Chapter 14: Math of Space: Surface Area and Volume (Surface Area and Volume)

$a^3=64$, so $a=4$ cm and TSA $=6a^2=6\times 16=96$ cm².

96 cm²

2 m = 200 cm, so $200÷20=10$ cubes fit along each edge.

$$10\times 10\times 10=1000$$

1000 cubes

The boxes fit exactly: $40÷2=20$, $25÷1.25=20$, $10÷1=10$.

$$20\times 20\times 10=4000\left(=\frac{40\times 25\times 10}{2\times 1.25\times 1}=\frac{10000}{2.5}\right)$$

4000 boxes

Each cube has side 5 cm, so the cuboid is 10 cm × 5 cm × 5 cm.

$$\text{TSA}=2(10\times 5+5\times 5+10\times 5)=2(50+25+50)=250{\text{cm}}^2$$

(Check: two cubes have 2 × 150 = 300 cm², minus the two hidden faces, 2 × 25 = 50 cm².)

250 cm²

Original: $6\times 16=96$ cm². There are $4^3=64$ small cubes, each with surface area 6 cm², total $384$ cm².

$$96:384=1:4$$

1 : 4 (the volume is unchanged, but the surface area becomes 4 times as much).

Let the edges be l, w, h with lw = 6, wh = 15, lh = 10. Multiplying:

$$(lwh{)}^2=6\times 15\times 10=900\Rightarrow V=lwh=30{\text{cm}}^3$$

(The edges are 2, 3 and 5 cm.)

30 cm³

There are $5^3=125$ small cubes.

• (i) Three faces: only the corner cubes: 8.
• (ii) Two faces: cubes along the edges, not at corners: each of the 12 edges has $5-2=3$, so $12\times 3=36$.
• (iii) One face: the middle part of each face, a 3 × 3 square: $6\times 9=54$.
• (iv) No face: the inner 3 × 3 × 3 cube: 27.

Check: $8+36+54+27=125$ ✓

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

We need $2(lw+wh+hl)=100$, i.e. $lw+wh+hl=50$, with whole numbers $l\le w\le h$.

Since each product is at least $l^2$, $3l^2\le 50$, so $l\le 4$.

• l = 1: $w+h+wh=50$, i.e. $(w+1)(h+1)=51=3\times 17$. With $w\ge 1$: w = 2, h = 16.
• l = 2: $2w+2h+wh=50$, i.e. $(w+2)(h+2)=54$. With $w\ge 2$, $w+2$ is between 4 and $\sqrt{54}\approx 7.3$: only 6 × 9 works, so w = 4, h = 7.
• l = 3: $(w+3)(h+3)=59$, which is prime: no solution.
• l = 4: $(w+4)(h+4)=66$ with $w+4$ between 8 and $\sqrt{66}\approx 8.1$; 8 does not divide 66: no solution.

(i) Yes. (ii)–(iii) Exactly two: 1 × 2 × 16 cm and 2 × 4 × 7 cm (the case analysis above shows there are no others).

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

$$\frac{{\text{CSA}}_A}{{\text{CSA}}_B}=\frac{2\pi rh}{2\pi (2r)\left(\frac{h}{2}\right)}=1,\frac{V_A}{V_B}=\frac{\pi r^{2}h}{\pi (2r{)}^2\left(\frac{h}{2}\right)}=\frac{1}{2}$$

CSA ratio 1 : 1; volume ratio 1 : 2

Take radii 2k, 3k and heights 3m, 2m.

$$\text{(a) }\frac{\pi (2k{)}^2(3m)}{\pi (3k{)}^2(2m)}=\frac{12}{18}=\frac{2}{3}\text{(b) }\frac{2\pi (2k)(3m)}{2\pi (3k)(2m)}=\frac{6}{6}=1$$

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

The largest cylinder stands on one face: its base is the circle fitting in the square face (radius $\frac{r}{2}$) and its height is r.

$$V=\pi {\left(\frac{r}{2}\right)}^{2}r=\frac{\pi r^3}{4}{\text{cm}}^3$$

(About 78.5% of the cube.)

$\frac{\pi r^3}{4}$ cm³

New volume $=\pi (1.1r{)}^2\times h\left(1-\frac{x}{100}\right)=\pi r^{2}h$, so

$$1.21\left(1-\frac{x}{100}\right)=1\Rightarrow 1-\frac{x}{100}=\frac{100}{121}\Rightarrow x=\frac{2100}{121}\approx 17.36$$

x = 2100/121 ≈ 17.36

(i) $12^3=1728$ cm³.

(ii) $\frac{22}{7}\times 4\times 12=\frac{1056}{7}\approx 150.86$ cm³.

(iii) $1728÷150.86\approx 11.45$, so 11 complete rods.

(i) 1728 cm³ (ii) ≈ 150.86 cm³ (iii) 11 complete rods

r = 12 m, l = 21 m.

$$\text{TSA}=\pi r(l+r)=\frac{22}{7}\times 12\times 33=\frac{8712}{7}\approx 1244.57{\text{m}}^2$$

About 1244.57 m²

$$\text{CSA}=\pi rl=\frac{22}{7}\times 7\times 10=220{\text{cm}}^2$$

220 cm²

$l=\sqrt{16^2+12^2}=\sqrt{400}=20$ cm. Using π = 3.14:

$$\text{CSA}=\pi rl=3.14\times 12\times 20=753.6{\text{cm}}^2,\text{TSA}=\pi r(l+r)=3.14\times 12\times 32=1205.76{\text{cm}}^2$$

CSA = 753.6 cm², TSA = 1205.76 cm² (π = 3.14)

$$\frac{1}{3}\times 3.14\times r^2\times 15=1570\Rightarrow 15.7r^2=1570\Rightarrow r^2=100\Rightarrow r=10\text{cm}$$

10 cm

(i) $\frac{22}{7}\times r\times 14=308\Rightarrow 44r=308\Rightarrow r=7$ cm.

(ii) TSA = CSA + base $=308+\frac{22}{7}\times 49=308+154=462$ cm².

(i) 7 cm (ii) 462 cm²

$l=\sqrt{24^2+7^2}=\sqrt{625}=25$ cm. One cap (curved surface only) $=\frac{22}{7}\times 7\times 25=550$ cm².

$$10\times 550=5500{\text{cm}}^2$$

5500 cm²

$l=\sqrt{8^2+6^2}=10$ m. Curved surface $=\pi rl=3.14\times 6\times 10=188.4$ m².

$$\text{Length}=\frac{188.4}{3}+0.2=62.8+0.2=63\text{m}$$

63 m (using π = 3.14)

Rotating about the 8 cm leg gives a cone with h = 8 cm, r = 6 cm and l = 10 cm.

$$V=\frac{1}{3}\pi r^{2}h=\frac{1}{3}\times \frac{22}{7}\times 36\times 8=96\pi \approx 301.71{\text{cm}}^3,\text{CSA}=\pi rl=60\pi \approx 188.57{\text{cm}}^2$$

A cone: V = 96π ≈ 301.71 cm³, CSA = 60π ≈ 188.57 cm²

The water forms a smaller cone with half the height. Its radius is also half (the shapes are similar), so

$$\frac{\frac{1}{3}\pi {\left(\frac{r}{2}\right)}^2\frac{h}{2}}{\frac{1}{3}\pi r^{2}h}=\frac{1}{8}$$

The water fills only $\frac{1}{8}$ of the cup.

$$4\pi r^2=4\times \frac{22}{7}\times 0.49=6.16{\text{cm}}^2$$

6.16 cm²

Same metal, so the weights are in the ratio of the volumes: $\frac{5920}{740}=8={\left(\frac{R}{r}\right)}^3$, so $\frac{R}{r}=2$. The smaller radius is 2.5 cm, so R = 5 cm.

The larger sphere has radius 5 cm.

Surface areas are in the ratio of the squares of the radii: ${\left(\frac{1}{4}\right)}^2=\frac{1}{16}$.

Moon : Earth = 1 : 16

$$\text{(i) }2\pi r^2=2\times \frac{22}{7}\times 441=2772{\text{cm}}^2\text{(ii) }3\pi r^2=4158{\text{cm}}^2$$

(i) 2772 cm² (ii) 4158 cm²

Box: $16\times 8\times 8=1024$ cm³. The spheres (diameter 4 cm) fit as 4 × 2 × 2 = 16.

$$16\times \frac{4}{3}\times \frac{22}{7}\times 2^3=\frac{11264}{21}\approx 536.38{\text{cm}}^3$$

$$\text{Liquid}=1024-536.38\approx 487.62\approx 488{\text{cm}}^3$$

About 488 cm³

$$\frac{\frac{4}{3}\pi (1.1r{)}^3}{\frac{4}{3}\pi r^3}=1.1^3=1.331$$

So the volume becomes 1.331 times as much: an increase of 33.1%.

(1.1)³ = 1.331, an increase of 33.1%.

$${\left(1+\frac{x}{100}\right)}^3=1.728=1.2^3\Rightarrow 1+\frac{x}{100}=1.2\Rightarrow x=20$$

x = 20

$2\pi r=35.2\Rightarrow r=\frac{35.2\times 7}{44}=5.6$ m. Curved surface:

$$2\pi r^2=2\times \frac{22}{7}\times 31.36=197.12{\text{m}}^2=1971200{\text{cm}}^2$$

$$\text{Cost}=\frac{1971200}{100}\times 10=₹1,97,120$$

₹1,97,120

Container: $10\times 15\times 20=3000$ cm³. Assume a scoop is a sphere of radius 2.5 cm: $\frac{4}{3}\times 3.14\times 2.5^3\approx 65$ cm³.

$$3000÷65\approx 46$$

(Ice cream gets pressed a little when scooped, so about 45 scoops. With a smaller scoop of radius 2 cm, about 33.5 cm³, we would get about 90.)

About 45 scoops (assuming a scoop of radius 2.5 cm; the answer depends on the scoop size).

$$(a+1{)}^3-a^3=3a^2+3a+1$$

(For example, from a 2-cube (8 cubes) to a 3-cube (27 cubes) we add $12+6+1=19$.)

$3a^2+3a+1$ unit cubes

(i) Assume a classroom of 9 m × 7 m × 3.5 m: about 220 m³ = 2,20,000 litres. A person drinks about 2.5 litres a day; in 70 years that is about $2.5\times 365\times 70\approx 64,000$ litres. That fills less than a third of the room.

(ii) Wall area: perimeter $2(9+7)=32$ m, height 3.5 m, so 112 m²; subtract about 12 m² for doors and windows: about 100 m². With walls 23 cm thick (one brick length), the brickwork is about $100\times 0.23=23$ m³. A brick with mortar takes about 20 cm × 10 cm × 10 cm = 0.002 m³, so about $23÷0.002\approx 11,500$ bricks.

(i) No: about 64,000 litres in a lifetime vs about 2,20,000 litres for the room. (ii) Roughly 11,000–12,000 bricks (for a 9 m × 7 m × 3.5 m room).

$$\text{A: }50^3=1,25,000{\text{mm}}^3,\text{B: }100\times 10^3=1,00,000{\text{mm}}^3$$

(A cube of side 50 mm is made of $5^3=125$ cubes of side 10 mm.)

Option A gives more chocolate (125 small cubes' worth instead of 100).

Assume about 820 crore (8.2 billion) people, each standing in about 0.5 m × 0.5 m = 0.25 m².

$$8.2\times 10^9\times 0.25=2.05\times 10^9{\text{m}}^2\approx 2050{\text{km}}^2$$

That is only about 45 km × 45 km, a little larger than Delhi (about 1,484 km²).

About 2,000 km² (roughly a 45 km × 45 km square).

One glass: $\frac{22}{7}\times 3.5^2\times 12=462$ cm³.

$$1600\times 462=7,39,200{\text{cm}}^3=739.2\text{litres}$$

739.2 litres

Sphere: $4\pi (25)=100\pi$. So the cone's curved surface is $20\pi =\pi \times 4\times l$, giving l = 5 cm. Then $h=\sqrt{5^2-4^2}=3$ cm.

$$V=\frac{1}{3}\pi \times 16\times 3=16\pi \approx 50.29{\text{cm}}^3$$

Height 3 cm; volume 16π ≈ 50.29 cm³

Volumes are in the ratio of the cubes of the radii.

$$\text{(i) }{\left(\frac{69900}{6370}\right)}^3\approx 10.97^3\approx 1321\text{(ii) }{\left(\frac{695700}{6370}\right)}^3\approx 109.2^3\approx 13,03,000$$

(i) about 1,321 (ii) about 13 lakh (1.3 million)

The smallest enclosing cylinder has radius r and height 2r, so its volume is $\pi r^2\times 2r=2\pi r^3$.

$$\frac{\frac{4}{3}\pi r^3}{2\pi r^3}=\frac{2}{3}$$

Sphere ÷ cylinder = (4/3)πr³ ÷ 2πr³ = 2/3.

If the radius is R and the gap is h:

$$2\pi (R+h)-2\pi R=1\Rightarrow 2\pi h=1\Rightarrow h=\frac{1}{2\pi }\approx 0.159\text{m}$$

R cancels out! So the gap is about 16 cm for the Earth, and also for (i) the Moon, (ii) Jupiter and (iii) a volleyball. It is surprising because 1 m seems tiny compared with the Earth's equator (about 40,000 km), yet it lifts the string by the same 16 cm as around a ball.

About 15.9 cm (1/(2π) m) in every case, whatever the size of the sphere.

The square's diagonal is the diameter 2r, so its side is $r\sqrt{2}$ and its area is $2r^2$.

$$\frac{\frac{1}{3}(2r^2)h}{\pi r^{2}h}=\frac{2}{3\pi }\approx 0.21$$

$\frac{2}{3\pi }$ (about 0.21)

(i) $(l+1)wh-lwh=wh$: option (c).

(ii) $\pi (r+1{)}^{2}h-\pi r^{2}h=\pi (2r+1)h=2\pi rh+\pi h$: option (e).

(iii) The volume decreases by

$$\frac{4}{3}\pi r^3-\frac{4}{3}\pi (r-1{)}^3=\frac{4}{3}\pi (3r^2-3r+1)\text{cubic units}$$

(i) (c) wh (ii) (e) 2πrh + πh (iii) a decrease of (4/3)π(3r² − 3r + 1) cubic units

Side $a=V^{1/3}$, so $S=6a^2=6V^{2/3}$.

$S=6V^{2/3}$

$6a^2=S$ gives $a=\sqrt{\frac{S}{6}}$, so $V={\left(\frac{S}{6}\right)}^{3/2}$.

$V={\left(\frac{S}{6}\right)}^{3/2}$

The water must rise 4 cm: the extra volume is $\pi \times 16\times 4=64\pi$ cm³. Each marble displaces $\frac{4}{3}\pi$ cm³.

$$64\pi ÷\frac{4}{3}\pi =48$$

48 marbles

The ink is in the thin refill tube, a cylinder. Typical measurements: inner diameter about 1.6 mm (r ≈ 0.08 cm) and ink length about 9 cm.

$$V=\pi r^{2}h\approx 3.14\times 0.0064\times 9\approx 0.18{\text{cm}}^3$$

So a new pen holds roughly 0.2 mL of ink. (Use your own measurements of the refill.)

Roughly 0.2 cm³ (about 0.2 mL), with the measurements shown.

(i) Dough: $\frac{4}{3}\times 3.14\times 216\approx 905$ cm³. A chapati is a thin cylinder, say radius 8 cm and thickness 0.3 cm: $3.14\times 64\times 0.3\approx 60$ cm³.

$$905÷60\approx 15$$

(ii) Treat each half as a hemispherical shell. Example measurements: outer radius of the flesh 6 cm and flesh thickness 1 cm (inner radius 5 cm). For both halves (a full spherical shell):

$$\frac{4}{3}\pi (6^3-5^3)=\frac{4}{3}\times 3.14\times 91\approx 381{\text{cm}}^3$$

(i) About 15 chapatis (ii) e.g. about 380 cm³ of flesh for a coconut with flesh from radius 5 cm to 6 cm (use your own measurements).

(i) The book has about 180 printed pages, i.e. about 90 sheets of about 20 cm × 27.5 cm (0.055 m² each). Laid out, the sheets cover about $90\times 0.055\approx 5$ m² (about 10 m² if every page is printed on a separate sheet). A classroom floor is about 60 m², so no.

(ii) Storeroom: $15\times 20\times 30=9000$ ft³ $\approx 9000\times 0.0283\approx 255$ m³. A book is about 20 cm × 27.5 cm × 1 cm $=550$ cm³ $=0.00055$ m³.

$$255÷0.00055\approx 4,60,000$$

Books are cuboids and pack with almost no gaps, so the maximum is about 4.6 lakh books (a little less in practice).

(i) No, the pages cover only about 5–10 m². (ii) About 4.6 lakh books.

A person of about 70 kg has a volume of about 0.07 m³ (people are roughly as dense as water). For 8.2 billion people:

$$8.2\times 10^9\times 0.07\approx 5.7\times 10^8{\text{m}}^3,\text{side}=\sqrt[3]{5.7\times 10^8}\approx 830\text{m}$$

(If people only stand packed shoulder to shoulder, not squeezed, the cube would be somewhat larger, around 1 km.)

Less than 1 km: about 830 m (packed tightly).

(i) If the thickness is t, the water is the shell between radii R and R + t:

$$\frac{4}{3}\pi [(R+t{)}^3-R^3]=V\Rightarrow t=\sqrt[3]{R^3+\frac{3V}{4\pi }}-R$$

Since t is tiny compared with R, a very good approximation is $t\approx \frac{V}{4\pi R^2}$ (volume ÷ surface area).

(ii)

$$t\approx \frac{1.38\times 10^9}{4\times 3.1416\times 6371^2}=\frac{1.38\times 10^9}{5.10\times 10^8}\approx 2.7\text{km}$$

(i) $t=\sqrt[3]{R^3+\frac{3V}{4\pi }}-R\approx \frac{V}{4\pi R^2}$ (ii) about 2.7 km

We need two cuboids A and B such that B has half the volume of A but twice its surface area. Volume and surface area are not tied to each other, so this is possible: flattening and stretching a cuboid adds surface area while reducing volume.

Example (all lengths in cm):

• A = 9 × 10 × 12: volume 1080 cm³, surface area $2(90+120+108)=636$ cm².
• B = 1 × 6 × 90: volume 540 cm³ (half), surface area $2(6+90+540)=1272$ cm² (double).

For example, a 9 × 10 × 12 cuboid (V = 1080, S = 636) becomes a 1 × 6 × 90 cuboid (V = 540, S = 1272).

A plan for the project:

• Draw a rough floor plan. Measure (or pace out) the length and breadth of each room, and the ceiling height.
• Treat each room as a cuboid: volume = l × b × h. Add all the rooms (and corridors, if any).
• Sloping roofs or domes can be treated as half-cuboids (triangular prisms) or hemispheres.
• Example: 3 rooms of 4 m × 3.5 m × 3 m (42 m³ each), a kitchen of 3 m × 2.5 m × 3 m (22.5 m³) and a hall of 5 m × 4 m × 3 m (60 m³) give about $126+22.5+60=208.5$ m³.
• Present the plan, the table of measurements, the calculation and a note on where you approximated.

Treat each room as a cuboid (length × breadth × height) and add them; e.g. about 210 m³ for the sample house above.