NCERT Solutions Class 9 Maths Ch 13: Two Variables, One Line

Surface area = 6a² and volume = a³, where a is the side.

1. VolumeThink of a cuboid as a stack of thin layers. For whole-number sides, each layer of height 1 holds l × w unit cubes, and there are h layers: volume = l × w × h = area of base × height. For a cube of side a: volume = a × a × a = a³.
2. Surface areaA cube has 6 square faces, each of area a². So the surface area is 6a².

There is no single right answer. Pool A holds about 2,26,400 litres and pool B about 2,54,700 litres; the choice depends on safety, land and water.

1. Pool A

   	Pool A	Pool B
   Depth × length × breadth	8 × 50 × 20 ft	6 × 100 × 15 ft
   Area of the base	1000 ft²	1500 ft²
   Volume	8000 ft³	9000 ft³
   Water needed (× 28.3 L)	≈ 2,26,400 litres	≈ 2,54,700 litres

2. ComparingB is 2 ft shallower and long and narrow, which is good for swimming laps, but it needs 500 ft² more land and about 28,300 litres more water. A is shorter, wider and deeper.
3. What to considerSafety and the depth for beginners, the age of the students, the land and water available, the cost of building and cleaning, and the use (laps or play). If land or water is limited, A may be more practical; if the pool is mainly for lap swimming and beginners, B (shallower) may suit better. Either choice is fine if it is explained with these reasons.

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

Both have volume 216 cm³. The cube has surface area 216 cm² and the cuboid 228 cm², so the cuboid has the greater surface area.

1. Cube AV = 6³ = 216 cm³   and   S = 6 × 6² = 216 cm²
2. Cuboid BV = 9 × 6 × 4 = 216 cm³S = 2(9 × 6 + 6 × 4 + 4 × 9) = 2(54 + 24 + 36) = 228 cm²
3. Use in real lifeFor the same volume, a cube needs less covering material (good for packaging), while a shape with more surface area exposes more area to its surroundings (good when heat has to be lost or gained quickly, as in a radiator).

550 cm² of chart paper.

1. Curved surface area of a cylinder2πrh = 2 × 227 × 3.5 × 25 = 550 cm²

The solid is a cone, and its volume is 100π ≈ 314.3 cm³.

1. The solid5² + 12² = 169 = 13², so the triangle is right-angled with legs 5 and 12. Rotating it about the leg of 12 cm gives a cone with height h = 12 cm and radius r = 5 cm.
2. Volume13πr²h = 13 × π × 25 × 12 = 100π ≈ 314.3 cm³

About 89.8 cm³.

1. Volume of a hemisphere23πr³ = 23 × 227 × 3.5 × 3.5 × 3.5 ≈ 89.8 cm³

About 900 to 1900 golgappas, depending on the size of a golgappa and how much pani goes in each (it is in the hundreds or low thousands).

1. Pani in the vesselV = π × 10² × 50 = 5000π ≈ 15,708 cm³
2. Estimate 1Radius of a golgappa 1.5 cm: volume ⁴⁄₃π(1.5)³ ≈ 14.14 cm³. If each is filled with 60% pani: 14.14 × 0.6 ≈ 8.5 cm³.15,708 ÷ 8.5 ≈ 1850 golgappas
3. Estimate 2Radius 1.75 cm: volume ≈ 22.4 cm³. If each is filled with 80% pani: ≈ 18 cm³.15,708 ÷ 18 ≈ 870 golgappas
4. ConclusionThe answer depends on the assumptions (size of a golgappa, how much pani it holds). A reasonable answer is between about 870 and 1850.

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

An upper limit of about 3.4 million balls; packing in an ordered cubic way gives about 1.7 million, and the best packing about 2.5 million.

1. Volume of a ball and of the roomBall: ⁴⁄₃π × 3³ ≈ 113 cm³Room: 30 × 30 × 15 = 13,500 ft³ ≈ 13,500 × 28,000 cm³ ≈ 3.78 × 10⁸ cm³
2. An upper bound3.78 × 10⁸ ÷ 113 ≈ 33,45,000This is only an upper limit: round balls cannot fill a room without gaps.
3. Placing balls in rowsTake 1 ft ≈ 30 cm, so the room is 900 cm × 900 cm × 450 cm and the diameter of a ball is 6 cm. Along each side: 900 ÷ 6 = 150, 150 and 450 ÷ 6 = 75.150 × 150 × 75 = 16,87,500 balls
4. Better packingsIn the other ordered ways (like B and C in the figure) the balls sit in the gaps of the layer below, so more balls fit. The best packing fills about 74% of the space instead of the 52% of the cubic way, which gives about 25 lakh (2.5 million) balls.

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

The total surface area is 96 cm².

1. Sidea³ = 64 ⇒ a = 4 cm
2. Surface area6a² = 6 × 16 = 96 cm²

1000 cubes.

1. VolumesBox: (200 cm)³ = 80,00,000 cm³   and   small cube: 20³ = 8000 cm³
2. Number80,00,000 ÷ 8000 = 1000
3. CheckAlong each edge 200 ÷ 20 = 10 cubes fit, and 10 × 10 × 10 = 1000.

4000 boxes.

1. VolumesGodown: 40 × 25 × 10 = 10,000 m³   and   one box: 2 × 1.25 × 1 = 2.5 m³
2. Number10,000 ÷ 2.5 = 4000

250 cm².

1. Side of each cubea³ = 125 ⇒ a = 5 cm
2. The cuboidJoined end to end, the cuboid is 10 cm × 5 cm × 5 cm.
3. Surface area2(10 × 5 + 10 × 5 + 5 × 5) = 2(125) = 250 cm²

1 : 4.

1. Original cubeS = 6 × 4² = 96 cm²
2. The small cubesNumber of cubes = 4³ ÷ 1³ = 64. Each has surface area 6 × 1² = 6 cm², so all of them have 64 × 6 = 384 cm².
3. Ratio96 : 384 = 1 : 4

The volume is 30 cm³.

1. The face areasThe three faces have areas lw, wh and lh.
2. Multiply them(lw)(wh)(lh) = (lwh)² = 6 × 15 × 10 = 900
3. Volumelwh = √900 = 30 cm³

(i) 8 (ii) 36 (iii) 54 (iv) 27.

1. Set upThere are 5 small cubes along each edge, 5 × 5 × 5 = 125 in all.
2. (i) Three faces paintedThese are the corner cubes. A cube has 8 corners: 8.
3. (ii) Two faces paintedThey lie along the edges, not counting the corners. Each edge has 5 − 2 = 3 of them, and there are 12 edges: 12 × 3 = 36.
4. (iii) One face paintedThey lie in the middle of a face. Each face has (5 − 2)² = 9 of them, and there are 6 faces: 6 × 9 = 54.
5. (iv) No face paintedThey form the inner cube of side 5 − 2 = 3: 3³ = 27.

Check: 8 + 36 + 54 + 27 = 125.

(i) Yes. (ii) 1 × 2 × 16 and 2 × 4 × 7. (iii) There are no others (shown below).

1. Set upLet the edges be a ≤ b ≤ c (whole numbers). Then 2(ab + bc + ca) = 100, so ab + bc + ca = 50.
2. Two cuboids1 × 2 × 16:   2(2 + 32 + 16) = 100 ✓2 × 4 × 7:   2(8 + 28 + 14) = 100 ✓
3. Limit for aSince a ≤ b ≤ c, each of ab, bc, ca is at least a², so 3a² ≤ 50 and a can only be 1, 2, 3 or 4.
4. Check each aWrite the equation as (b + a)(c + a) = 50 + a².

   a	(b + a)(c + a)	Possible factor pairs with both factors at least 2a	Result
   1	51 = 3 × 17	(3, 17)	b = 2, c = 16
   2	54	(6, 9) [(3, 18) gives b = 1 < a]	b = 4, c = 7
   3	59 (a prime)	none	no cuboid
   4	66	none (6 × 11 has 6 < 8)	no cuboid

5. ConclusionExactly two cuboids are possible (apart from turning them round): 1 × 2 × 16 and 2 × 4 × 7.

Curved surface areas 1 : 1, volumes 1 : 2.

1. Set upA has radius r and height h. Then B has radius 2r and height h ÷ 2.
2. Curved surface areasA: 2πrh   and   B: 2π(2r)(h ÷ 2) = 2πrh ⇒ 1 : 1
3. VolumesA: πr²h   and   B: π(2r)²(h ÷ 2) = 2πr²h ⇒ 1 : 2

(a) 2 : 3. (b) 1 : 1.

1. (a) Volumesπr₁²h₁ : πr₂²h₂ = (2² × 3) : (3² × 2) = 12 : 18 = 2 : 3
2. (b) Curved surface areas2πr₁h₁ : 2πr₂h₂ = (2 × 3) : (3 × 2) = 1 : 1

The volume is πr³ ÷ 4 cm³.

1. The cylinderThe base of the cylinder fits exactly inside a square face of the cube, so its diameter is r and its radius is r ÷ 2. Its height is r.
2. Volumeπ(r ÷ 2)² × r = πr³4 cm³

Here r is the edge of the cube, not the radius of the cylinder.

x ≈ 17.36.

1. New dimensionsNew radius = 1.1r and new height = (1 − x ÷ 100)h.
2. Equal volumesπ(1.1r)²(1 − x ÷ 100)h = πr²h ⇒ 1.21(1 − x ÷ 100) = 1
3. Solve1 − x100 = 100121 ⇒ x100 = 21121 ⇒ x = 2100121 ≈ 17.36

(i) 1728 cm³ (ii) about 150.86 cm³ (iii) 11 rods.

1. (i)12³ = 1728 cm³
2. (ii)πr²h = 227 × 2² × 12 = 10567 ≈ 150.86 cm³
3. (iii)1728 ÷ 10567 = 12611 ≈ 11.45Only 11 complete rods can be made. We do not round up, because there is not enough metal for a twelfth complete rod.

396π ≈ 1244.57 m².

1. Radiusr = 24 ÷ 2 = 12 m and l = 21 m.
2. Total surface areaπr(l + r) = π × 12 × 33 = 396π ≈ 396 × 227 ≈ 1244.57 m²

220 cm².

1. Formulaπrl = 227 × 7 × 10 = 220 cm²

Curved surface area 240π ≈ 754.29 cm². Total surface area 384π ≈ 1206.86 cm².

1. Slant heightl = √(16² + 12²) = √400 = 20 cm
2. Curved surface areaπrl = π × 12 × 20 = 240π ≈ 754.29 cm²
3. Total surface areaπr(l + r) = π × 12 × 32 = 384π ≈ 1206.86 cm²

The radius is 10 cm (using π = 3.14).

1. Volume13πr²h = 13 × 3.14 × r² × 15 = 15.7r² = 1570
2. Solver² = 100 ⇒ r = 10 cm

(i) 7 cm. (ii) 462 cm².

1. (i)πrl = 227 × r × 14 = 44r = 308 ⇒ r = 7 cm
2. (ii)Total = curved + base = 308 + 227 × 7² = 308 + 154 = 462 cm²

5500 cm².

1. Slant heightl = √(7² + 24²) = √625 = 25 cm
2. One capA cap is open at its base, so only the curved surface is needed.πrl = 227 × 7 × 25 = 550 cm²
3. 10 caps10 × 550 = 5500 cm²

About 63 m.

1. Slant heightl = √(8² + 6²) = 10 m
2. Area of the tentThe tent has no floor, so only the curved surface is needed: π × 6 × 10 = 60π m².
3. Length of tarpaulinLength = area ÷ width = 60π ÷ 3 = 20π ≈ 62.8 mAdd 20 cm (0.2 m) for margins and wastage: 62.8 + 0.2 ≈ 63 m.

Volume 96π ≈ 301.71 cm³ and curved surface area 60π ≈ 188.57 cm².

1. The solidIt is a cone with height h = 8 cm, radius r = 6 cm and slant height l = 10 cm.
2. Volume13πr²h = 13 × π × 36 × 8 = 96π ≈ 301.71 cm³
3. Curved surface areaπrl = π × 6 × 10 = 60π ≈ 188.57 cm²

The water fills 1/8 of the cup.

1. Similar conesLet the cup have radius r and height h, with its point down. At half the height, the water surface has half the radius, r ÷ 2, because the water forms a smaller cone of the same shape.
2. Volume of the water13π(r ÷ 2)²(h ÷ 2) = 18 × 13πr²h
3. FractionVolume of water ÷ volume of cup = 1/8. Half the depth is only one-eighth of the volume.

6.16 cm².

1. Surface area of a sphere4πr² = 4 × 227 × 0.7² = 6.16 cm²

The radius of the larger sphere is 5 cm.

1. VolumesSame metal, so volumes are in the ratio of the weights: 5920 : 740 = 8 : 1.
2. RadiiVolume is proportional to r³, so the ratio of radii is ∛8 : ∛1 = 2 : 1.
3. Radius of the larger sphereThe smaller radius is 5 ÷ 2 = 2.5 cm, so the larger radius is 2 × 2.5 = 5 cm.

Moon : Earth = 1 : 16.

1. RadiiThe ratio of the radii is 1 : 4.
2. Surface areas4πr²:   1² : 4² = 1 : 16

(i) 2772 cm². (ii) 4158 cm².

1. (i) Curved surface area2πr² = 2 × 227 × 21² = 2772 cm²
2. (ii) Total surface areaThe total surface area also includes the flat circular base.3πr² = 3 × 227 × 21² = 4158 cm²

About 488 cm³.

1. Volume of the box16 × 8 × 8 = 1024 cm³
2. Volume of 16 spheres16 × 43π × 2³ = 512π3 ≈ 536.2 cm³
3. Liquid1024 − 536.2 ≈ 487.8 ≈ 488 cm³

The volume increases by 33.1%.

1. New volumeV′ = 43π(1.1r)³ = 1.331 × 43πr³ = 1.331V
2. Increase(1.331 − 1) × 100 = 33.1%

x = 20.

1. Volume ratioV′ ÷ V = 1 + 72.8100 = 1.728
2. Since V is proportional to r³(1 + x ÷ 100)³ = 1.728 = 1.2³ ⇒ 1 + x ÷ 100 = 1.2 ⇒ x = 20

The cost is ₹1,97,120.

1. Radius2πr = 35.2 ⇒ r = 35.2 ÷ (2 × 227) = 5.6 m
2. Area to paintOnly the curved surface of the dome is painted.2πr² = 2 × 227 × 5.6² = 197.12 m²
3. Cost1 m² = 10,000 cm².197.12 × 10,000 = 19,71,200 cm²Cost = (19,71,200 ÷ 100) × ₹10 = ₹1,97,120

About 90 scoops (with a scoop of radius 2 cm).

1. First guessAbout 100 scoops.
2. AssumptionsA full container, no wastage, and each scoop is a sphere of radius 2 cm.
3. CalculationContainer: 10 × 15 × 20 = 3000 cm³One scoop: 43π × 2³ = 32π3 ≈ 33.5 cm³3000 ÷ 33.5 ≈ 89.5 ≈ 90 scoops
4. RemarkThe answer changes with the size of the scoop: a bigger scoop gives fewer scoops.

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

3a² + 3a + 1 unit cubes.

1. CountNeeded: (a + 1)³ − a³ = a³ + 3a² + 3a + 1 − a³ = 3a² + 3a + 1
2. CheckFor a = 3: (4³ − 3³) = 64 − 27 = 37 and 3(9) + 3(3) + 1 = 37.

No: with these assumptions the water would fill only about 35% of a classroom.

1. AssumptionsA person drinks 2 litres a day for 70 years. The classroom is 8 m × 6 m × 3 m.
2. Water drunk2 × 365 × 70 = 51,100 litres = 51.1 m³
3. The room8 × 6 × 3 = 144 m³ = 1,44,000 litres
4. Compare51,100 ÷ 1,44,000 ≈ 0.355 = 35.5%So one person could not fill the classroom.

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

About 8000 bricks (for the assumed room).

1. First guessAbout 8000 bricks.
2. AssumptionsInside dimensions 8 m × 6 m, wall height 3 m, wall thickness 0.2 m (outside 8.4 m × 6.4 m).
3. Volume of the walls(8.4 × 6.4 − 8 × 6) × 3 = (53.76 − 48) × 3 = 17.28 m³
4. Take away the door and windowsOne door 1 m × 2 m and two windows 1.5 m × 1.2 m: (2 + 3.6) × 0.2 = 1.12 m³. Net wall volume = 17.28 − 1.12 = 16.16 m³.
5. Number of bricksA brick with its mortar takes 0.2 m × 0.1 m × 0.1 m = 0.002 m³.16.16 ÷ 0.002 = 8080 ≈ 8100 bricks

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

Option A gives more chocolate (25% more).

1. VolumesA: 50³ = 1,25,000 mm³   and   B: 100 × 10³ = 1,00,000 mm³
2. CompareA gives 25,000 mm³ more, which is 25% more than B.

About 2000 km² (with the assumptions below).

1. First guessA few thousand square kilometres.
2. AssumptionsAbout 8 billion people (an assumed round number), and each person needs 0.5 m × 0.5 m = 0.25 m².
3. Area8,000,000,000 × 0.25 = 2,000,000,000 m²1 km² = 10,00,000 m², so the area ≈ 2000 km²
4. RemarkThis has no space for paths or other facilities; with more space per person the area is larger.

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

739.2 litres.

1. Milk in one glassRadius = 7 ÷ 2 = 3.5 cm.πr²h = 227 × 3.5² × 12 = 462 cm³
2. 1600 students462 × 1600 = 7,39,200 cm³ = 739.2 litres

Height 3 cm and volume 16π ≈ 50.29 cm³.

1. Slant height4π × 5² = 100π = 5 × π × 4 × l = 20πl ⇒ l = 5 cm
2. Heighth = √(5² − 4²) = 3 cm
3. Volume13πr²h = 13 × π × 16 × 3 = 16π ≈ 50.29 cm³

(i) about 1321 : 1. (ii) about 13,02,710 : 1.

1. The ratioFor spheres V = ⁴⁄₃πr³, so the factor ⁴⁄₃π cancels and the ratio of volumes is the cube of the ratio of the radii.
2. (i) Jupiter : Earth(69,900 ÷ 6370)³ ≈ 10.973³ ≈ 1321
3. (ii) Sun : Earth(6,95,700 ÷ 6370)³ ≈ 109.2³ ≈ 13,02,710

The sphere is ⅔ of its enclosing cylinder.

1. The cylinderA sphere of radius r fits in the smallest cylinder of radius r and height 2r.
2. VolumesSphere: 43πr³   and   cylinder: πr² × 2r = 2πr³
3. Ratio(4/3)πr³2πr³ = 23

The second string is 1 ÷ (2π) m ≈ 0.159 m ≈ 15.9 cm above the ground, for the Earth, the Moon, Jupiter and a volleyball alike.

1. Set upLet the radius be R metres and the extra height x metres. The first string has length 2πR and the second 2π(R + x).
2. Subtract2π(R + x) − 2πR = 1 ⇒ 2πx = 1 ⇒ x = 12π m ≈ 0.159 m ≈ 15.9 cm
3. The four answersEarth, Moon, Jupiter and volleyball: about 15.9 cm each.
4. Are we surprised?Many people are, because the objects are of very different sizes. But R cancels in the calculation: the gap depends only on the extra length of the string, not on the original radius.

The ratio is 2 : 3π (about 7 : 33 if π = 22/7).

1. The base of the pyramidThe corners of the square are on the circle, so the diagonal of the square is the diameter 2r. If the side is a, then a² + a² = (2r)², so 2a² = 4r² and the area of the square is a² = 2r².
2. Volume of the pyramid13 × base area × height = 13 × 2r² × h = 23r²h
3. Volume of the cylinderπr²h
4. Ratio(2/3)r²hπr²h = 23π ⇒ 2 : 3π (≈ 7 : 33)

(i) option (c) wh. (ii) option (e) 2πrh + πh. (iii) the volume decreases by ⁴⁄₃π(3r² − 3r + 1).

1. (i) CuboidOld volume = lwh, new volume = (l + 1)wh.Change = (l + 1)wh − lwh = wh cubic unitsThe answer is (c).
2. (ii) CylinderOld volume = πr²h, new volume = π(r + 1)²h.Change = πh[(r + 1)² − r²] = πh(2r + 1) = 2πrh + πhThe answer is (e).
3. (iii) SphereLet the radius be r (with r ≥ 1). Old volume = ⁴⁄₃πr³, new volume = ⁴⁄₃π(r − 1)³.Decrease = 43π[r³ − (r − 1)³] = 43π(3r² − 3r + 1) cubic units

S = 6V^(2/3) = 6(∛V)².

1. SideV = a³ ⇒ a = ∛V
2. Surface areaS = 6a² = 6(∛V)² = 6V^(2/3)

V = (S ÷ 6)^(3/2).

1. SideS = 6a² ⇒ a = √(S ÷ 6)
2. VolumeV = a³ = (√(S ÷ 6))³ = (S ÷ 6)^(3/2)

48 marbles.

1. Rise of the waterThe water must rise 20 − 16 = 4 cm. The marbles must take up a volume equal to this layer of water.
2. Volume of that layerπ × 4² × 4 = 64π cm³
3. One marble43π × 1³ = 4π3 cm³
4. Number of marbles64π ÷ 4π3 = 48

About 0.25 mL for the sample measurements below; yours will differ.

1. First guessAbout 0.3 mL.
2. Measurements (sample)The ink-filled part of the refill is a cylinder with internal diameter 2 mm (radius 1 mm = 0.1 cm) and length 8 cm.
3. Volumeπr²h = π × 0.1² × 8 = 0.08π ≈ 0.25 cm³Since 1 cm³ = 1 mL, there is about 0.25 mL of ink (ignoring the small tip).

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

About 40 chapatis (for the assumed thickness).

1. AssumptionsEach uncooked chapati is a thin cylinder of radius 6 cm and thickness 0.2 cm, and no dough is wasted.
2. VolumesDough ball: 43π × 6³ = 288π cm³One chapati: π × 6² × 0.2 = 7.2π cm³
3. Number288π ÷ 7.2π = 40

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

About 128 cm³ in one half for the sample measurements below.

1. ModelTreat half the coconut as a hemispherical layer. Sample measurements: outer radius of the flesh R = 5 cm and radius of the empty cavity r = 4 cm.
2. Volume of the flesh in one half23π(R³ − r³) = 23π(125 − 64) = 122π3 ≈ 127.8 cm³
3. Both halvesAbout 256 cm³ for the whole coconut with these measurements.

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

(i) No: about 5 m² of paper against about 48 m² of floor. (ii) About 2.43 lakh textbooks.

1. (i) AssumptionsEach sheet is 25 cm × 20 cm and the book has 200 numbered pages, that is 100 sheets (ignore the covers). The classroom floor is 8 m × 6 m.
2. (i) AreaOne sheet: 25 × 20 = 500 cm²   ⇒   100 sheets: 50,000 cm² = 5 m²The floor is 8 × 6 = 48 m². Since 5 < 48, the pages would not cover the floor.
3. (ii) AssumptionsA closed book measures 25 cm × 20 cm × 2 cm. Take 1 ft ≈ 30 cm, so the storeroom is 450 cm × 600 cm × 900 cm.
4. (ii) PackingPut the 25 cm side along 450 cm, the 20 cm side along 600 cm and the 2 cm thickness along 900 cm.(450 ÷ 25) × (600 ÷ 20) × (900 ÷ 2) = 18 × 30 × 450 = 2,43,000This fills the room without gaps, so it is the most that can fit (no space is left to walk).

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

A cube about 2 km on each side (with the assumptions below).

1. AssumptionsAbout 8 billion people (an assumed round number), and 1 m³ of space for each person, including some empty space.
2. Volume8 × 10⁹ m³
3. Sidea³ = 8 × 10⁹ ⇒ a = ∛(8 × 10⁹) = 2000 m = 2 km

Estimation question: the answer depends on the assumptions you make, so your numbers may be different. What matters is that the assumptions are reasonable and clearly stated.

(i) t = ∛[R³ + 3W ÷ (4π)] − R. (ii) t ≈ 2.70 km.

1. (i) The expressionLet R = 6371 km and W = 1.38 × 10⁹ km³. The outer surface has radius R + t, and the water is the space between the two spheres:W = 43π[(R + t)³ − R³] ⇒ (R + t)³ = R³ + 3W4πt = ∛[R³ + 3W4π] − R = ∛[6371³ + 3 × 1.38 × 10⁹4π] − 6371 km
2. (ii) SimplifyWith a calculator, t ≈ 2.7044 km, that is, about 2.70 km (about 2700 m).

One example: a cuboid 8 cm × 14 cm × 14 cm becomes 1 cm × 28 cm × 28 cm. The dimensions are not unique.

1. The ideaMake the cuboid flat: squash one side and stretch the other two.
2. Check the example

   	Original	New
   Dimensions	8 × 14 × 14 cm	1 × 28 × 28 cm
   Volume	8 × 14 × 14 = 1568 cm³	1 × 28 × 28 = 784 cm³ (half)
   Surface area	2(112 + 112 + 196) = 840 cm²	2(28 + 28 + 784) = 1680 cm² (double)

This is a project, so your answer will be your own. A way to do it is shown below, with sample numbers.

1. PlanBreak the house into rooms (cuboids). Measure the length, breadth and height of each room, and add the volumes. Make a note of every approximation.
2. Sample (made-up numbers)

   Room	Length × breadth × height (m)	Volume (m³)
   Hall	5 × 4 × 3	60
   Bedroom 1	4 × 3.5 × 3	42
   Bedroom 2	4 × 3 × 3	36
   Kitchen	3 × 3 × 3	27
   Bathroom	2 × 1.5 × 3	9
   Total		174

3. Presenting itDraw a plan of the house, show the measurements, the calculation for each room, and the total. Say what you ignored (walls, furniture, the roof) and how that may change the answer.

This is a project: use measurements of your own house. The numbers in the table are only an example.