

Example 7

A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 2 cm and the diameter of the base is 4 cm. Determine the volume of the toy. If a right circular cylinder circumscribes the toy, find the difference of the volumes of the cylinder and the toy. (Take π = 3.14)

Fig. 12.14 — the toy: a hemisphere of radius 2 cm surmounted by a cone of the same radius and height, circumscribed by a cylinder EFGH of radius 2 cm and height 4 cm.

Solution

Step 1 — what is given.

\(r = \frac{4}{2} = 2\ \text{cm}\ \text{(shared radius)},\ h_{\text{cone}} = 2\ \text{cm}\)

Step 2 — the volume of the toy, by simple addition.

\(\text{Volume of toy} = \frac{2}{3}\pi r^{3} + \frac{1}{3}\pi r^{2} h_{\text{cone}}\)

\(\text{Hemisphere: } \frac{2}{3}\times 3.14\times 2^{3} = 16.75\ \text{cm}^{3}\ \text{(approx.)}\)

\(\text{Cone: } \frac{1}{3}\times 3.14\times 2^{2}\times 2 = 8.37\ \text{cm}^{3}\ \text{(approx.)}\)

\(\text{Volume of toy} = 16.75+8.37 \approx 25.12\ \text{cm}^{3}\)

Step 3 — the circumscribing cylinder. A cylinder that just wraps around the whole toy has the SAME radius as the hemisphere/cone, and a height that reaches from the flat bottom of the hemisphere to the tip of the cone — that is, the hemisphere's radius PLUS the cone's own height.

\(r_{\text{cyl}} = r = 2\ \text{cm}\)

\(h_{\text{cyl}} = r + h_{\text{cone}} = 2+2 = 4\ \text{cm}\)

Step 4 — the volume of that cylinder.

\(\text{Volume of cylinder} = \pi r_{\text{cyl}}^{2} h_{\text{cyl}}\)

\(= 3.14\times 2^{2}\times 4\ \text{cm}^{3} = 50.24\ \text{cm}^{3}\)

Step 5 — the difference.

\(\text{Difference} = 50.24 - 25.12\)

\(= 25.12\ \text{cm}^{3}\)

So the volume of the toy is 25.12 cm³, and the difference between the cylinder's volume and the toy's volume is 25.12 cm³. (These two numbers coming out equal is a real coincidence of \(r=h_{\text{cone}}\) here, not a general rule — it happens because a cylinder of radius \(r\) and height \(2r\) always has EXACTLY TWICE the volume of a hemisphere-plus-cone of the same radius when the cone's height also equals \(r\).)

Exercise 12.2

A solid is in the shape of a cone standing on a hemisphere with both their radii being equal to 1 cm and the height of the cone is equal to its radius. Find the volume of the solid in terms of π.

Answer

Step 1 — same construction as Example 7. A cone standing on a hemisphere of the same radius — only the numbers differ, and here the answer is wanted symbolically, "in terms of pi", so no numeric value of \(\pi\) is substituted at all.

\(\text{Volume} = \frac{2}{3}\pi r^{3} + \frac{1}{3}\pi r^{2} h\)

Step 2 — what is given.

\(r = 1\ \text{cm},\ h = r = 1\ \text{cm}\)

Step 3 — substitute and simplify.

\(\text{Volume} = \frac{2}{3}\pi(1)^{3} + \frac{1}{3}\pi(1)^{2}(1)\)

\(= \frac{2}{3}\pi + \frac{1}{3}\pi\)

\(= \pi\ \text{cm}^{3}\)

So the volume of the solid is \(\pi\) cm³ exactly.

Rachel, an engineering student, was asked to make a model shaped like a cylinder with two cones attached at its two ends by using a thin aluminium sheet. The diameter of the model is 3 cm and its length is 12 cm. If each cone has a height of 2 cm, find the volume of air contained in the model that Rachel made. (Assume the outer and inner dimensions of the model to be nearly the same.)

Answer

Step 1 — what is given.

\(r = \frac{3}{2} = 1.5\ \text{cm},\ \text{total length} = 12\ \text{cm},\ h_{\text{cone}} = 2\ \text{cm}\)

Step 2 — find the cylindrical portion's length. The two cones use up their own heights from the total length.

\(h_{\text{cyl}} = 12 - 2\times 2\)

\(h_{\text{cyl}} = 8\ \text{cm}\)

Step 3 — add the three volumes. Since the aluminium sheet is thin, the air inside is (to a very good approximation) the full volume enclosed — cylinder plus TWO cones.

\(\text{Volume of air} = \pi r^{2} h_{\text{cyl}} + 2\times \frac{1}{3}\pi r^{2} h_{\text{cone}}\)

\(= \frac{22}{7}\times (1.5)^{2}\times \left[8 + \frac{2}{3}\times 2\right]\text{cm}^{3}\)

\(= \frac{22}{7}\times 2.25\times \frac{28}{3}\ \text{cm}^{3} = 66\ \text{cm}^{3}\)

So the volume of air contained in the model is 66 cm³.

A gulab jamun contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends, with length 5 cm and diameter 2.8 cm.

Answer

Step 1 — the shape. Same construction as Exercise 12.1 Q6's capsule — a cylinder with a hemisphere stuck on each end — here with different numbers and a different question (a percentage of the volume, not a surface area).

\(r = \frac{2.8}{2} = 1.4\ \text{cm},\ \text{length} = 5\ \text{cm}\)

Step 2 — find the cylindrical portion's length.

\(h = 5 - 2\times 1.4\)

\(h = 2.2\ \text{cm}\)

Step 3 — volume of ONE gulab jamun.

\(\text{Volume of one} = \pi r^{2} h + \frac{4}{3}\pi r^{3}\)

\(= \frac{22}{7}\times (1.4)^{2}\times \left[2.2 + \frac{4}{3}\times 1.4\right]\text{cm}^{3}\)

\(\approx 25.05\ \text{cm}^{3}\)

Step 4 — volume of all 45.

\(\text{Volume of 45} = 45\times 25.05\ \text{cm}^{3}\)

\(\approx 1127.28\ \text{cm}^{3}\)

Step 5 — 30% of that is syrup.

\(\text{Syrup} = 30\%\ \text{of}\ 1127.28\)

\(\approx 338.18\ \text{cm}^{3}\)

So the syrup in 45 gulab jamuns is approximately 338.18 cm³.

A pen stand made of wood is in the shape of a cuboid with four conical depressions to hold pens. The dimensions of the cuboid are 15 cm by 10 cm by 3.5 cm. The radius of each of the depressions is 0.5 cm and the depth is 1.4 cm. Find the volume of wood in the entire stand.

Fig. 12.16 — a pen stand: a cuboid 15 cm × 10 cm × 3.5 cm with 4 conical depressions, each of radius 0.5 cm and depth 1.4 cm.

Answer

Step 1 — what is given.

\(l=15\ \text{cm},\ b=10\ \text{cm},\ h=3.5\ \text{cm}\)

\(r=0.5\ \text{cm},\ d=1.4\ \text{cm (each depression)}\)

Step 2 — the volume of wood is the block MINUS the four holes. Boring a conical depression removes exactly a cone's worth of wood, regardless of what shape the hole's opening looks like on the surface — volume, unlike surface area, does not care about the seam.

\(\text{Volume of wood} = \text{volume of cuboid}\)

\(\phantom{\text{Volume of wood}} {}- 4\times \text{volume of one cone}\)

Step 3 — volume of the solid cuboid.

\(\text{Volume of cuboid} = 15\times 10\times 3.5 = 525\ \text{cm}^{3}\)

Step 4 — volume of ONE conical depression.

\(\text{Volume of one cone} = \frac{1}{3}\pi r^{2} d = \frac{1}{3}\times \frac{22}{7}\times (0.5)^{2}\times 1.4\ \text{cm}^{3}\)

\(= \frac{11}{30}\ \text{cm}^{3} \approx 0.367\ \text{cm}^{3}\)

Step 5 — subtract all four.

\(4\times \text{one cone} = 4\times \frac{11}{30} = \frac{44}{30} \approx 1.467\ \text{cm}^{3}\)

\(\text{Volume of wood} = 525 - 1.467\)

\(\approx 523.53\ \text{cm}^{3}\)

So the volume of wood in the pen stand is approximately 523.53 cm³.

A vessel is in the form of an inverted cone. Its height is 8 cm and the radius of its top, which is open, is 5 cm. It is filled with water up to the brim. When lead shots, each of which is a sphere of radius 0.5 cm, are dropped into the vessel, one-fourth of the water flows out. Find the number of lead shots dropped in the vessel.

An inverted cone (vessel), height 8 cm, radius of the open top 5 cm, full of water; small lead shots of radius 0.5 cm are dropped in.

Answer

Step 1 — what is given.

\(h = 8\ \text{cm},\ r = 5\ \text{cm}\ \text{(top radius)},\ r_{\text{shot}} = 0.5\ \text{cm}\)

Step 2 — the water displaced equals the total volume of the shots. Each lead shot pushes its own volume of water out over the brim, so the volume that OVERFLOWED (one-fourth of the full cone) exactly equals the combined volume of however many shots went in.

\(\text{Volume overflowed} = n\times \text{volume of one shot}\)

Step 3 — volume of the cone, and one-fourth of it.

\(\text{Volume of cone} = \frac{1}{3}\pi r^{2} h = \frac{1}{3}\pi (5)^{2}(8) = \frac{200}{3}\pi\ \text{cm}^{3}\)

\(\text{Volume overflowed} = \frac{1}{4}\times \frac{200}{3}\pi = \frac{50}{3}\pi\ \text{cm}^{3}\)

Step 4 — volume of one lead shot.

\(\text{Volume of one shot} = \frac{4}{3}\pi (0.5)^{3} = \frac{1}{6}\pi\ \text{cm}^{3}\)

Step 5 — divide. Notice \(\pi\) cancels completely out of this division — the count does not depend on which value of \(\pi\) is used at all.

\(n = \frac{\text{volume overflowed}}{\text{volume of one shot}} = \frac{\tfrac{50}{3}\pi}{\tfrac{1}{6}\pi}\)

\(= \frac{50}{3}\times 6\)

\(= 100\)

So the number of lead shots dropped into the vessel is 100.

A solid iron pole consists of a cylinder of height 220 cm and base diameter 24 cm, which is surmounted by another cylinder of height 60 cm and radius 8 cm. Find the mass of the pole, given that 1 cm³ of iron has approximately 8g mass. (Use π = 3.14)

A solid iron pole: a cylinder of height 220 cm, radius 12 cm, surmounted by a narrower cylinder of height 60 cm, radius 8 cm.

Answer

Step 1 — what is given.

\(r_{1} = \frac{24}{2} = 12\ \text{cm},\ h_{1} = 220\ \text{cm}\)

\(r_{2} = 8\ \text{cm},\ h_{2} = 60\ \text{cm}\)

Step 2 — volumes simply add, whatever the radii. Unlike a surface-area question, it makes no difference here that the two cylinders have different radii — each contributes its own full volume regardless of what happens at the join.

\(\text{Volume of pole} = \pi r_{1}^{2} h_{1} + \pi r_{2}^{2} h_{2}\)

Step 3 — substitute and simplify.

\(= 3.14\times \left[(12)^{2}\times 220 + (8)^{2}\times 60\right]\text{cm}^{3}\)

\(= 3.14\times \left[31680+3840\right]\text{cm}^{3}\)

\(= 3.14\times 35520\ \text{cm}^{3} = 111532.8\ \text{cm}^{3}\)

Step 4 — convert volume to mass.

\(\text{Mass} = 111532.8\times 8\ \text{g}\)

\(= 892262.4\ \text{g} \approx 892.26\ \text{kg}\)

So the mass of the iron pole is about 892.26 kg.

A solid consisting of a right circular cone of height 120 cm and radius 60 cm standing on a hemisphere of radius 60 cm is placed upright in a right circular cylinder full of water such that it touches the bottom. Find the volume of water left in the cylinder, if the radius of the cylinder is 60 cm and its height is 180 cm.

Answer

Step 1 — the shape sitting inside the water. This is the same hemisphere-and-cone construction as Example 7, with different numbers and now standing inside a filled cylinder rather than being circumscribed by one.

\(r = 60\ \text{cm},\ h_{\text{cone}} = 120\ \text{cm}\)

Step 2 — volume of the solid (cone + hemisphere).

\(\text{Volume of solid} = \frac{1}{3}\pi r^{2} h_{\text{cone}} + \frac{2}{3}\pi r^{3}\)

\(= \frac{1}{3}\pi (60)^{2}(120) + \frac{2}{3}\pi (60)^{3} = 144000\pi + 144000\pi = 288000\pi\ \text{cm}^{3}\)

Step 3 — volume of the full cylinder.

\(\text{Volume of cylinder} = \pi (60)^{2}(180) = 648000\pi\ \text{cm}^{3}\)

Step 4 — water left, by displacement. The solid sits touching the bottom of the cylinder, so it displaces exactly its own volume of water — the water left is the cylinder's full volume minus the solid's volume.

\(\text{Water left} = 648000\pi - 288000\pi = 360000\pi\ \text{cm}^{3}\)

\(= 360000\times \frac{22}{7}\ \text{cm}^{3}\)

\(= \frac{7920000}{7}\ \text{cm}^{3} \approx 1131428.6\ \text{cm}^{3}\)

So the volume of water left in the cylinder is \(\dfrac{7920000}{7}\) cm³, which is about 1131428.6 cm³.

A spherical glass vessel has a cylindrical neck 8 cm long, 2 cm in diameter; the diameter of the spherical part is 8.5 cm. By measuring the amount of water it holds, a child finds its volume to be 345 cm³. Check whether she is correct, taking the above as the inside measurements, and π = 3.14.

A spherical glass vessel of diameter 8.5 cm with a cylindrical neck 8 cm long and 2 cm in diameter.

Answer

Step 1 — what is given.

\(r = \frac{2}{2} = 1\ \text{cm},\ h = 8\ \text{cm}\ \text{(neck)}\)

Step 2 — the true volume is the sphere plus the neck.

\(\text{Volume} = \pi r^{2} h + \frac{4}{3}\pi R^{3}\)

Step 3 — volume of the neck.

\(\text{Volume of neck} = \pi (1)^{2}(8) = 3.14\times 8 = 25.12\ \text{cm}^{3}\)

Step 4 — volume of the spherical part.

\(R = \frac{8.5}{2} = 4.25\ \text{cm}\)

\(\text{Volume of sphere} = \frac{4}{3}\pi R^{3} = \frac{4}{3}\times 3.14\times (4.25)^{3}\ \text{cm}^{3} \approx 321.39\ \text{cm}^{3}\)

Step 5 — add, and compare with the child's claim.

\(\text{Total volume} = 25.12 + 321.39\)

\(\approx 346.51\ \text{cm}^{3}\)

The correct volume is about 346.51 cm³, not 345 cm³ — so the child is not correct (though her measurement of 345 cm³ is close, it is off by about 1.51 cm³).

Chapter 11: Areas Related to Circles

All Chapters

Chapter 13: Statistics

Surface Areas and Volumes

Worked Examples — Surface Area of a Combination of Solids (Examples 1–4)

Exercise 12.1

Worked Examples — Volume of a Combination of Solids (Examples 5–7)

Exercise 12.2

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