Math of Space: Surface Area and Volume

1. Let side = a. Then a³ = 64.
2. Take the cube root: a = 4 cm.
3. TSA = 6a² = 6 × 4² = 6 × 16 = 96 cm².

1. Convert 2 m = 200 cm.
2. Number along each edge = 200 ÷ 20 = 10.
3. As the cubes fit exactly in three directions, total = 10 × 10 × 10 = 1000.

1. Along length: 40 ÷ 2 = 20 boxes.
2. Along width: 25 ÷ 1.25 = 20 boxes.
3. Along height: 10 ÷ 1 = 10 boxes.
4. Total = 20 × 20 × 10 = 4000.

This is a valid packing arrangement, not just a volume comparison.

1. Side of each cube = ∛125 = 5 cm.
2. Joining along one whole face makes a cuboid of 10 cm × 5 cm × 5 cm.
3. TSA = 2(10 × 5 + 5 × 5 + 10 × 5) = 2(50 + 25 + 50) = 250 cm².

1. Original surface area = 6 × 4² = 96 cm².
2. Number of small cubes = 4³ = 64.
3. Area of each small cube = 6 × 1² = 6 cm².
4. Combined area = 64 × 6 = 384 cm².
5. Ratio = 96 : 384 = 1 : 4.

The total volume is unchanged; cutting exposes new surfaces.

1. Let edges be a, b, c. We know ab = 6, bc = 15 and ca = 10.
2. Multiply: (ab)(bc)(ca) = (abc)² = 6 × 15 × 10 = 900.
3. Hence abc = √900 = 30 cm³.

1. Three painted faces: only corner cubes; a cube has 8 corners → 8.
2. Two painted faces: along edges but not corners. 12 edges × (5 − 2) = 36.
3. One painted face: interiors of 6 faces, each with (5 − 2)² cubes → 6 × 9 = 54.
4. No painted face: internal cube (5 − 2)³ = 27.

Check: 8 + 36 + 54 + 27 = 125 = 5³, so every small cube was counted once.

1. TSA = 2(ab + bc + ca) = 100, so ab + bc + ca = 50.
2. Take a ≤ b ≤ c so different orientations are not counted repeatedly.
3. Rearrange: c = (50 − ab)/(a + b), which must be a positive integer at least b.
4. Because a ≤ b ≤ c, 3a² ≤ 50, so a ≤ 4. Check a = 1, 2, 3, 4 and b ≥ a subject to ab + b² + ab ≤ 50 (as c ≥ b).
5. For a = 1 the possible values of b are 1 through 6. Since c = 51/(1 + b) − 1, only b = 2 gives an integer. For a = 2, b ranges from 2 through 5 and c = 54/(2 + b) − 2 is integral only for b = 4. For a = 3, b is 3 or 4, giving c = 41/6 or 38/7. For a = 4, only b = 4 is possible and c = 17/4. Thus the only solutions are (a,b,c) = (1,2,16) and (2,4,7).
6. Verification: 1 × 2 + 2 × 16 + 1 × 16 = 50; 2 × 4 + 4 × 7 + 2 × 7 = 50.

Option A volume = 8 × 50 × 20 = 8000 ft³ ≈ 226,400 L.

Option B volume = 6 × 100 × 15 = 9000 ft³ ≈ 254,700 L.

Option B offers a shallower depth and longer length for swimming; it also requires about 28,300 L more water. A school must consider pupil ages, supervision, usable plot, construction costs, water consumption and safety regulations. No single choice is forced by the data.

Let A have radius r and height h. B has radius 2r and height h/2.

1. CSA(A) = 2πrh; CSA(B) = 2π × 2r × h/2 = 2πrh → 1 : 1.
2. V(A) = πr²h; V(B) = π(2r)²(h/2) = 2πr²h → 1 : 2.

1. Volume is proportional to r²h, so volume ratio = (2² × 3) : (3² × 2) = 12 : 18 = 2 : 3.
2. CSA is proportional to rh, so CSA ratio = (2 × 3) : (3 × 2) = 6 : 6 = 1 : 1.

1. The largest circular base fits exactly in the square face: its diameter equals r, so its radius is r/2.
2. The cylinder height is the cube edge r.
3. Volume = π(r/2)² × r = πr³/4.

1. New radius = 1.1r, new height = h(1 − x/100).
2. Equate volumes: π(1.1r)²h(1 − x/100) = πr²h.
3. Thus 1.21(1 − x/100) = 1.
4. 1 − x/100 = 1/1.21 = 100/121.
5. x/100 = 21/121, giving x = 2100/121 ≈ 17.36.

1. Cube volume = 12³ = 1728 cm³.
2. Rod volume = π × 2² × 12 = 48π = 1056/7 ≈ 150.86 cm³.
3. Number by volume = 1728/(48π) = 36/π ≈ 11.45. Only 11 complete rods can be formed; some metal is left over.

1. Radius r = 24 ÷ 2 = 12 m.
2. TSA = πr(l + r) = π × 12 × (21 + 12) = 396π m².
3. Using π ≈ 22/7, TSA ≈ 1244.57 m².

CSA = πrl = (22/7) × 7 × 10 = 220 cm².

1. Slant height l = √(12² + 16²) = √400 = 20 cm.
2. CSA = π × 12 × 20 = 240π cm² ≈ 753.6 cm² (π = 3.14).
3. TSA = π × 12 × (20 + 12) = 384π cm² ≈ 1205.76 cm².

1. V = ⅓πr²h, so 1570 = ⅓ × 3.14 × r² × 15.
2. 1570 = 15.7r², therefore r² = 100.
3. Since radius is positive, r = 10 cm.

1. CSA = πrl → 308 = (22/7) × r × 14 = 44r; hence r = 7 cm.
2. TSA = πr(l + r) = (22/7) × 7 × (14 + 7) = 462 cm².

1. Slant height l = √(7² + 24²) = √625 = 25 cm.
2. Only the curved surface is made from the sheet; the bottom is open.
3. One cap: CSA = (22/7) × 7 × 25 = 550 cm².
4. Ten caps need 10 × 550 = 5500 cm².

1. Slant height l = √(8² + 6²) = 10 m.
2. Curved area needed = πrl = 60π m²; no circular floor is needed.
3. With fabric width 3 m, basic length = 60π ÷ 3 = 20π m.
4. Add 20 cm = 0.20 m → length = 20π + 0.2 m.
5. Using π = 3.14, length = 62.8 + 0.2 = 63.0 m.

1. A right circular cone forms with radius r = 6 cm, height h = 8 cm and slant height l = 10 cm.
2. Volume = ⅓π × 6² × 8 = 96π cm³ ≈ 301.44 cm³.
3. CSA = π × 6 × 10 = 60π cm² ≈ 188.4 cm².

1. Assume the cup is upright with its pointed end at the bottom, as in the textbook diagram.
2. The water forms a smaller cone similar to the whole cup.
3. Water depth = h/2; by similarity, water surface radius = r/2.
4. Fraction = [⅓π(r/2)²(h/2)] / [⅓πr²h] = (1/2)³ = 1/8.

Half the height does not mean half the volume.

Surface area of sphere = 4πr² = 4 × (22/7) × (0.7)² = 6.16 cm².

1. Same metal means same density, so mass is proportional to volume.
2. Mass ratio = 5920 : 740 = 8 : 1.
3. Since V ∝ r³, the radii ratio is ∛8 : ∛1 = 2 : 1.
4. Smaller radius = 5/2 = 2.5 cm. Larger radius = 2 × 2.5 = 5 cm.

The radius ratio is also 1 : 4. Surface area is proportional to radius squared, so Moon : Earth = 1² : 4² = 1 : 16.

1. CSA = 2πr² = 2 × (22/7) × 21² = 2772 cm².
2. TSA = 3πr² = 3 × (22/7) × 21² = 4158 cm².

1. Box capacity = 16 × 8 × 8 = 1024 cm³.
2. One sphere volume = ⁴⁄₃π × 2³ = 32π/3 cm³.
3. Sixteen spheres volume = 16 × 32π/3 = 512π/3 cm³.
4. Liquid fills the gaps: V = 1024 − 512π/3 ≈ 487.62 cm³ (π = 22/7), rounding to 488 cm³.

The spheres fit as a 4 × 2 × 2 array because each diameter is 4 cm.

1. Old volume = ⁴⁄₃πr³.
2. New radius = 1.1r, so new volume = ⁴⁄₃π(1.1r)³ = 1.331 × old volume.
3. Percentage increase = (1.331 − 1) × 100 = 33.1%.

1. New volume/old volume = 1 + 72.8/100 = 1.728.
2. (1 + x/100)³ = 1.728 = 1.2³.
3. Therefore 1 + x/100 = 1.2 and x = 20.

1. Circumference = 2πr = 35.2 m; with π = 22/7, r = 35.2 × 7/44 = 5.6 m.
2. Only the curved dome is painted: CSA = 2πr² = 2 × (22/7) × 5.6² = 197.12 m².
3. Convert: 197.12 m² = 1,971,200 cm².
4. Cost = (1,971,200 ÷ 100) × ₹10 = ₹197,120.

1. Ice-cream volume = 10 × 15 × 20 = 3000 cm³.
2. Assume each scoop is a complete sphere of radius 3 cm. Its volume = ⁴⁄₃π × 3³ = 36π ≈ 113.1 cm³.
3. 3000 ÷ 113.1 ≈ 26.5, so about 26 complete spherical scoops.

These are scoops formed from the ice cream, not whole spheres packed into the tub. If the scoop is a hemisphere of radius 3 cm, its volume is 18π and the answer is about 53 scoops. Since the question gives no scoop size or shape, both are valid estimates with stated assumptions.

1. Original cube contains a³ unit cubes.
2. New cube contains (a + 1)³ unit cubes.
3. Additional cubes = (a + 1)³ − a³ = a³ + 3a² + 3a + 1 − a³.

(i) Sample model: Drink 2 litres of water daily for 70 years. Lifetime consumption ≈ 2 × 365 × 70 = 51,100 litres = 51.1 m³. A classroom of 8 m × 6 m × 3 m holds 144 m³. Under these assumptions, the person cannot fill the room. Water consumption and room size can change this estimate.

(ii) Sample model: Room 8 m × 6 m, wall height 3 m. Gross wall area = perimeter × height = 2(8 + 6) × 3 = 84 m². Subtract 8 m² for doors/windows → 76 m². For 0.20 m wall thickness, brickwork volume ≈ 15.2 m³. Taking effective brick-with-mortar volume ≈ 0.20 × 0.10 × 0.10 = 0.002 m³ gives 15.2/0.002 ≈ 7600 bricks. Actual count depends on wall thickness, bricks, mortar and openings.

1. A: V = 50³ = 125,000 mm³.
2. B: V = 100 × 10³ = 100,000 mm³.
3. Difference = 25,000 mm³. Option A contains 25% more chocolate than B.

We must state our assumptions. Suppose there are 8 billion people, and each person standing closely takes 0.25 m² of ground area.

Estimated area = 8,000,000,000 × 0.25 = 2,000,000,000 m² = 2000 km². This is roughly a square about 44.7 km per side.

Real crowd density varies greatly; this is a mathematical estimate, not a safe crowd-capacity recommendation.

1. Radius = 7/2 = 3.5 cm.
2. Milk per glass = πr²h = (22/7) × 3.5² × 12 = 462 cm³.
3. For 1600 glasses = 462 × 1600 = 739,200 cm³.
4. Convert to litres: 739,200/1000 = 739.2 L.

1. Sphere area = 4π × 5² = 100π cm².
2. Cone CSA = 100π/5 = 20π cm².
3. πrl = π × 4 × l = 20π → l = 5 cm.
4. For a right cone, h = √(l² − r²) = √(25 − 16) = 3 cm.
5. Volume = ⅓π × 4² × 3 = 16π cm³ ≈ 50.24 cm³ (π = 3.14).

For spheres, V = ⁴⁄₃πr³, so volume ratios are cubes of the radius ratios.

1. Jupiter/Earth = (69,900/6370)³ ≈ 1321.34 : 1.
2. Sun/Earth = (695,700/6370)³ ≈ 1,302,709.90 : 1.

These values follow the given textbook radii and the ideal-sphere assumption.

1. For a sphere of radius r, the tightest cylinder has radius r and height 2r.
2. V(cylinder) = πr²(2r) = 2πr³.
3. V(sphere) = ⁴⁄₃πr³.
4. V(sphere)/V(cylinder) = (⁴⁄₃πr³)/(2πr³) = 2/3.

1. First circumference C = 2πR.
2. Longer circumference C + 1 = 2π(R + h).
3. Subtract: 1 = 2πh, so h = 1/(2π) metres.
4. With π ≈ 3.14, h ≈ 0.159 m = 15.9 cm.

The starting radius R cancels out. Therefore the lift is the same 15.9 cm for Earth, Moon, Jupiter and a volleyball (ideal circular models).

1. The square is inscribed in a circle of radius r, so square diagonal = circle diameter = 2r.
2. For square side s, s√2 = 2r, hence s = r√2 and base area = s² = 2r².
3. Pyramid volume = ⅓ × 2r² × h = 2r²h/3.
4. Cylinder volume = πr²h.
5. Ratio = (2r²h/3)/(πr²h) = 2/(3π).

(i) Original V = lwh. New V = (l + 1)wh. Increase = (l + 1)wh − lwh = wh cubic units → option (c).

(ii) Original V = πr²h. New V = π(r + 1)²h. Increase = πh[(r + 1)² − r²] = πh(2r + 1) = 2πrh + πh cubic units → option (e).

(iii) For r ≥ 1, new V = ⁴⁄₃π(r − 1)³. Volume decreases by ⁴⁄₃π[r³ − (r − 1)³] = ⁴⁄₃π(3r² − 3r + 1) cubic units. The signed change (new minus old) is negative of this expression.

1. Let side = a. Then V = a³.
2. Therefore 𝑎 =3√𝑉 =𝑉1/3.
3. Surface area S = 6a² = 6⁢(𝑉1/3)2 = 6⁢𝑉2/3.

1. Let side = a. Surface area S = 6a².
2. a² = S/6, so a = √(S/6).
3. Volume V = a³ = [√𝑆/6]3 = (𝑆/6)3/2.

1. Required water level increase = 20 − 16 = 4 cm.
2. Required displaced volume = cylinder base area × rise = π × 4² × 4 = 64π cm³.
3. One marble volume = ⁴⁄₃π × 1³ = 4π/3 cm³.
4. Number = 64π ÷ (4π/3) = 48 marbles.

The original water is deep enough to submerge every marble. The calculation assumes marbles can be put in and that no water spills.

Sample estimate: treat the ink chamber as a narrow cylinder with inner radius 0.10 cm and ink-filled length 10 cm. Then V ≈ π × (0.10)² × 10 = 0.1π ≈ 0.314 cm³ = 0.314 mL.

To answer for your actual pen, measure the refill’s inside diameter and ink column length, then substitute into V = πr²h. A non-cylindrical chamber or air gaps require adjustment.

(i) Dough volume ≈ ⁴⁄₃π × 6³ = 288π ≈ 904.8 cm³. Assume each chapati is approximately a cylindrical disc with radius 9 cm and thickness 0.25 cm, so V ≈ π × 9² × 0.25 = 20.25π cm³. Number ≈ 288π/(20.25π) ≈ 14 chapatis. Real size and thickness vary.

(ii) For example, suppose the flesh in a coconut half extends from an inner cavity of radius r = 4 cm to the inside of the hard shell at R = 6 cm. Approximate edible flesh as a hemispherical shell: V = ⅔π(R³ − r³) = ⅔π(216 − 64) = 304π/3 ≈ 318 cm³ for that half. Use your own measurements; a muskmelon’s seeded core and rind must be excluded from edible flesh.

(i) Sample assumptions: 160 printed pages, each 0.20 × 0.26 = 0.052 m². Their total page-face area ≈ 160 × 0.052 = 8.32 m². An 8 × 6 m classroom floor has area 48 m²; so they would not cover it. If placing physical sheets rather than individual page faces, a two-sided book has fewer sheets and covers even less area.

(ii) Convert storeroom to metres: 15 ft ≈ 4.572 m; 20 ft ≈ 6.096 m; 30 ft ≈ 9.144 m. Suppose each textbook measures 0.25 × 0.20 × 0.01 m. With one fixed alignment and no gaps, complete-book counts are ⌊4.572/0.25⌋ = 18, ⌊6.096/0.20⌋ = 30 and ⌊9.144/0.01⌋ = 914; total 18 × 30 × 914 = 493,560. This arrangement fits 493,560 books. The room volume divided by one book’s volume gives an upper bound: (4.572 × 6.096 × 9.144)/(0.25 × 0.20 × 0.01) ≈ 509,703.24. Even with mixed orientations, at most 509,703 whole books could fit. Under these sample dimensions, roughly half a million is a reasonable estimate for an empty room before allowing for access and shelf space.

Assume 8 billion people and an average human body volume of about 0.07 m³ (70 litres). Total body volume ≈ 8 × 10⁹ × 0.07 = 5.6 × 10⁸ m³.

Cube side = ∛(5.6 × 10⁸) ≈ 824 m. This assumes no air gaps and treats bodies as perfectly compressible packing units; a real crowd would need more space and would not physically fit safely in such a cube.

1. Let R = 6371 km, water volume W = 1.38 × 10⁹ km³ and thickness = t km.
2. Water shell volume = volume of outer sphere − Earth sphere = ⁴⁄₃π[(R + t)³ − R³] = W.
3. Rearrange: (R + t)³ = R³ + 3W/(4π). Thus t = ∛[R³ + 3W/(4π)] − R.
4. Substitute values to get t ≈ 2.704 km (about 2704 m).

A useful thin-shell approximation is t ≈ W/(4πR²) ≈ 2.706 km, very close because the thickness is tiny compared with Earth’s radius.

This asks for an example of two cuboids (before and after); changing every dimension by the same factor would not work.

Take the original dimensions as 20 cm × 20 cm × 6 cm:

• Original volume = 20 × 20 × 6 = 2400 cm³.
• Original total surface area = 2(400 + 120 + 120) = 1280 cm².

Change the dimensions to 20 cm × 60 cm × 1 cm:

• New volume = 20 × 60 × 1 = 1200 cm³, which is half of 2400.
• New TSA = 2(1200 + 60 + 20) = 2560 cm², which is twice 1280.

So these dimensions satisfy both conditions exactly.

Divide the house into simple cuboidal rooms. As a sample, suppose all ceilings are 3 m high:

Total enclosed room volume ≈ 36 + 36 + 22.5 + 12 + 9 = 115.5 m³. Do not double-count overlapping spaces. For irregular rooms, divide them into rectangles; for sloping roofs, estimate separately. The result changes with actual floor dimensions, wall boundaries and ceiling heights.