Chapter 12 · Surface Areas and Volumes

12.1 Surface Area of a Combination of Solids

A tent, a capsule, an ice-cream cone: each is two solids joined. The paint or canvas you need is the area of the faces that stay on the outside — and the faces where the solids touch do not count.

A company wants to paint a thousand wooden toys, each a cone sitting on a half-ball. A tailor must cut canvas for a tent that has a cylindrical wall and a conical roof. A pharmacist wants to know how much coating a capsule needs. In each case the answer is the surface area of a solid that is a combination of two simpler solids, and the cost depends on getting it right: the paint goes only on the outside.

The key idea of this section is easy to state and easy to forget. When two solids are joined, the faces that touch are hidden inside, so the surface area of the combination is not the sum of the two surface areas. We first recall the surface area of the six basic solids and then learn to add only the faces you can see. Throughout we say which value of $\pi$ is used: $\tfrac{22}{7}$ when the radius is a multiple of $7$ or $3.5$ (or $10.5$), and $3.14$ otherwise. Answers are given to two decimal places with units.

The six basic solids

A solid has a curved surface (a surface that is not flat) or flat faces or both. The curved surface area (CSA) is the area of the curved part only; the total surface area (TSA) is the area of the whole outside, flat faces included.

📘 Definition 12.1 — Curved and total surface area
The curved surface area of a solid is the area of its curved surface alone (for a solid with only flat faces, the area of its side faces is called the lateral surface area). The total surface area is the sum of the areas of all the faces of the solid, curved and flat, that lie on its outer boundary. Both are measured in square units (cm², m²).

A right circular cylinder and a right circular cone have a circular base and an axis at right angles to it; their radius is $r$ and their height $h$. A cone also has a slant height $l$, the length from the apex to any point of the edge of the base. A hemisphere is half a sphere, cut by a plane through its centre, so it has a curved top and one flat circular face.

📐 Theorem 12.1 — Surface areas of the basic solids
SolidCurved (or lateral) surface areaTotal surface area
Cuboid $l\times b\times h$$2h(l+b)$$2(lb+bh+hl)$
Cube of edge $a$$4a^2$$6a^2$
Cylinder, radius $r$, height $h$$2\pi rh$$2\pi rh+2\pi r^2=2\pi r(h+r)$
Cone, radius $r$, height $h$, slant height $l$$\pi rl$$\pi rl+\pi r^2=\pi r(l+r)$
Sphere, radius $r$$4\pi r^2$$4\pi r^2$
Hemisphere, radius $r$$2\pi r^2$$2\pi r^2+\pi r^2=3\pi r^2$

Why they hold. Unroll the curved surface of a cylinder: it is a rectangle of length $2\pi r$ (the circumference of the base) and width $h$, so its area is $2\pi rh$. Cut a cone along a slant line and unroll it: it becomes a sector of a circle of radius $l$ whose arc has length $2\pi r$ (the circumference of the base). Using $\text{area of a sector}=\tfrac12\times\text{radius}\times\text{arc length}$, which follows from the sector formulas of Section 11.1, the area is $\tfrac12\cdot l\cdot2\pi r=\pi rl$. The sphere formula $4\pi r^2$ is the one to remember, and the hemisphere is half of it, plus its flat circular face $\pi r^2$ for the total.

📐 Theorem 12.2 — Slant height of a cone
For a right circular cone of base radius $r$ and height $h$, the slant height is $l=\sqrt{r^2+h^2}$, that is, $l^2=r^2+h^2$.

Proof. The axis, the radius to the edge of the base and the slant line form a right-angled triangle whose right angle is at the centre of the base. Its hypotenuse is $l$, so by the Pythagoras theorem (Class 9) $l^2=r^2+h^2$. $\blacksquare$ Always find $l$ before using $\pi rl$: the height $h$ is not the slant height.

Joining two solids

📘 Definition 12.2 — Combination of solids
A combination of solids is a solid made by joining two solids (here, any two of the cube, cuboid, sphere, hemisphere, cylinder and cone) face to face. A solid with a piece scooped out of another, a hollowed solid, is also called a combination.

Think of gluing a cone on top of a hemisphere to make a toy. The circular base of the cone and the flat circular face of the hemisphere are pressed together and vanish from the outside. Every other face stays. This gives a rule that works for every combination in this chapter.

📐 Theorem 12.3 — Surface area of a combination
(a) Joined solids. If two solids are joined so that they touch over a region of area $A_c$ (the contact area), then $$\text{TSA of the combination}=\text{TSA}_1+\text{TSA}_2-2A_c.$$ Equivalently, add the areas of exactly those faces that lie on the outside.
(b) A hollow scooped out. If a hollow whose opening has area $A_o$ is scooped out of a solid, then $$\text{TSA}=\text{TSA of the original solid}-A_o+\text{area of the inside surface of the hollow}.$$

Reason. In (a) each of the two touching regions was counted once in $\text{TSA}_1$ and once in $\text{TSA}_2$ but is now inside, so we remove it twice. In (b) the opening is a piece of the outside that disappears, while the inner surface of the hollow becomes new outside. Check on a capsule: a cylinder (TSA $2\pi rh+2\pi r^2$) and two hemispheres (TSA $3\pi r^2$ each), joined along two circles of area $\pi r^2$: $2\pi rh+2\pi r^2+6\pi r^2-4\pi r^2=2\pi rh+4\pi r^2$. That agrees with adding the cylinder's curved wall to two hemispherical surfaces, $2\pi rh+2\cdot2\pi r^2$.

🧭 Problem-solving strategy — surface area of a combination
  1. Sketch the solid and name its two parts. Mark $r$, $h$ and, for a cone, find $l=\sqrt{r^2+h^2}$.
  2. Ask of every face: can the paint reach it? A face where two solids touch, or a face covered by the other solid, is not counted.
  3. List the faces that stay: curved surfaces, free flat faces, and the part of a flat face that is left uncovered.
  4. Write the area of each with its formula, substitute with the stated value of $\pi$, and add (or, for a hollow, use Theorem 12.3(b)).
  5. Attach units (cm², m²). For a cost, multiply the area by the rate and check whether the base is painted or stitched at all.
⚠️ Three mistakes that cost marks
1. Adding the two total surface areas, which counts the hidden joined faces. 2. Using the height of the cone in $\pi rl$ instead of the slant height. 3. Forgetting that a hemisphere scooped from a solid adds an inside surface $2\pi r^2$ while it removes a circle $\pi r^2$ from the outside.
✏️ Example 1 — A toy: cone on a hemisphere
A toy is a cone of radius $7$ cm mounted on a hemisphere of the same radius. The total height of the toy is $31$ cm. Find its total surface area. (Use $\pi=\tfrac{22}{7}$.)
Figure Hemisphere radius $r=7$ cm, so the cone has height $h=31-7=24$ cm.
Slant height $l=\sqrt{7^2+24^2}=\sqrt{49+576}=\sqrt{625}=25$ cm.
Faces on the outside Only the curved surface of the cone and the curved surface of the hemisphere. The cone's base and the hemisphere's flat face are joined and hidden.
Add $\pi rl+2\pi r^2=\tfrac{22}{7}\times7\times25+2\times\tfrac{22}{7}\times49=550+308=858$ cm².
Check Theorem 12.3(a): $\text{TSA}_1+\text{TSA}_2-2A_c=(550+154)+462-2\times154=704+462-308=858$ ✓.
✏️ Example 2 — A capsule-shaped tank
A fuel tank is a cylinder with a hemisphere at each end. The radius is $0.7$ m and the total length is $5.4$ m. Find the area to be painted and the cost at ₹ $150$ per m². (Use $\pi=\tfrac{22}{7}$.)
Figure $r=0.7$ m. The two hemispheres take $2r=1.4$ m of the length, so the cylinder has height $h=5.4-1.4=4$ m.
Outside faces The curved wall of the cylinder, $2\pi rh$, and two hemispherical surfaces, together $2\times2\pi r^2=4\pi r^2$.
Area $2\times\tfrac{22}{7}\times0.7\times4+4\times\tfrac{22}{7}\times0.49=17.6+6.16=23.76$ m².
Cost $23.76\times150=3564$, so the cost is ₹ $3564$.
Check A capsule of total length $L$ has TSA $2\pi rL$: $2\times\tfrac{22}{7}\times0.7\times5.4=23.76$ ✓ (Exercise IV, part 2).
Interactive 1 — Build It and See Which Faces Count

Pick a combination, then change the radius $r$ and the height $h$ with the sliders (or press Play). Faces that count are tinted red, blue or green; the dashed faces are joined and hidden. The tilt slider turns the view to show more of the top. The list on the right adds only the faces that count. Switch $\pi$ to see the effect.

3.58.0
part 1—
part 2—
hidden (not counted)—
total surface area (cm²)—
✏️ Example 3 — Canvas for a tent
A tent has a cylindrical wall of radius $6$ m and height $4$ m, with a conical roof of the same radius and slant height $10$ m. Find the area of canvas needed and its cost at ₹ $85$ per m² (the ground is not covered with canvas). (Use $\pi=3.14$.)
Figure $r=6$ m, wall height $h=4$ m, roof slant height $l=10$ m (so the roof is $8$ m high, since $6^2+8^2=10^2$).
Canvas faces The curved wall $2\pi rh$ and the conical roof $\pi rl$. The floor and the circle where roof meets wall are not canvas.
Area $2\times3.14\times6\times4+3.14\times6\times10=150.72+188.4=339.12$ m².
Cost $339.12\times85=28825.2$, so the cost is ₹ $28{,}825.20$.
Check Factor: $\pi r(2h+l)=3.14\times6\times18=339.12$ ✓.
✏️ Example 4 — A cube with a hemisphere on top
A wooden block is a cube of edge $14$ cm with a hemisphere of the largest possible radius fixed on one face. Find the total surface area of the solid. (Use $\pi=\tfrac{22}{7}$.)
Figure The largest hemisphere on a $14$ cm face has diameter $14$ cm, so $r=7$ cm.
Faces All six faces of the cube, but a circle of area $\pi r^2$ on the top face is covered by the hemisphere; and the curved surface of the hemisphere, $2\pi r^2$.
Add $6a^2-\pi r^2+2\pi r^2=6a^2+\pi r^2=6\times196+\tfrac{22}{7}\times49=1176+154=1330$ cm².
Remark The surface area is more than the cube's $1176$ cm², because the dome has area $308$ cm² and replaces only $154$ cm².
✏️ Example 5 — A cylinder with a conical hollow
From a solid wooden cylinder of radius $7$ cm and height $24$ cm, a cone of the same base radius and the same height is drilled out, its base being the top face of the cylinder. Find the total surface area of what remains. (Use $\pi=\tfrac{22}{7}$.)
Figure The cone's base is the whole top face, so the top face disappears completely. Slant height of the cone: $l=\sqrt{7^2+24^2}=25$ cm.
Faces Outer curved wall $2\pi rh$; bottom face $\pi r^2$; inner conical surface $\pi rl$.
Add $2\times\tfrac{22}{7}\times7\times24+\tfrac{22}{7}\times49+\tfrac{22}{7}\times7\times25=1056+154+550=1760$ cm².
Check Theorem 12.3(b): original TSA $=1056+308=1364$; remove the opening $154$; add the cone's surface $550$: $1364-154+550=1760$ ✓.
Interactive 2 — Add It, Join It or Scoop It

A cylinder of radius $r$ and height $h$ meets a hemisphere of radius $r$. Choose Apart (two separate solids), Joined (hemisphere on top) or Scooped (the hemisphere is cut out of the top). Vary $r$ and $h$ (slider, or Play). Compare the two surface areas: joined and scooped give the same total, and each is exactly $2\pi r^2$ less than the two solids apart. Here $\pi=3.14$.

3.08.0
cylinder TSA $2\pi rh+2\pi r^2$ (cm²)—
hemisphere TSA $3\pi r^2$ (cm²)—
this solid's TSA (cm²)—
saved by joining $2\pi r^2$ (cm²)—
✏️ Example 6 — A cylinder on a cuboid
A trophy stand is a cuboid $14$ cm $\times$ $14$ cm $\times$ $4$ cm with a solid cylinder of radius $3.5$ cm and height $10$ cm standing on the middle of its top face. Find the total surface area. (Use $\pi=\tfrac{22}{7}$.)
Figure The cylinder covers a circle of area $\pi r^2$ of the top face of the cuboid. Its own top face $\pi r^2$ is free.
Cuboid $2(14\times14+14\times4+14\times4)=2\times308=616$ cm².
Cylinder Curved wall $2\pi rh=2\times\tfrac{22}{7}\times3.5\times10=220$ cm². Top face $\pi r^2=38.5$ cm². Covered circle on the cuboid: $-38.5$ cm².
Add $616-38.5+220+38.5=836$ cm².
Remark The covered circle and the cylinder's top circle cancel, so a cylinder of the same radius placed on a flat face adds exactly its curved wall $2\pi rh$.

Exercises

I
The basic solids. State the value of $\pi$ you use ($\tfrac{22}{7}$ unless told otherwise); give units.
  • 1 A cone has base radius $7$ cm and height $24$ cm. Find its slant height, curved surface area and total surface area.
  • 2 A cylinder has radius $3.5$ m and height $10$ m. Find its curved and total surface area.
  • 3 Find the curved and the total surface area of a hemisphere of radius $10.5$ cm.
  • 4 Find the surface area of a sphere of diameter $21$ cm.
  • 5 Find the total surface area of (a) a cube of edge $6$ cm; (b) a cuboid $12$ cm $\times$ $5$ cm $\times$ $4$ cm.
  • 6 A cone has base radius $5$ cm and slant height $13$ cm. Find its height and its curved and total surface area. (Use $\pi=3.14$.)
  • 7 The curved surface area of a cylinder of radius $7$ cm is $440$ cm². Find its height.
  • 8 The total surface area of a solid hemisphere is $462$ cm². Find its radius.
1 $l=\sqrt{49+576}=25$ cm. CSA $=\tfrac{22}{7}\times7\times25=550$ cm². TSA $=550+\tfrac{22}{7}\times49=550+154=704$ cm².
2 CSA $=2\times\tfrac{22}{7}\times3.5\times10=220$ m². TSA $=220+2\times\tfrac{22}{7}\times12.25=220+77=297$ m².
3 $r^2=110.25$. CSA $=2\pi r^2=2\times\tfrac{22}{7}\times110.25=693$ cm². TSA $=3\pi r^2=1039.5$ cm².
4 $r=10.5$ cm: $4\pi r^2=4\times\tfrac{22}{7}\times110.25=1386$ cm².
5 (a) $6\times36=216$ cm². (b) $2(60+20+48)=256$ cm².
6 $h=\sqrt{169-25}=12$ cm. CSA $=3.14\times5\times13=204.10$ cm². TSA $=204.10+3.14\times25=204.10+78.50=282.60$ cm².
7 $2\times\tfrac{22}{7}\times7\times h=44h=440$, so $h=10$ cm.
8 $3\pi r^2=462$ gives $r^2=\dfrac{462\times7}{3\times22}=49$, so $r=7$ cm.
II
Combinations: add only the faces on the outside.
  • 1 An ice-cream cone has a hemisphere of ice cream of radius $3.5$ cm on a cone of the same radius and height $12$ cm. Find the area of the outer surface of cone and ice cream together. (Use $\pi=\tfrac{22}{7}$.)
  • 2 A solid is a cylinder with a hemisphere at each end. The radius is $3.5$ cm and the total length is $20$ cm. Find its total surface area. (Use $\pi=\tfrac{22}{7}$.)
  • 3 A cube of edge $10$ cm has a hemisphere of radius $5$ cm scooped out of its top face. Find the total surface area of the remaining solid. (Use $\pi=3.14$.)
  • 4 A solid cylinder of radius $7$ cm and height $10$ cm has a hemispherical hollow of radius $7$ cm scooped from its top. Find its total surface area. (Use $\pi=\tfrac{22}{7}$.)
  • 5 A model rocket is a cylinder of radius $3$ cm and height $10$ cm with a cone of the same radius and height $4$ cm on top. Find its total surface area including the flat base. (Use $\pi=3.14$.)
  • 6 A cuboid $10$ cm $\times$ $10$ cm $\times$ $5$ cm carries a solid cylinder of radius $3$ cm and height $8$ cm on the middle of its top face. Find the total surface area. (Use $\pi=3.14$.)
  • 7 A spinning top is a cone of radius $6$ cm and slant height $10$ cm on a hemisphere of radius $6$ cm. Find its total surface area. (Use $\pi=3.14$.)
  • 8 A cube of edge $14$ cm has a hemispherical hollow of radius $7$ cm scooped out of each of two opposite faces. Find the total surface area of the solid. (Use $\pi=\tfrac{22}{7}$.)
1 $l=\sqrt{3.5^2+12^2}=\sqrt{156.25}=12.5$ cm. Cone $\pi rl=\tfrac{22}{7}\times3.5\times12.5=137.5$; hemisphere $2\pi r^2=77$. Total $214.5$ cm².
2 $h=20-7=13$ cm. $2\pi rh=286$; $4\pi r^2=154$. Total $440$ cm².
3 $6a^2-\pi r^2+2\pi r^2=600+3.14\times25=600+78.5=678.50$ cm².
4 Wall $2\pi rh=440$; base $\pi r^2=154$; inside of the hollow $2\pi r^2=308$ (the top face disappears completely). Total $902$ cm².
5 Cone: $l=\sqrt{9+16}=5$, $\pi rl=47.1$. Wall $2\pi rh=188.4$. Base $\pi r^2=28.26$. Total $263.76$ cm².
6 Cuboid $2(100+50+50)=400$. Cylinder wall $2\pi rh=150.72$. The covered circle and the cylinder's top circle cancel. Total $550.72$ cm².
7 Cone $\pi rl=3.14\times6\times10=188.4$; hemisphere $2\pi r^2=226.08$. Total $414.48$ cm².
8 $6a^2=1176$. Each hollow removes $\pi r^2=154$ and adds $2\pi r^2=308$, a net $+154$; two hollows: $+308$. Total $1484$ cm².
III
Applications with cost.
  • 1 A hemispherical dome of radius $7$ m is painted on the outside (not the floor) at ₹ $12$ per m². Find the cost. (Use $\pi=\tfrac{22}{7}$.)
  • 2 A tanker is a cylinder of radius $1$ m and length $8$ m with a hemisphere at each end. Find the cost of painting it at ₹ $40$ per m². (Use $\pi=3.14$.)
  • 3 A tent has a cylindrical wall of radius $3.5$ m and height $4$ m with a conical roof of the same radius and height $12$ m. Find the area of canvas and its cost at ₹ $200$ per m². (Use $\pi=\tfrac{22}{7}$.)
  • 4 The toy of Example 1 is polished at $50$ paise per cm². Find the cost of polishing one toy.
  • 5 A pencil holder is a solid wooden cylinder of radius $3.5$ cm and height $12$ cm with a conical hollow of the same base and height drilled from the top. Find the area to be polished, inside and outside. (Use $\pi=\tfrac{22}{7}$.)
  • 6 A closed container is a cylinder of radius $7$ cm and height $20$ cm with a hemisphere attached at one end. Find its total surface area. (Use $\pi=\tfrac{22}{7}$.)
1 $2\pi r^2=2\times\tfrac{22}{7}\times49=308$ m². Cost $308\times12=$ ₹ $3696$.
2 $h=8-2=6$ m. $2\pi rh=37.68$; $4\pi r^2=12.56$. Area $50.24$ m². Cost $50.24\times40=$ ₹ $2009.60$.
3 $l=\sqrt{3.5^2+12^2}=12.5$ m. Wall $2\pi rh=88$; roof $\pi rl=137.5$. Canvas $225.5$ m². Cost $225.5\times200=$ ₹ $45{,}100$.
4 Area $858$ cm², cost $858\times0.50=$ ₹ $429$.
5 Outside wall $2\pi rh=264$; bottom $\pi r^2=38.5$; inside cone $\pi rl=137.5$ (with $l=12.5$). Total $440$ cm².
6 Base $\pi r^2=154$; wall $2\pi rh=880$; dome $2\pi r^2=308$. Total $1342$ cm².
IV
Show that…
  • 1 A sphere of radius $r$ fits exactly inside a cylinder of radius $r$ and height $2r$. Show that the surface area of the sphere equals the curved surface area of the cylinder.
  • 2 A capsule is a cylinder with a hemisphere at each end, radius $r$ and total length $L$. Show that its total surface area is $2\pi rL$.
  • 3 A cone of radius $r$ and height $r\sqrt3$ is placed on a hemisphere of radius $r$. Show that the total surface area of the toy is $4\pi r^2$.
  • 4 A hemisphere of the largest possible radius is fixed on a face of a cube of edge $a$. Show that the total surface area is $6a^2+\dfrac{\pi a^2}{4}$ and check it for $a=14$ cm.
1 Cylinder CSA $=2\pi r\times2r=4\pi r^2$, which is the surface area of the sphere.
2 $h=L-2r$. TSA $=2\pi r(L-2r)+4\pi r^2=2\pi rL-4\pi r^2+4\pi r^2=2\pi rL$.
3 $l=\sqrt{r^2+3r^2}=2r$. Cone $\pi r\cdot2r=2\pi r^2$; hemisphere $2\pi r^2$. Total $4\pi r^2$.
4 The hemisphere's radius is $\tfrac a2$. TSA $=6a^2-\pi\dfrac{a^2}{4}+2\pi\dfrac{a^2}{4}=6a^2+\dfrac{\pi a^2}{4}$. For $a=14$: $1176+\tfrac{22}{7}\times49=1176+154=1330$ cm², as in Example 4.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksBoard · 5 marksBoard · Case study

How this is examined: a surface-area question on combinations is worth $2$ to $5$ marks. Credit goes to naming the faces that are counted, the slant height, each area with its formula, the sum and the unit; a cost or a "how much canvas" part is often added. MCQs test whether you add hidden faces by mistake. State the value of $\pi$ you use.

E1. Board · 1 mark A toy is a cone of base radius $3$ cm and height $4$ cm joined to a hemisphere of radius $3$ cm along their bases. Its total surface area is  (A) $33\pi$ cm²  (B) $42\pi$ cm²  (C) $27\pi$ cm²  (D) $24\pi$ cm²

Solve $l=\sqrt{9+16}=5$ cm. Cone $\pi rl=15\pi$, hemisphere $2\pi r^2=18\pi$. Total $33\pi$ cm².
Trap (B) $42\pi$ adds the hidden base $9\pi$. (D) $24\pi$ is the total surface area of the cone alone; (C) $27\pi$ that of the hemisphere alone.
Answer (A).

E2. Board · 1 mark Assertion (A): A hemisphere of radius $r$ is fixed on one flat end of a cylinder of radius $r$. The total surface area of the combination is not the sum of the total surface areas of the two solids. Reason (R): The two circular faces that are joined are no longer on the outside.
(A) Both A and R are true and R is the correct explanation of A.  (B) Both A and R are true but R is not the correct explanation of A.  (C) A is true but R is false.  (D) A is false but R is true.

Check The joined circles are each of area $\pi r^2$; the true total is $\text{TSA}_1+\text{TSA}_2-2\pi r^2$, which is smaller than the sum, so A is true. R is true: the joined faces are inside.
Link R gives exactly the reason for A.
Answer (A).

E3. Board · 2 marks A toy is a cone of radius $5$ cm and height $12$ cm on a hemisphere of radius $5$ cm. Find its total surface area. (Use $\pi=3.14$.)

Slant height (½ mark) $l=\sqrt{25+144}=13$ cm.
Areas (1 mark) Cone $\pi rl=3.14\times5\times13=204.1$ cm²; hemisphere $2\pi r^2=2\times3.14\times25=157$ cm².
Total (½ mark) $204.1+157=361.10$ cm².

E4. Board · 3 marks From a solid cylinder of radius $5$ cm and height $12$ cm a conical cavity of the same base radius and height is hollowed out. Find the total surface area of the remaining solid. (Use $\pi=3.14$.)

Faces (1 mark) Outer wall $2\pi rh$, bottom $\pi r^2$ and inner cone $\pi rl$ (the top face is entirely removed). $l=\sqrt{25+144}=13$ cm.
Areas (1 mark) $2\pi rh=120\pi$, $\pi r^2=25\pi$, $\pi rl=65\pi$.
Total (1 mark) $210\pi=210\times3.14=659.40$ cm².

E5. Board · 5 marks A reception tent is cylindrical to a height of $3$ m, with diameter $14$ m, and has a conical roof of slant height $12.5$ m. Find the area of canvas used and the cost of the canvas at ₹ $80$ per m². (Use $\pi=\tfrac{22}{7}$.)

Figure (1 mark) $r=7$ m, wall height $h=3$ m, $l=12.5$ m. Canvas covers the wall and the roof only.
Wall (1 mark) $2\pi rh=2\times\tfrac{22}{7}\times7\times3=132$ m².
Roof (1 mark) $\pi rl=\tfrac{22}{7}\times7\times12.5=275$ m².
Area (1 mark) $132+275=407$ m².
Cost (1 mark) $407\times80=32560$, so ₹ $32{,}560$.

E6. Board · Case study A designer makes a table lamp: a wooden cube of edge $21$ cm with a hemispherical glass dome of radius $7$ cm fixed on the middle of its top face. All outside surfaces, including the bottom, are to be polished. (Use $\pi=\tfrac{22}{7}$.)
(i) Find the curved surface area of the dome and the area of the top face of the cube that remains visible. (2 marks)
(ii) Find the total area of the other five faces of the cube. (1 mark)
(iii) Find the total area to be polished. (1 mark)

(i) Dome $2\pi r^2=2\times\tfrac{22}{7}\times49=308$ cm². Top face of cube $441$ cm² minus the covered circle $\pi r^2=154$ cm²: $287$ cm². (1 + 1 marks)
(ii) $5\times21^2=2205$ cm². (1 mark)
(iii) $2205+287+308=2800$ cm². (1 mark)
🔬 Research–Teaching NexusMini project · optional, about 45 minutes

Research question. If you build a paper tent or rocket from a cylinder and a cone, does the paper you cut have the area that the formulas predict?

Why it matters. Tent makers, packaging designers and sheet-metal workers must order material before they cut. Too little wastes the whole piece; too much wastes money. They compute the area of every exposed face from a plan, exactly as in this section, and then add a margin for joins.

What to do.

  1. Choose a radius $r$ (for example $4$ cm), a cylinder height $h$ (for example $8$ cm) and a cone slant height $l$ (for example $6$ cm). Compute the predicted paper area $2\pi rh+\pi rl$ with $\pi=3.14$.
  2. On chart paper draw a rectangle $2\pi r$ by $h$ (the wall) and a sector of radius $l$ with arc length $2\pi r$ (the roof; its angle is $\theta=\dfrac{2\pi r}{2\pi l}\times360^\circ$, so for these values $240^\circ$). Cut both out and tape them into a tent.
  3. Cut a $10$ cm $\times$ $10$ cm square from the same paper. Weigh it, then weigh your two pieces on a kitchen scale (or a postal scale) to estimate their area as $100\times\dfrac{\text{mass of pieces}}{\text{mass of square}}$ cm².
  4. Compare the two areas and compute the percentage difference.
  5. Repeat for a different $r$ and explain which faces you did not cut (the floor and the circle where the roof meets the wall).

Tools: Chart paper, scissors, ruler, compass, tape, a scale that reads in grams (or $0.1$ g)

Hand in: Your tent, a table of predicted and measured areas, and a paragraph on the main source of error.

Going further. Add a hemisphere "dome" made of paper strips and discuss why a sphere cannot be flattened without stretching.

Teacher/parent note: for a thin sheet the mass is proportional to the area, so the scale gives the area without any measuring. A small error in the sector's angle changes the cone noticeably — a good moment to talk about measurement error.

🧪 Self-check quiz

0/5 correct

Q1. A cone has base radius $7$ cm and height $24$ cm. Its curved surface area (with $\pi=\tfrac{22}{7}$) is

Q2. A hemisphere of radius $r$ is fixed on one flat end of a cylinder of radius $r$ and height $h$ (the other end is a free flat face). The total surface area is

Q3. A hemisphere of radius $5$ cm is scooped out of the top face of a cube of edge $10$ cm. The total surface area of the remaining solid ($\pi=3.14$) is

Q4. A capsule is a cylinder of radius $7$ cm and height $10$ cm with a hemisphere at each end. Its total surface area ($\pi=\tfrac{22}{7}$) is

Q5. Two solids are joined face to face. Compared with the sum of their separate total surface areas, the total surface area of the combination is

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