Key Concepts at a Glance
| Solid | Surface area | Volume |
|---|---|---|
| Cuboid (l, w, h) | 2(wl + hw + lh) | lwh |
| Cube (side a) | 6a² | a³ |
| Right circular cylinder | CSA 2πrh; TSA 2πr(h + r); one end closed πr(2h + r) | πr²h |
| Right circular cone | CSA πrl; TSA πr(l + r) | ⅓πr²h |
| Pyramid | sum of the areas of its faces | ⅓ × base area × height |
| Sphere | 4πr² | ⁴⁄₃πr³ |
| Hemisphere | CSA 2πr²; TSA 3πr² | ⅔πr³ |
- For a cone, the slant height satisfies l² = h² + r².
- A cone holds exactly one-third of a cylinder with the same base and height.
- A sphere's surface area equals the curved surface of the cylinder that fits tightly around it (Archimedes).
- Scaling all lengths by k multiplies surface area by k² and volume by k³.
The chapter works out surface areas and volumes of the basic solids — cuboids, cylinders, cones, pyramids, spheres and hemispheres — explaining each formula by a model: stacked cards for a cuboid, stacked khakhras for a cylinder, an unrolled sector for a cone, and Archimedes' wrapping cylinder for a sphere.
It then turns to guesstimates: real questions (how many golgappas, how many tennis balls in a room) where you choose sensible measurements, state your assumptions and estimate. For these, there is no single right number — a clear, well-reasoned estimate is the goal.
In-Text Questions & Think and Reflect
Answers to both Think and Reflect boxes and the questions the chapter asks inside its explanations, in the order they appear in the textbook.
Think and Reflect
TRTry to work out for yourself why this model explains the formula for the volume of a cuboid, i.e., volume = area of base × height.
Write the formula for the surface area and the volume of a cube.
Think of the cuboid as a stack of many identical thin rectangular sheets, each of length l and width w, like cards in a pack. Each sheet covers an area lw.
If each sheet is t units thick and there are n sheets, the stack is h = nt high. Each sheet has volume (area × thickness) = lw × t, so the whole stack has volume
i.e. volume = area of base × height. Making the sheets thinner and thinner does not change this.
Cube of side a: all six faces are squares of area a², so
Total surface area = 6a², Volume = a × a × a = a³.
Think and Reflect
TRSuppose a swimming pool is being built in your school. Two choices are available for the dimensions of the pool: (A) 8 ft (depth) × 50 ft × 20 ft and (B) 6 ft (depth) × 100 ft × 15 ft. Which option would be suitable/appropriate for your school? Why? What parameters/aspects did you consider to make the choice? [1 cubic foot ≈ 28.3 litres]
Volumes (water needed):
(A) 8 × 50 × 20 = 8000 ft³ ≈ 8000 × 28.3 = 2,26,400 litres.
(B) 6 × 100 × 15 = 9000 ft³ ≈ 9000 × 28.3 = 2,54,700 litres.
Other aspects to consider:
• Safety: 8 ft (about 2.4 m) is too deep for most school students; 6 ft (about 1.8 m) is safer.
• Use for swimming laps: B is 100 ft (about 30 m) long, good for lap swimming and races; A is only 50 ft long.
• Area of floor and walls (cost of tiles): A: 50 × 20 + 2(8 × 50 + 8 × 20) = 2120 ft²; B: 100 × 15 + 2(6 × 100 + 6 × 15) = 2880 ft², so B costs more to tile.
• Space available on the campus, water supply and running cost also matter.
A reasonable choice: B, because it is safer for students and better for swimming, even though it needs about 28,000 litres more water and more tiling. (Other choices are fine if justified.)
ITBut you may be able to find your own derivation of the formula by connecting it with the formula for volume of a cone; namely, by writing it as \(V = \frac{1}{3} \times (4\pi r^2) \times r\), i.e., Volume of sphere = \(\frac{1}{3}\) × surface area of sphere × radius of sphere. Try to work out the connecting links on your own.
Divide the whole surface of the sphere into a very large number of tiny pieces. Join every piece to the centre O. This splits the solid sphere into many thin "cones" (or pyramids), each with its tiny base on the surface and its apex at O.
Every one of these thin cones has the same height: the radius r.
Volume of one thin cone = \(\frac13\) × (its base area) × r.
Adding all of them:
since the bases together make up the whole surface area 4πr².
ITUsing the other configurations, we might be able to fit more balls because of higher packing efficiency. Make an estimate based on the second and third configurations.
Room ≈ 900 cm × 900 cm × 450 cm; ball diameter 6 cm. The first configuration (balls directly on top of each other) gave 150 × 150 × 75 = 16,87,500 balls.
Layers sitting in the hollows of the layer below. In the other two arrangements, each layer rests in the gaps of the one below instead of directly on top, so the layers are closer together.
Square layers, next layer in the hollows: a ball resting in the hollow between 4 balls is raised only \(\frac{6}{\sqrt2} \approx 4.24\) cm above the layer below. Number of layers ≈ 450 ÷ 4.24 ≈ 106. The layers alternate between 150 × 150 = 22,500 and 149 × 149 = 22,201 balls, so the total ≈ 53 × 22,500 + 53 × 22,201 ≈ 23,70,000 balls.
Triangular (hexagonal) layers: in each layer, rows are \(6 \times \frac{\sqrt3}{2} \approx 5.2\) cm apart, so about 900 ÷ 5.2 ≈ 173 rows of about 150 balls ≈ 25,900 balls per layer. The next layer, sitting in the hollows, is about 4.9 cm higher, giving about 450 ÷ 4.9 ≈ 91 layers. Total ≈ 91 × 25,900 ≈ 23,60,000 balls.
Both ways fill about 74% of the space, compared with about 52% for the first configuration, so roughly 40% more balls fit. Still fewer than the upper bound of 33,45,132.
Exercise Set 14.1
Cuboids and cubes.
1The volume of a cube is 64 cm³. What is its total surface area?
a³ = 64 ⇒ a = 4 cm.
TSA = 6a² = 6 × 16 = 96 cm².
2How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?
2 m = 200 cm. Along each edge, 200 ÷ 20 = 10 small cubes fit.
Number of cubes = 10 × 10 × 10 = 1000.
3The dimensions of a godown are 40 m × 25 m × 10 m. If it is filled with cuboidal boxes, each of dimensions 2 m × 1.25 m × 1 m, then find the number of boxes.
Volume of godown = 40 × 25 × 10 = 10000 m³. Volume of one box = 2 × 1.25 × 1 = 2.5 m³.
Number of boxes = 10000 ÷ 2.5 = 4000.
(They fit exactly: 40 ÷ 2 = 20, 25 ÷ 1.25 = 20, 10 ÷ 1 = 10, and 20 × 20 × 10 = 4000.)
4Two cubes each of volume 125 cm³ are joined end to end. Find the surface area of the resulting cuboid.
Side of each cube = ∛125 = 5 cm. The cuboid is 10 cm × 5 cm × 5 cm.
TSA = 2(10 × 5 + 5 × 5 + 10 × 5) = 2(50 + 25 + 50) = 250 cm².
(Check: two cubes have 2 × 150 = 300 cm², minus the two joined faces, 2 × 25 = 50 cm².)
5A cube of side 4 cm is cut into cubes of side 1 cm. What is the ratio of the surface areas of the original cube and all the cut-out cubes? (Note that there is no change in volume but a big change in the surface area. This property has major consequences in the biological world.)
Original cube: 6 × 4² = 96 cm².
Number of 1 cm cubes = 4³ = 64; total surface area = 64 × 6 = 384 cm².
Ratio = 96 : 384 = 1 : 4.
6The surface areas of the three faces of a cuboid that meet at one of the corners of the cuboid are 6 cm², 15 cm², and 10 cm² respectively. What is the volume of the cuboid?
Let the edges be l, w, h with lw = 6, wh = 15, hl = 10.
Multiplying: (lw)(wh)(hl) = (lwh)² = 6 × 15 × 10 = 900, so lwh = 30.
(The edges are 2 cm, 3 cm and 5 cm.)
7*A cube of side 5 cm is painted on all its faces. If it is sliced into 1 cm³ cubes, how many of these 1 cm³ cubes have
(i) exactly three faces painted?
(ii) exactly two faces painted?
(iii) exactly one face painted?
(iv) no face painted?
There are 5 × 5 × 5 = 125 small cubes.
(i) Only the corner cubes: 8.
(ii) Cubes along an edge, excluding the corners: 5 − 2 = 3 per edge, 12 edges: 36.
(iii) Cubes in the middle of each face: (5 − 2)² = 9 per face, 6 faces: 54.
(iv) The inner cube of side 3: 3³ = 27.
Check: 8 + 36 + 54 + 27 = 125 ✓.
8*Find a cuboid with edges whose lengths are integers (in cm), given that it has a total surface area of exactly 100 cm².
(i) Is there more than one such cuboid?
(ii) Can you find them all?
(iii) Show that you have found them all.
2(lw + wh + hl) = 100 ⇒ lw + wh + hl = 50. List the edges in order l ≤ w ≤ h.
Example: 1 × 2 × 16: 2 + 32 + 16 = 50 ✓.
(i) Yes — 2 × 4 × 7 also works: 8 + 28 + 14 = 50 ✓.
(ii) These are the only two: 1 × 2 × 16 and 2 × 4 × 7.
(iii) Since l is the smallest, lw + wh + hl ≥ 3l², so 3l² ≤ 50 and l ≤ 4. Add l² to both sides of lw + wh + hl = 50 to factorise: (w + l)(h + l) = 50 + l².
l = 1: (w + 1)(h + 1) = 51 = 3 × 17 (w ≥ 1 rules out 1 × 51) ⇒ w = 2, h = 16.
l = 2: (w + 2)(h + 2) = 54 with w + 2 ≥ 4 and w + 2 ≤ h + 2 ⇒ 6 × 9 ⇒ w = 4, h = 7.
l = 3: (w + 3)(h + 3) = 59, a prime — no solution with w ≥ 3.
l = 4: (w + 4)(h + 4) = 66 with 8 ≤ w + 4 ≤ √66 ≈ 8.1 ⇒ w + 4 = 8, but 8 does not divide 66 — no solution.
Exercise Set 14.2
Right circular cylinders.
1Two cylinders, A and B, are given. The radius of cylinder B is twice that of cylinder A, and the height of cylinder B is half that of cylinder A. Find the ratio of the curved surface area of A to the curved surface area of B. Also find the ratio of the volume of A to the volume of B.
Let A have radius r and height h; then B has radius 2r and height h/2.
CSA: \(\dfrac{2\pi rh}{2\pi (2r)(h/2)} = \dfrac{2\pi rh}{2\pi rh} = 1\), so 1 : 1.
Volume: \(\dfrac{\pi r^2 h}{\pi (2r)^2 (h/2)} = \dfrac{\pi r^2 h}{2\pi r^2 h} = \dfrac12\), so 1 : 2.
2The radii of two cylinders are in the ratio 2:3, and their heights are in the ratio 3:2. Find (a) the ratio of their volumes, and (b) the ratio of their curved surface areas.
Take radii 2k, 3k and heights 3m, 2m.
(a) \(\dfrac{\pi (2k)^2 (3m)}{\pi (3k)^2 (2m)} = \dfrac{12}{18} = \dfrac23\) ⇒ 2 : 3.
(b) \(\dfrac{2\pi (2k)(3m)}{2\pi (3k)(2m)} = \dfrac{6}{6} = 1\) ⇒ 1 : 1.
3The edge of a cube measures r cm. The largest possible right circular cylinder is cut out of the cube. What do you think is the volume of the cylinder (in cm³)?
The largest cylinder stands on one face: its base circle fits inside the square face, so its diameter is r (radius r/2), and its height is r.
That is about 78.5% of the cube's volume.
4*The radius of the base of a cylinder is increased by 10%. At the same time, the height of the cylinder is decreased by x%. Given that the volume of the cylinder remains unchanged, find the value of x.
New radius = 1.1r, new height = \(h\left(1 - \frac{x}{100}\right)\).
\(\pi (1.1r)^2 h\left(1 - \frac{x}{100}\right) = \pi r^2 h\) ⇒ \(1 - \frac{x}{100} = \frac{1}{1.21} = \frac{100}{121}\).
\(\frac{x}{100} = \frac{21}{121}\) ⇒ \(x = \frac{2100}{121} \approx 17.36\).
5A solid metallic cube of side 12 cm is melted and recast into solid cylindrical rods, each having radius 2 cm and height 12 cm. Find:
(i) the volume of the cube,
(ii) the volume of one cylindrical rod,
(iii) the approximate number of complete cylindrical rods that can be formed. \(\left(\pi \approx \frac{22}{7}\right)\)
(i) 12³ = 1728 cm³.
(ii) \(\frac{22}{7} \times 2^2 \times 12 = \frac{1056}{7} \approx 150.86\) cm³.
(iii) 1728 ÷ 150.86 ≈ 11.45, so 11 complete rods (with some metal left over).
Exercise Set 14.3
For π, use one of the following approximations: \(\pi \approx \frac{22}{7}\) or π ≈ 3.14
1Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m.
r = 12 m, l = 21 m.
TSA = πr(l + r) = \(\frac{22}{7}\) × 12 × 33 = \(\frac{8712}{7}\) ≈ 1244.57 m².
(With π ≈ 3.14: 3.14 × 396 = 1243.44 m².)
2Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.
CSA = πrl = \(\frac{22}{7}\) × 7 × 10 = 220 cm².
3The height of a cone is 16 cm, and its base radius is 12 cm. Find the curved surface area and the total surface area of the cone.
\(l = \sqrt{16^2 + 12^2} = \sqrt{400} = 20\) cm.
CSA = πrl = 3.14 × 12 × 20 = 753.6 cm².
TSA = πr(l + r) = 3.14 × 12 × 32 = 1205.76 cm².
4A cone has a height of 15 cm. If its volume is 1570 cm³, find the radius of the base.
\(1570 = \frac13 \times 3.14 \times r^2 \times 15 = 15.7r^2\) ⇒ r² = 100 ⇒ r = 10 cm.
5The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find (i) the radius of the base, (ii) the total surface area of the cone.
(i) \(\frac{22}{7} \times r \times 14 = 308\) ⇒ 44r = 308 ⇒ r = 7 cm.
(ii) TSA = 308 + πr² = 308 + \(\frac{22}{7}\) × 49 = 308 + 154 = 462 cm².
6A joker’s cap is in the form of a right circular cone of base radius 7 cm and height 24 cm. Find the area of the sheet required to make 10 such caps.
\(l = \sqrt{24^2 + 7^2} = \sqrt{625} = 25\) cm.
CSA of one cap = \(\frac{22}{7}\) × 7 × 25 = 550 cm². For 10 caps: 5500 cm².
7*What length of tarpaulin 3 m wide is required to make a conical tent of height 8 m and base radius 6 m? Assume that the extra length of material that is required for stitching margins and wastage in cutting is 20 cm.
\(l = \sqrt{8^2 + 6^2} = 10\) m. CSA = πrl = 3.14 × 6 × 10 = 188.4 m².
Length needed = 188.4 ÷ 3 = 62.8 m. Adding 20 cm = 0.2 m: 63 m.
8A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through 360° about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.
The solid is a cone with h = 8 cm, r = 6 cm, l = 10 cm.
Volume = \(\frac13\)π × 36 × 8 = 96π ≈ 301.44 cm³.
CSA = π × 6 × 10 = 60π ≈ 188.4 cm².
9*Suppose you have a cup in the shape of a right circular cone. Fill it with water to half the depth of the cone. What fraction of the volume of the cup is occupied by the water?
Cross-section of the cup (vertex down). The water forms a smaller cone with half the height and half the radius.
Let the cup have height H and top radius R, with the vertex at the bottom. The water forms a smaller cone with height H/2.
In the cross-section, the water surface joins the midpoints of the two slant sides, so by the Midpoint Theorem its width is half the top width: the water's radius is R/2.
Exercise Set 14.4
Spheres and hemispheres.
1A ball bearing has a radius of 0.7 cm. Find its surface area.
4πr² = 4 × \(\frac{22}{7}\) × 0.49 = 6.16 cm².
2*Two solid spheres made of the same metal have weights 5920 g and 740 g. Determine the radius of the larger sphere, if the diameter of the smaller one is 5 cm.
Same metal, so weight is proportional to volume, i.e. to r³.
\(\frac{R^3}{r^3} = \frac{5920}{740} = 8\) ⇒ \(\frac{R}{r} = 2\).
r = 2.5 cm, so R = 5 cm.
3The diameter of the moon is approximately one fourth the diameter of the earth. Given that the moon and earth are both roughly spherical, find the ratio of their surface areas.
Radii are in the ratio 1 : 4, so surface areas (∝ r²) are in the ratio 1² : 4² = 1 : 16.
4Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.
(i) 2πr² = 2 × \(\frac{22}{7}\) × 441 = 2772 cm².
(ii) 3πr² = 3 × \(\frac{22}{7}\) × 441 = 4158 cm².
5Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions 16 cm × 8 cm × 8 cm. When 16 spheres are packed, the box is filled with preservative liquid. Find the volume of this liquid. Round your answer to the nearest integer.
Box: 16 × 8 × 8 = 1024 cm³.
One sphere: \(\frac43\pi(2)^3 = \frac{32\pi}{3}\). Sixteen spheres: \(\frac{512\pi}{3} \approx 536.2\) cm³.
Liquid = 1024 − 536.2 ≈ 487.8 cm³.
6The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.
New volume = \(\frac43\pi(1.1r)^3 = 1.331 \times \frac43\pi r^3\).
Increase = 1.331 − 1 = 0.331 of the original, i.e. 33.1%.
7*The radius of a sphere is increased by x%. The volume of the sphere increases by 72.8%. Find the value of x.
\(\left(1 + \frac{x}{100}\right)^3 = 1.728 = (1.2)^3\) ⇒ \(1 + \frac{x}{100} = 1.2\) ⇒ x = 20.
8The hemispherical dome of a building needs to be painted (see Fig. 14.16). If the circumference of the base of the dome is 35.2 m, find the cost of painting it, given the cost of painting is ₹10 per 100 cm².
2πr = 35.2 ⇒ \(r = \frac{35.2 \times 7}{44} = 5.6\) m.
Curved surface area = 2πr² = 2 × \(\frac{22}{7}\) × 31.36 = 197.12 m² = 19,71,200 cm².
Cost = \(\frac{1971200}{100}\) × 10 = ₹1,97,120.
End-of-Chapter Exercises
For guesstimate problems, make a guess first before solving. All 22 End-of-Chapter Exercises are solved below; answers to guesstimates depend on the assumptions made. Starred (*) questions are the more challenging ones.
1Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions 10 cm × 15 cm × 20 cm.
Guess first, then estimate.
Container: 10 × 15 × 20 = 3000 cm³.
Assume a scoop is a ball of radius about 2.5 cm: \(\frac43 \times 3.14 \times 2.5^3 \approx 65\) cm³.
Number of scoops ≈ 3000 ÷ 65 ≈ 46.
Ice cream gets pressed while scooping, so a sensible answer is about 40–45 scoops. A different scoop size gives a different answer — what matters is the reasoning.
2A cube of integer side length a is made of unit cubes. Write an expression giving the number of unit cubes to be added to make a cube of side length a + 1.
\((a+1)^3 - a^3 = 3a^2 + 3a + 1\).
Picture: three slabs a × a × 1 on three faces (3a²), three rods a × 1 × 1 along the edges between them (3a), and one corner cube (1).
3*Solve the following:
(i) Could a person drink enough water in a lifetime to fill an entire room the size of your classroom?
(ii) Estimate the number of bricks used to build the walls of your classroom.
(i) Assume a classroom of 9 m × 7 m × 3.5 m ≈ 220 m³ = 2,20,000 litres.
A person drinks about 2 litres a day: 2 × 365 × 75 years ≈ 55,000 litres.
So no — a lifetime of drinking fills only about a quarter of the room.
(ii) Walls: perimeter 2(9 + 7) = 32 m, height 3.5 m, thickness about 0.23 m (one brick length). Volume ≈ 32 × 3.5 × 0.23 ≈ 25.8 m³. Subtract about 15% for doors and windows: ≈ 22 m³.
A brick with its mortar takes about 20 cm × 10 cm × 10 cm = 0.002 m³.
Number of bricks ≈ 22 ÷ 0.002 = 11,000 (roughly 10,000–12,000).
4You are given two options: (A) one chocolate cube of side 50 mm, (B) hundred chocolate cubes of side 10 mm. Which option gives you more chocolate?
(A) 50³ = 1,25,000 mm³.
(B) 100 × 10³ = 1,00,000 mm³.
Option A gives more (it is equal to 125 small cubes).
5If all the people in the world are crowded together in one location, how much area would it cover?
World population ≈ 8.2 billion. In a tight crowd, each person needs about 0.5 m × 0.5 m = 0.25 m².
Area ≈ 8.2 × 10⁹ × 0.25 = 2.05 × 10⁹ m² ≈ 2050 km².
That is a square about 45 km on each side — only a little bigger than Delhi (about 1500 km²).
6A school provides milk to its students in cylindrical glasses, each with a diameter of 7 cm. If each glass is filled with milk to a height of 12 cm, how many litres of milk are needed to serve 1600 students? Use 1000 cm³ = 1 litre.
One glass: \(\frac{22}{7} \times 3.5^2 \times 12 = 462\) cm³.
1600 glasses: 462 × 1600 = 7,39,200 cm³ = 739.2 litres.
7The surface area of a sphere of radius 5 cm is five times the area of the curved surface of a cone of radius 4 cm. Find the height and the volume of the cone.
4π(5)² = 100π = 5 × π × 4 × l ⇒ 20πl = 100π ⇒ l = 5 cm.
\(h = \sqrt{5^2 - 4^2} = 3\) cm.
Volume = \(\frac13\)π × 16 × 3 = 16π ≈ 50.29 cm³.
8Take Earth to be a perfect sphere with radius 6370 km. Take Jupiter to be a perfect sphere with radius 69,900 km. Take the Sun to be a perfect sphere with radius 6,95,700 km. Compute approximately:
(i) the ratio of the volume of Jupiter to the volume of the Earth;
(ii) the ratio of the volume of the Sun to the volume of the Earth.
(Calculator can be used.)
Volumes are proportional to r³, so the ratio is (ratio of radii)³.
(i) \(\left(\frac{69900}{6370}\right)^3 \approx (10.97)^3 \approx 1321\).
(ii) \(\left(\frac{695700}{6370}\right)^3 \approx (109.2)^3 \approx 13,00,000\).
9Show that the volume of a sphere is equal to \(\frac{2}{3}\) of the volume of the smallest cylinder which encloses it.
The smallest cylinder enclosing a sphere of radius r has radius r and height 2r (drawing not to scale).
The smallest cylinder enclosing a sphere of radius r has radius r and height 2r. Its volume is πr² × 2r = 2πr³.
10*Suppose the Earth is perfectly spherical. A string is wrapped tightly around the equator of the Earth. Another string, 1 metre longer, is placed around the Earth so that it forms a larger circle, staying the same distance above the ground everywhere. How high above the ground is the second string? Repeat this for: (i) the Moon (ii) Jupiter (iii) a volleyball. Are you surprised by the three answers? Why or why not?
Let the radius be R and the gap be h. Then
So the string is about 16 cm above the ground.
R cancels out, so the answer is the same 16 cm for (i) the Moon, (ii) Jupiter and (iii) a volleyball.
It is surprising at first — one metre seems tiny compared with 40,000 km around the Earth. But circumference grows at a fixed rate 2π per unit of radius, whatever the size of the circle.
11*We have a cylinder with a base radius of r cm and height h cm. A square pyramid is fitted inside it. The square base of the pyramid lies on the base of the cylinder, its corners on the boundary of the cylinder. The apex of the pyramid lies on the top of the cylinder. Find the ratio of the volume of this pyramid to the volume of the cylinder.
Top view: the square base of the pyramid has its corners on the circle, so its diagonal is 2r.
The square's diagonal is a diameter, 2r. Area of a square = \(\frac12\)(diagonal)² = \(\frac12(2r)^2 = 2r^2\).
Pyramid: \(\frac13 \times 2r^2 \times h = \frac23 r^2h\). Cylinder: πr²h.
12What is the change in volume when:
(i) the length of a cuboid with dimensions l, w, h is increased by 1 unit?
(a) 1 cubic unit (b) (lwh + 1) cubic unit (c) wh cubic units (d) lw cubic units (e) lh cubic units
(ii) the radius of a cylinder with dimensions r, h is increased by 1 unit?
(a) 1 cubic unit (b) πr²h cubic units (c) πr² cubic units (d) 2πrh + 2πh cubic units (e) 2πrh + πh cubic units
*(iii) the radius of a sphere is decreased by 1 unit?
(i) (l + 1)wh − lwh = wh. Answer: (c).
(ii) π(r + 1)²h − πr²h = π(2r + 1)h = 2πrh + πh. Answer: (e).
(iii) \(\frac43\pi r^3 - \frac43\pi (r-1)^3 = \frac43\pi(3r^2 - 3r + 1)\). The volume decreases by \(\frac43\pi(3r^2 - 3r + 1)\) cubic units.
13*Given a cube with volume V, express its total surface area S in terms of V.
Side \(a = V^{1/3}\), so \(S = 6a^2 = 6V^{2/3}\).
14*Given a cube with total surface area S, express its volume V in terms of S.
\(6a^2 = S\) ⇒ \(a = \sqrt{\frac{S}{6}}\), so \(V = a^3 = \left(\frac{S}{6}\right)^{3/2}\).
15A cylindrical glass of height 25 cm and radius 4 cm has water up to a height of 16 cm. A crow wants to drink water from this glass. The water must be at a height of 20 cm for the crow to reach it. There are some marbles lying around. How many marbles of radius 1 cm should the crow drop into the glass to make the water reach the required height?
Volume needed to raise the water 4 cm: π × 4² × 4 = 64π cm³.
One marble: \(\frac43\pi(1)^3 = \frac43\pi\) cm³.
Number = \(64\pi \div \frac43\pi = 48\).
16Find the volume of ink in a new ball point pen. Take the necessary measurements and make approximations, as needed.
The ink fills the refill tube, which is a thin cylinder.
Typical measurements: inner diameter about 2 mm (radius 0.1 cm), ink column about 10 cm long.
Volume ≈ 3.14 × 0.1² × 10 ≈ 0.31 cm³, i.e. about 0.3 mL.
Your own pen may give a slightly different value; the method is what matters.
17Solve:
(i) A ball of chapati dough of radius 6 cm is prepared. Estimate how many chapatis can be made from it?
*(ii) Cut a coconut/muskmelon in half and find out the approximate volume of edible coconut flesh/fruit by taking the necessary measurements.
(i) Dough: \(\frac43 \times 3.14 \times 6^3 \approx 904\) cm³.
A rolled chapati is a thin cylinder, radius about 8 cm and thickness about 0.2 cm: 3.14 × 64 × 0.2 ≈ 40 cm³.
Number ≈ 904 ÷ 40 ≈ 22 chapatis.
(ii) Model the flesh as a hollow sphere (a spherical shell). Measure the inner radius of the hard shell (say R ≈ 5.5 cm) and the flesh thickness (say t ≈ 1 cm).
Flesh ≈ \(\frac43\pi\left(R^3 - (R - t)^3\right) = \frac43 \times 3.14 \times (5.5^3 - 4.5^3) \approx \frac43 \times 3.14 \times 75.25 \approx 315\) cm³.
For a muskmelon, measure the outer radius and the radius of the seed cavity, and subtract the rind in the same way.
18Looking at the Ganita Manjari, Grade 9, Part 2 textbook, Sheela wonders:
(i) If all the pages of this textbook were laid out side-by-side on the floor would they cover the entire classroom floor?
(ii) What is the maximum number of textbooks that can fit in an empty storeroom of dimensions 15 ft × 20 ft × 30 ft?
Measure the book: about 16 cm × 23.5 cm, about 1 cm thick, with about 80 sheets (160 pages).
(i) One sheet ≈ 16 × 23.5 ≈ 376 cm². 80 sheets ≈ 30,000 cm² = 3 m² (even counting both sides separately, only about 6 m²). A classroom floor is about 60 m², so no.
(ii) Storeroom: 15 × 20 × 30 = 9000 ft³ ≈ 9000 × 28,317 ≈ 25,48,50,000 cm³. One book ≈ 16 × 23.5 × 1 = 376 cm³.
Maximum ≈ 25,48,50,000 ÷ 376 ≈ 6,78,000 books (an upper limit; in practice fewer, because of gaps).
19*If the entire human population decided to climb into one giant cube, how long would the side have to be?
Population ≈ 8.2 billion. A standing person, packed in, needs about 0.5 m × 0.5 m of floor and 2 m of height: 0.5 m³.
Total ≈ 8.2 × 10⁹ × 0.5 = 4.1 × 10⁹ m³.
Side = ∛(4.1 × 10⁹) ≈ 1600 m ≈ 1.6 km.
(If people were squeezed together as tightly as their bodies' own volume — about 0.07 m³ each — the side would be only about 830 m.)
20*The Earth’s surface has an estimated volume of 1.38 billion km³ of water. Suppose the Earth is a perfect sphere and all this water forms a uniform layer completely covering the Earth’s surface, like a thin water bubble. Estimate the thickness of this water layer. The Earth’s radius is ~6371 km.
(i) Write an expression that gives the thickness of this water layer.
(ii) Simplify the expression in (i) using a calculator.
(i) If the thickness is t, the water is the shell between radius R and R + t:
Because the layer is very thin compared with R, a simpler approximation is \(t \approx \dfrac{V}{4\pi R^2}\) (volume ≈ surface area × thickness).
(ii) 4πR² = 4 × 3.1416 × 6371² ≈ 5.10 × 10⁸ km².
t ≈ 1.38 × 10⁹ ÷ 5.10 × 10⁸ ≈ 2.7 km. The exact formula also gives about 2.70 km.
21*Give the dimension of a cuboid whose volume is halved when its surface area is doubled.
We want two cuboids, where the second has half the volume but twice the surface area of the first.
Example: Cuboid 1: 2 cm × 3 cm × 4 cm. Volume = 24 cm³; surface area = 2(6 + 12 + 8) = 52 cm².
Cuboid 2: 4 cm × 12 cm × \(\frac14\) cm. Volume = 4 × 12 × \(\frac14\) = 12 cm³ (half); surface area = 2(48 + 3 + 1) = 104 cm² (double).
The idea: flattening a solid into a thin slab reduces the volume but greatly increases the surface area. Another example: 4 × 7 × 7 (V = 196, S = 210) and 14 × 14 × \(\frac12\) (V = 98, S = 420).
22Project: Find the volume of your house making necessary approximations. Present how you solved it.
Sample approach.
1. Split the house into simple shapes — each room a cuboid (length × breadth × ceiling height), and a sloping roof as a triangular prism or pyramid if needed.
2. Measure each room with a tape (or count floor tiles of known size).
3. Example: 2 bedrooms 4 m × 3.5 m, hall 5 m × 4 m, kitchen 3 m × 2.5 m, bathroom 2 m × 1.5 m, all 3 m high. Floor area = 14 + 14 + 20 + 7.5 + 3 = 58.5 m². Volume ≈ 58.5 × 3 ≈ 175 m³.
4. Add wall thickness if you want the outside volume, and state all your assumptions.
Extra Practice Questions
Seven extra questions for independent practice once you have gone through the solved questions above. Try each one, then tap to check your answer.
1Find the total surface area and volume of a cuboid 8 cm × 6 cm × 5 cm.
TSA = 2(48 + 30 + 40) = 236 cm². Volume = 8 × 6 × 5 = 240 cm³.
2Find the curved surface area and volume of a cylinder of radius 7 cm and height 10 cm. (π = 22/7)
CSA = 2 × \(\frac{22}{7}\) × 7 × 10 = 440 cm². Volume = \(\frac{22}{7}\) × 49 × 10 = 1540 cm³.
3A cone has radius 3 cm and height 4 cm. Find its slant height, curved surface area and volume. (π = 3.14)
l = 5 cm. CSA = 3.14 × 3 × 5 = 47.1 cm². Volume = ⅓ × 3.14 × 9 × 4 = 37.68 cm³.
4Find the surface area and volume of a sphere of radius 3.5 cm. (π = 22/7)
SA = 4 × \(\frac{22}{7}\) × 12.25 = 154 cm². Volume = \(\frac43 \times \frac{22}{7} \times 42.875 \approx 179.67\) cm³.
5If the edge of a cube is doubled, how do its surface area and volume change?
Surface area ∝ a², so it becomes 4 times; volume ∝ a³, so it becomes 8 times.
6Find the total surface area and volume of a solid hemisphere of radius 7 cm. (π = 22/7)
TSA = 3 × \(\frac{22}{7}\) × 49 = 462 cm². Volume = \(\frac23 \times \frac{22}{7} \times 343 \approx 718.67\) cm³.
7A cone and a cylinder have the same base radius and the same height. What is the ratio of their volumes?
Cone: ⅓πr²h; cylinder: πr²h.