Question 1

The volume of a cube is $64 \text{ cm}^3$. What is its total surface area?

Accepted Answer

Let the side be $a$. The volume of a cube is $a^3$. $a^3 = 64$ $a = \sqrt[3]{64} = 4 \text{ cm}$ The total surface area of a cube is $6a^2$ (six equal square faces). $\text{TSA} = 6 \times 4^2$ $= 6 \times 16 = 96 \text{ cm}^2$ The total surface area is $96 \text{ cm}^2$.

Question 2

How many small cubes with side 20 cm can be packed tight in a cubical box with side 2 m?

Accepted Answer

First bring both lengths to the same unit: $2 \text{ m} = 200 \text{ cm}$.

Along one edge of the box the number of small cubes that fit is

$\dfrac{200}{20} = 10$

The box holds 10 cubes along its length, 10 along its width and 10 along its height, so

$10 \times 10 \times 10 = 1000$

Check by volume:

$\dfrac{200^3}{20^3}$

$= \left(\dfrac{200}{20}\right)^3$

$= 10^3$

$= 1000$

1000 small cubes can be packed.

Dividing the two volumes gives the right count here only because 20 divides 200 exactly, so the cubes fill the box with no gaps. If the box were 2.1 m, the volume division would give more cubes than can actually fit.

Question 3

The dimensions of a godown are $40 \text{ m} \times 25 \text{ m} \times 10 \text{ m}$. If it is filled with cuboidal boxes, each of dimensions $2 \text{ m} \times 1.25 \text{ m} \times 1 \text{ m}$, then find the number of boxes.

Accepted Answer

Count how many boxes fit along each edge of the godown, matching the edges in the given order:

$\dfrac{40}{2} = 20$

$\dfrac{25}{1.25} = 20$

$\dfrac{10}{1} = 10$

So the number of boxes is

$20 \times 20 \times 10 = 4000$

Check by volume:

$\text{volume of godown} = 40 \times 25 \times 10$

$= 10000 \text{ m}^3$

$\text{volume of one box} = 2 \times 1.25 \times 1$

$= 2.5 \text{ m}^3$

$\dfrac{10000}{2.5} = 4000$

The godown holds 4000 boxes. Both methods agree because each edge of the godown is an exact multiple of the matching edge of a box, so the boxes fill it completely.

Question 4

Two cubes each of volume $125 \text{ cm}^3$ are joined end to end. Find the surface area of the resulting cuboid.

Accepted Answer

Side of each cube: $a^3 = 125$, so $a = 5 \text{ cm}$.Joining two cubes end to end gives a cuboid $10 \text{ cm} \times 5 \text{ cm} \times 5 \text{ cm}$, i.e. $l = 10$, $w = 5$, $h = 5$. $\text{TSA} = 2(lw + wh + hl)$ $= 2(10 \times 5 + 5 \times 5 + 5 \times 10)$ $= 2(50 + 25 + 50) = 250 \text{ cm}^2$ The surface area of the cuboid is $250 \text{ cm}^2$.Check another way: the two cubes have $2 \times 6 = 12$ faces of $25 \text{ cm}^2$ each, i.e. $300 \text{ cm}^2$. Two faces are pressed together and disappear, so $300 - 2 \times 25 = 250 \text{ cm}^2$.

Question 5

A 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.)

Accepted Answer

Surface area of the original cube: $6 \times 4^2 = 96 \text{ cm}^2$ Number of small cubes: $4 \times 4 \times 4 = 64$ (the volume $64 \text{ cm}^3$ is unchanged). Each small cube has surface area $6 \times 1^2 = 6 \text{ cm}^2$, so all of them together have $64 \times 6 = 384 \text{ cm}^2$ $96 : 384 = 1 : 4$ The ratio is 1 : 4 — the same material now has four times the surface.This is why living things are made of tiny cells, why our lungs and intestines are folded into huge numbers of small pockets, and why a small animal loses heat much faster than a large one: cutting the same volume into smaller pieces multiplies the surface through which food, oxygen and heat can pass.

Question 6

The surface areas of the three faces of a cuboid that meet at one of the corners of the cuboid are $6 \text{ cm}^2$, $15 \text{ cm}^2$, and $10 \text{ cm}^2$ respectively. What is the volume of the cuboid?

Accepted Answer

Let the edges be $l$, $w$, $h$. The three faces at a corner are $l \times w$, $w \times h$ and $h \times l$: $lw = 6$ $wh = 15$ $hl = 10$ Multiply the three equations: $(lw)(wh)(hl) = 6 \times 15 \times 10$ $(lwh)^2 = 900$ $lwh = \sqrt{900} = 30$ The volume is $30 \text{ cm}^3$.We can even recover the edges: $l = \dfrac{lwh}{wh} = \dfrac{30}{15} = 2$ $w = \dfrac{lwh}{hl} = \dfrac{30}{10} = 3$ $h = \dfrac{lwh}{lw} = \dfrac{30}{6} = 5$ Check: $2 \times 3 = 6$, $3 \times 5 = 15$, $5 \times 2 = 10$ ✓.

Question 7

* A cube of side 5 cm is painted on all its faces. If it is sliced into $1 \text{ cm}^3$ cubes, how many of these $1 \text{ cm}^3$ cubes have (i) exactly three faces painted? (ii) exactly two faces painted? (iii) exactly one face painted? (iv) no face painted?

Accepted Answer

There are $5 \times 5 \times 5 = 125$ small cubes. How many faces of a small cube are painted depends only on where it sits in the big cube.(i) Three faces painted: only the cubes at the corners. A cube has 8 corners, so 8 cubes.(ii) Two faces painted: cubes along an edge but not at a corner. Each edge has 5 cubes; removing the 2 corner cubes leaves 3. A cube has 12 edges: $12 \times (5 - 2) = 12 \times 3 = 36$ (iii) One face painted: cubes in the middle of a face, away from its edges. On each face that is a $3 \times 3$ square, and there are 6 faces: $6 \times 3 \times 3 = 54$ (iv) No face painted: the hidden inner cube of side $5 - 2 = 3$: $3 \times 3 \times 3 = 27$ Check: $8 + 36 + 54 + 27 = 125$ ✓. Answers: 8, 36, 54, 27.For a cube of side $n$ the same reasoning gives $8$, $12(n-2)$, $6(n-2)^2$ and $(n-2)^3$ — and they always add up to $n^3$.

Question 8

* Find a cuboid with edges whose lengths are integers (in cm), given that it has a total surface area of exactly $100 \text{ cm}^2$. (i) Is there more than one such cuboid? (ii) Can you find them all? (iii) Show that you have found them all.

Accepted Answer

Let the edges be whole numbers $l \le w \le h$ (we may name the shortest edge $l$). $2(lw + wh + hl) = 100$ $lw + wh + hl = 50$ How big can the shortest edge be? Since $l \le w \le h$, each of the three products is at least $l^2$, so $3l^2 \le 50$, i.e. $l^2 \le 16$. So $l$ can only be 1, 2, 3 or 4.A factorising trick. Add $l^2$ to both sides of $wh + l(w + h) = 50$: $wh + lw + lh + l^2 = 50 + l^2$ $(w + l)(h + l) = 50 + l^2$ So for each value of $l$ we only need to split $50 + l^2$ into two factors, the smaller one at least $2l$ (because $w \ge l$).$l = 1$: $(w+1)(h+1) = 51 = 3 \times 17$ (the split $1 \times 51$ would make $w = 0$). So $w = 2$, $h = 16$.$l = 2$: $(w+2)(h+2) = 54$. Factor pairs of 54 are $1 \times 54$, $2 \times 27$, $3 \times 18$, $6 \times 9$; only $6 \times 9$ has its smaller factor at least 4. So $w = 4$, $h = 7$.$l = 3$: $(w+3)(h+3) = 59$, which is prime — no cuboid.$l = 4$: $(w+4)(h+4) = 66$. The smaller factor must be at least 8 and at most $\sqrt{66} \approx 8.1$, so it would have to be 8, which does not divide 66 — no cuboid. $1 \times 2 \times 16:\ 2(2 + 32 + 16) = 100$ \(2 \time

Question 9

Two 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.

Accepted Answer

Let cylinder A have radius $r$ and height $h$. Then B has radius $2r$ and height $\dfrac{h}{2}$.Curved surface areas (CSA $= 2\pi rh$): $\text{CSA}_A = 2\pi rh$ $\text{CSA}_B = 2\pi (2r)\left(\dfrac{h}{2}\right) = 2\pi rh$ So $\text{CSA}_A : \text{CSA}_B = $ 1 : 1.Volumes ($V = \pi r^2 h$): $V_A = \pi r^2 h$ $V_B = \pi (2r)^2 \left(\dfrac{h}{2}\right) = 2\pi r^2 h$ So $V_A : V_B = $ 1 : 2.The curved surface depends on the radius only once (the circumference), so doubling the radius and halving the height cancel. The volume depends on the radius twice (the area of the base), so doubling the radius wins.

Question 10

The 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.

Accepted Answer

Take the radii as $2k$ and $3k$, and the heights as $3m$ and $2m$.(a) Volumes: $\dfrac{V_1}{V_2}$ $= \dfrac{\pi (2k)^2 (3m)}{\pi (3k)^2 (2m)}$ $= \dfrac{12k^2m}{18k^2m} = \dfrac{2}{3}$ The volumes are in the ratio 2 : 3.(b) Curved surface areas: $\dfrac{\text{CSA}_1}{\text{CSA}_2}$ $= \dfrac{2\pi (2k)(3m)}{2\pi (3k)(2m)}$ $= \dfrac{6km}{6km} = 1$ The curved surface areas are in the ratio 1 : 1.

Question 11

The 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 $\text{cm}^3$)?

Accepted Answer

The cylinder stands on one face of the cube. Its height can be at most the edge, $r$. Its base circle must fit inside a square face of side $r$; the largest such circle touches all four sides, so its diameter is $r$ and its radius is $\dfrac{r}{2}$.

Left: the largest cylinder inside a cube of edge r. Right: seen from above, its circle just touches the four sides of the square face, so its radius is r/2; its height is the full edge r.

$V = \pi \left(\dfrac{r}{2}\right)^2 \times r$

$= \pi \times \dfrac{r^2}{4} \times r = \dfrac{\pi r^3}{4}$

The volume of the cylinder is $\dfrac{\pi r^3}{4} \text{ cm}^3$.

Since $\dfrac{\pi}{4} \approx 0.785$, the cylinder keeps about 78.5% of the cube. It does not matter which face it stands on — all faces of a cube are the same.

Question 12

* 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 $x$.

Accepted Answer

Original cylinder: radius $r$, height $h$, volume $\pi r^2 h$.New radius $= r + 10\% \text{ of } r = 1.1r$. New height $= h\left(1 - \dfrac{x}{100}\right)$. $\pi (1.1r)^2 \, h\left(1 - \dfrac{x}{100}\right)$ $= \pi r^2 h$ Divide both sides by $\pi r^2 h$: $1.21\left(1 - \dfrac{x}{100}\right) = 1$ $1 - \dfrac{x}{100} = \dfrac{1}{1.21} = \dfrac{100}{121}$ $\dfrac{x}{100} = 1 - \dfrac{100}{121} = \dfrac{21}{121}$ $x = \dfrac{2100}{121} \approx 17.36$ The height must be decreased by $\dfrac{2100}{121}\% \approx 17.36\%$.It is not 10% and not 21%. A 10% bigger radius makes the base 21% bigger (the base is 1.21 times as large), so the height must shrink to $\dfrac{1}{1.21}$ of itself, which is a cut of about 17.36%.

Question 13

A 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

Accepted Answer

(i) Volume of the cube: $12^3 = 12 \times 12 \times 12 = 1728 \text{ cm}^3$ (ii) Volume of one rod, with $\pi = \dfrac{22}{7}$: $\pi r^2 h = \dfrac{22}{7} \times 2 \times 2 \times 12$ $= \dfrac{1056}{7} \approx 150.86 \text{ cm}^3$ (iii) Melting does not change the volume. Put both volumes in as products and cancel before multiplying ($1728 = 12 \times 12 \times 12$, and $2 \times 2 \times 12 = 48$ goes into it 36 times): $\text{number of rods} = \dfrac{12 \times 12 \times 12}{\frac{22}{7} \times 2 \times 2 \times 12}$ $= \dfrac{36 \times 7}{22} = \dfrac{252}{22} \approx 11.45$ Only whole rods count, so 11 complete rods can be formed (with about $1728 - 11 \times 150.86 \approx 68.6 \text{ cm}^3$ of metal left over — not enough for a twelfth rod).

Question 14

Find the total surface area of a cone, if its slant height is 21 m and diameter of its base is 24 m. (The book allows $\pi \approx \dfrac{22}{7}$ or $\pi \approx 3.14$.)

Accepted Answer

Radius $r = \dfrac{24}{2} = 12 \text{ m}$, slant height $l = 21 \text{ m}$. $\text{TSA} = \pi r l + \pi r^2 = \pi r(l + r)$ $= \pi \times 12 \times (21 + 12) = 396\pi$ With $\pi = \dfrac{22}{7}$: $396 \times \dfrac{22}{7}$ $= \dfrac{8712}{7} \approx 1244.57 \text{ m}^2$ The total surface area is $396\pi \approx 1244.57 \text{ m}^2$. (With $\pi = 3.14$ you get $396 \times 3.14 = 1243.44 \text{ m}^2$; both are accepted.)

Question 15

Find the curved surface area of a right circular cone whose slant height is 10 cm and base radius is 7 cm.

Accepted Answer

$r = 7 \text{ cm}$, $l = 10 \text{ cm}$. Since the radius is a multiple of 7, take $\pi = \dfrac{22}{7}$. $\text{CSA} = \pi r l = \dfrac{22}{7} \times 7 \times 10$ $= 220 \text{ cm}^2$ The curved surface area is $220 \text{ cm}^2$.

Question 16

The 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.

Accepted Answer

First find the slant height from $l^2 = h^2 + r^2$: $l^2 = 16^2 + 12^2$ $= 256 + 144$ $= 400$ $l = 20 \text{ cm}$ Curved surface area, with $\pi = 3.14$: $\text{CSA} = \pi r l = \pi \times 12 \times 20 = 240\pi$ $= 240 \times 3.14 = 753.6 \text{ cm}^2$ Total surface area (add the base): $\text{TSA} = \pi r(l + r)$ $= \pi \times 12 \times 32$ $= 384\pi$ $= 384 \times 3.14 = 1205.76 \text{ cm}^2$ CSA $= 240\pi \approx 753.6 \text{ cm}^2$; TSA $= 384\pi \approx 1205.76 \text{ cm}^2$.

Question 17

A cone has a height of 15 cm. If its volume is $1570 \text{ cm}^3$, find the radius of the base.

Accepted Answer

The number 1570 is a multiple of 3.14 ($1570 = 500 \times 3.14$), so take $\pi = 3.14$. $\dfrac{1}{3}\pi r^2 h = 1570$ $\dfrac{1}{3} \times 3.14 \times r^2 \times 15 = 1570$ $5 \times 3.14 \times r^2 = 1570$ $15.7 r^2 = 1570$ $r^2 = 100$ $r = 10 \text{ cm}$ The radius of the base is 10 cm. Check: $\dfrac{1}{3} \times 3.14 \times 100 \times 15 = 1570$ ✓.

Question 18

The curved surface area of a cone is $308 \text{ cm}^2$ and its slant height is 14 cm. Find (i) the radius of the base, (ii) the total surface area of the cone.

Accepted Answer

(i) Take $\pi = \dfrac{22}{7}$ (308 and 14 are both multiples of 7). $\pi r l = 308$ $\dfrac{22}{7} \times r \times 14 = 308$ $44r = 308$ $r = 7 \text{ cm}$ (ii) Total surface area = curved surface + base: $\text{TSA} = 308 + \pi r^2$ $= 308 + \dfrac{22}{7} \times 49$ $= 308 + 154 = 462 \text{ cm}^2$ Radius 7 cm; total surface area $462 \text{ cm}^2$.

Question 19

A 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.

Accepted Answer

A cap has no base, so only the curved surface is made of sheet. Slant height:

$l = \sqrt{h^2 + r^2} = \sqrt{24^2 + 7^2}$

$= \sqrt{576 + 49} = \sqrt{625} = 25 \text{ cm}$

$\text{CSA of one cap} = \pi r l$

$= \dfrac{22}{7} \times 7 \times 25 \text{ cm}^2$

$\text{sheet for 10 caps} = 10 \times \dfrac{22}{7} \times 7 \times 25$

$= 10 \times 22 \times 25 = 5500 \text{ cm}^2$

The sheet required is $5500 \text{ cm}^2$ (that is, $0.55 \text{ m}^2$), ignoring wastage.

Question 20

* 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.

Accepted Answer

The tarpaulin covers only the curved surface (the ground is the base). Slant height:

$l = \sqrt{8^2 + 6^2} = \sqrt{100} = 10 \text{ m}$

Curved surface area, with $\pi = 3.14$:

$\text{CSA} = \pi r l = 3.14 \times 6 \times 10 \text{ m}^2$

Keep this product as it is. The tarpaulin is a long rectangle 3 m wide, so its length is area $\div$ width; cancel the 6 by the 3 before multiplying:

$\text{length} = \dfrac{3.14 \times 6 \times 10}{3}$

$= 3.14 \times 2 \times 10 = 62.8 \text{ m}$

Add the 20 cm (0.2 m) for margins and wastage:

$62.8 + 0.2 = 63 \text{ m}$

A length of 63 m of tarpaulin is required.

Question 21

A right triangle with sides 6 cm, 8 cm and 10 cm is rotated through $360^\circ$ about the side of 8 cm. Find the volume and the curved surface area of the solid so formed.

Accepted Answer

Since $6^2 + 8^2 = 36 + 64 = 100 = 10^2$, the triangle is right-angled with legs 6 cm and 8 cm and hypotenuse 10 cm.Rotating it about the 8 cm leg gives a cone with height $h = 8 \text{ cm}$, base radius $r = 6 \text{ cm}$ (the other leg sweeps out the base) and slant height $l = 10 \text{ cm}$ (the hypotenuse sweeps out the curved surface). $V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \times 36 \times 8 = 96\pi$ $= 96 \times 3.14 = 301.44 \text{ cm}^3$ $\text{CSA} = \pi r l = \pi \times 6 \times 10 = 60\pi$ $= 60 \times 3.14 = 188.4 \text{ cm}^2$ Volume $96\pi \approx 301.44 \text{ cm}^3$; curved surface area $60\pi \approx 188.4 \text{ cm}^2$.

Question 22

* 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?

Accepted Answer

Hold the cup point-down. Let its rim have radius $R$ and its depth be $H$. The water is itself a cone, with depth $\dfrac{H}{2}$. Its slanting sides lie along the cup’s sides, so the water cone and the cup are similar: halving the depth also halves the radius, so the water surface has radius $\dfrac{R}{2}$.

Water filled to half the depth forms a smaller cone with half the radius and half the height of the cup, so it holds (1/2)³ = 1/8 of the cup.

$V_{\text{cup}} = \dfrac{1}{3}\pi R^2 H$

$V_{\text{water}}$

$= \dfrac{1}{3}\pi \left(\dfrac{R}{2}\right)^2 \left(\dfrac{H}{2}\right)$

$= \dfrac{1}{8} \times \dfrac{1}{3}\pi R^2 H$

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

Half the depth is only one-eighth of the volume, because the narrow part is at the bottom. To fill the cup half by volume you would have to fill it to $\dfrac{H}{\sqrt[3]{2}} \approx 0.79H$ — almost four-fifths of the way up.

Question 23

A ball bearing has a radius of 0.7 cm. Find its surface area.

Accepted Answer

Surface area of a sphere $= 4\pi r^2$. With $r = 0.7$ (a multiple of 7), take $\pi = \dfrac{22}{7}$. $4\pi r^2 = 4 \times \dfrac{22}{7} \times 0.7 \times 0.7$ $= 4 \times 22 \times 0.1 \times 0.7 = 6.16 \text{ cm}^2$ The surface area is $6.16 \text{ cm}^2$.

Question 24

* 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.

Accepted Answer

Both spheres are of the same metal, so weight is proportional to volume. The ratio of weights is

$\dfrac{5920}{740} = 8$

Let the radii be $R$ (larger) and $r$ (smaller). Then

$\dfrac{\frac{4}{3}\pi R^3}{\frac{4}{3}\pi r^3} = 8$

$\left(\dfrac{R}{r}\right)^3 = 8$

$\dfrac{R}{r} = 2$

The smaller sphere has diameter 5 cm, so $r = 2.5 \text{ cm}$ and

$R = 2 \times 2.5 = 5 \text{ cm}$

The radius of the larger sphere is 5 cm (its diameter is 10 cm).

Eight times the weight is only twice the radius: volume grows as the cube of the radius.

Question 25

The 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.

Accepted Answer

If the earth’s radius is $r$, the moon’s radius is $\dfrac{r}{4}$ (halving both diameters keeps the ratio). $\dfrac{\text{SA of moon}}{\text{SA of earth}}$ $= \dfrac{4\pi (r/4)^2}{4\pi r^2}$ $= \dfrac{1}{16}$ The surface areas of the moon and the earth are in the ratio 1 : 16.

Question 26

Find (i) the curved surface area and (ii) the total surface area of a hemisphere of radius 21 cm.

Accepted Answer

Take $\pi = \dfrac{22}{7}$; $r^2 = 21^2 = 441$.(i) The curved surface is half a sphere: $\text{CSA} = 2\pi r^2 = 2 \times \dfrac{22}{7} \times 441$ $= 2 \times 22 \times 63 = 2772 \text{ cm}^2$ (ii) Add the flat circular base: $\text{TSA} = 2\pi r^2 + \pi r^2 = 3\pi r^2$ $= 3 \times \dfrac{22}{7} \times 441 = 4158 \text{ cm}^2$ CSA $2772 \text{ cm}^2$; TSA $4158 \text{ cm}^2$.

Question 27

Metal spheres, each of radius 2 cm, are packed into a rectangular box of internal dimensions $16 \text{ cm} \times 8 \text{ cm} \times 8 \text{ cm}$. 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.

Accepted Answer

Each sphere has diameter 4 cm, so the number that fit in neat rows is

$\dfrac{16}{4} \times \dfrac{8}{4} \times \dfrac{8}{4} = 4 \times 2 \times 2 = 16$

— the 16 spheres do fit.

$\text{volume of box} = 16 \times 8 \times 8$

$= 1024 \text{ cm}^3$

$\text{volume of 16 spheres} = 16 \times \dfrac{4}{3}\pi \times 2^3$

$= \dfrac{512}{3}\pi$

With $\pi = 3.14$:

$\dfrac{512}{3} \times 3.14$

$= \dfrac{1607.68}{3} \approx 535.89 \text{ cm}^3$

$\text{liquid} = 1024 - 535.89$

$= 488.11 \text{ cm}^3$

Rounded to the nearest integer, the liquid is $488 \text{ cm}^3$. (With $\pi = \dfrac{22}{7}$ the liquid is $487.62 \text{ cm}^3$, which also rounds to 488.)

Question 28

The radius of a sphere is increased by 10%. Show that the volume increases by approximately 33.1%.

Accepted Answer

Original radius $r$, volume $V = \dfrac{4}{3}\pi r^3$. New radius $1.1r$. $V_{\text{new}} = \dfrac{4}{3}\pi (1.1r)^3$ $= (1.1)^3 \times \dfrac{4}{3}\pi r^3 = 1.331V$ $\text{increase} = 1.331V - V$ $= 0.331V$ An increase of $0.331V$ is 33.1% of $V$, as required.$1.1^3 = 1.1 \times 1.1 \times 1.1 = 1.21 \times 1.1 = 1.331$. A 10% change in a length becomes about a 21% change in an area and a 33.1% change in a volume.

Question 29

* The radius of a sphere is increased by $x$%. The volume of the sphere increases by 72.8%. Find the value of $x$.

Accepted Answer

New radius $= r\left(1 + \dfrac{x}{100}\right)$. The volume is multiplied by the cube of this factor, and a 72.8% increase means the volume becomes $1.728$ times as large: $\left(1 + \dfrac{x}{100}\right)^3 = 1.728$ Since $1.2^3 = 1.2 \times 1.2 \times 1.2 = 1.44 \times 1.2 = 1.728$, $1 + \dfrac{x}{100} = 1.2$ $x = 20$ The radius was increased by 20%.

Question 30

The 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 $100 \text{ cm}^2$.

Accepted Answer

Find the radius from the circumference, with $\pi = \dfrac{22}{7}$:

$2\pi r = 35.2$

$r = \dfrac{35.2 \times 7}{2 \times 22}$

$= \dfrac{246.4}{44}$

$= 5.6 \text{ m}$

Only the curved surface of the dome is painted:

$\text{CSA} = 2\pi r^2$

$= 2 \times \dfrac{22}{7} \times 5.6 \times 5.6 \text{ m}^2$

Keep this as it is. Since $1 \text{ m}^2 = 10000 \text{ cm}^2$ and the rate is ₹10 per $100 \text{ cm}^2$, multiply the area by $\dfrac{10000}{100} \times 10$. Then cancel before multiplying: $5.6 \div 7 = 0.8$ and $10000 \div 100 \times 10 = 1000$.

$\text{cost} = 2 \times \dfrac{22}{7} \times 5.6 \times 5.6 \times \dfrac{10000}{100} \times 10$

$= 2 \times 22 \times 0.8 \times 5.6 \times 1000$

$= 197120$

The cost of painting is ₹1,97,120.

A rate of ₹10 per $100 \text{ cm}^2$ is the same as ₹1000 per $\text{m}^2$ — that is why the answer is so large. Always convert the area to the unit in which the rate is given before multiplying.

Question 31

Estimate how many scoops of ice cream can be obtained from a cuboidal container of dimensions $10 \text{ cm} \times 15 \text{ cm} \times 20 \text{ cm}$. (For guesstimate problems, make a guess first before solving.)

Accepted Answer

$\text{volume of container} = 10 \times 15 \times 20$ $= 3000 \text{ cm}^3$ Assumption: a scoop is roughly a ball about 5 cm across, i.e. radius $2.5 \text{ cm}$. $\text{one scoop} = \dfrac{4}{3} \times 3.14 \times 2.5 \times 2.5 \times 2.5 \text{ cm}^3$ Divide the container by this product as it stands: $\text{scoops} = \dfrac{10 \times 15 \times 20 \times 3}{4 \times 3.14 \times 15.625}$ $= \dfrac{9000}{196.25} \approx 45.9$ So the container gives about 45 scoops.This is an estimate, not a fixed answer. A bigger scoop (radius 3 cm, about $113 \text{ cm}^3$) gives only about 26; ice cream is also pressed together when scooped. Any answer of roughly 25–50 with the reasoning shown is sensible.

Question 32

A 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 $a + 1$.

Accepted Answer

The new cube has $(a + 1)^3$ unit cubes and the old one has $a^3$. The number to be added is $(a + 1)^3 - a^3$ $= a^3 + 3a^2 + 3a + 1 - a^3$ $= 3a^2 + 3a + 1$ $3a^2 + 3a + 1$ unit cubes must be added.You can see each term: three slabs of $a \times a$ cubes on three faces ($3a^2$), three rods of $a$ cubes along the three edges where the slabs meet ($3a$), and one cube in the corner ($1$). Check with $a = 2$: $27 - 8 = 19$ and $3(4) + 3(2) + 1 = 19$ ✓.

Question 33

* 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.

Accepted Answer

Measure your own classroom; here we use a typical one, 8 m long, 6 m wide and 3.5 m high.

(i) Volume of the room:

$8 \times 6 \times 3.5 = 168 \text{ m}^3 = 168000 \text{ L}$

(1 m$^3$ = 1000 L.) Assumption: a person drinks about 2 litres a day for 70 years:

$2 \times 365 \times 70 = 51100 \text{ L} \approx 51 \text{ m}^3$

$\dfrac{51}{168} \approx 0.3$

No. A lifetime of drinking fills only about a third of the classroom. Even at 3 litres a day for 80 years (about $88 \text{ m}^3$) it would fill only about half.

(ii) Area of the four walls (perimeter of the floor $\times$ height):

$2(8 + 6) \times 3.5 = 98 \text{ m}^2$

Take off about $8 \text{ m}^2$ for a door and three windows, leaving $90 \text{ m}^2$. Assumptions: the wall is one brick thick, about 0.2 m; a modular brick is $19 \text{ cm} \times 9 \text{ cm} \times 9 \text{ cm}$, which with 1 cm of mortar takes up $20 \times 10 \times 10 = 2000 \text{ cm}^3 = 0.002 \text{ m}^3$.

$\text{volume of walls} = 90 \times 0.2$

$= 18 \text{ m}^3$

$\text{bricks} = \dfrac{18}{0.002} = 9000$

Roughly 9000 bricks.

This is an estimate, not a fixed answer. Your answer will depend on your classroom’s size, the thickness of its walls and the bricks used. What matters is the model: wall area $\times$ thickness $\div$ the space one brick takes up.

Question 34

You 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?

Accepted Answer

$\text{(A)}\ \ 50^3 = 125000 \text{ mm}^3$ $\text{(B)}\ \ 100 \times 10^3$ $= 100000 \text{ mm}^3$ Option (A) gives more chocolate — 25% more. (A cube of side 50 mm could be cut into $5^3 = 125$ cubes of side 10 mm, not just 100.)Option (B) has far more surface: $100 \times 6 \times 10^2 = 60000 \text{ mm}^2$ against $6 \times 50^2 = 15000 \text{ mm}^2$ — four times the wrapping for less chocolate.

Question 35

If all the people in the world are crowded together in one location, how much area would it cover?

Accepted Answer

Assumptions: the world has about 8.2 billion people ($8.2 \times 10^9$), and in a tightly packed standing crowd each person needs a square about 50 cm by 50 cm, i.e. $0.25 \text{ m}^2$. $8.2 \times 10^9 \times 0.25$ $= 2.05 \times 10^9 \text{ m}^2$ Since $1 \text{ km}^2 = 1000 \text{ m} \times 1000 \text{ m} = 10^6 \text{ m}^2$, $\dfrac{2.05 \times 10^9}{10^6} = 2050 \text{ km}^2$ That is about $2000 \text{ km}^2$ — a square only about $\sqrt{2050} \approx 45 \text{ km}$ on each side, a little more than the area of Delhi (about $1500 \text{ km}^2$).This is an estimate, not a fixed answer. With 1 m$^2$ per person (room to stand comfortably) the area becomes about 8200 km$^2$; either way it is smaller than many single Indian districts.

Question 36

A 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 $1000 \text{ cm}^3 = 1$ litre.

Accepted Answer

Radius $r = 3.5 \text{ cm}$, height of milk $h = 12 \text{ cm}$, $\pi = \dfrac{22}{7}$. $\text{milk in one glass} = \dfrac{22}{7} \times 3.5 \times 3.5 \times 12 \text{ cm}^3$ Keep it as a product. For 1600 students, and dividing by 1000 to get litres ($3.5 \div 7 = 0.5$, $1600 \div 1000 = 1.6$): $\dfrac{22}{7} \times 3.5 \times 3.5 \times 12 \times \dfrac{1600}{1000}$ $= 22 \times 0.5 \times 3.5 \times 12 \times 1.6$ $= 739.2 \text{ litres}$ About 739.2 litres of milk are needed.

Question 37

The 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.

Accepted Answer

$\text{SA of sphere} = 4\pi \times 5^2 = 100\pi$ Let the cone’s slant height be $l$. Its curved surface is $\pi \times 4 \times l = 4\pi l$. $100\pi = 5 \times 4\pi l = 20\pi l$ $l = 5 \text{ cm}$ $h = \sqrt{l^2 - r^2} = \sqrt{25 - 16} = 3 \text{ cm}$ $V = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3}\pi \times 16 \times 3 = 16\pi$ $= 16 \times \dfrac{22}{7} \approx 50.29 \text{ cm}^3$ Height 3 cm; volume $16\pi \approx 50.29 \text{ cm}^3$ ($50.24 \text{ cm}^3$ with $\pi = 3.14$).

Question 38

Take 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

Accepted Answer

For two spheres the $\dfrac{4}{3}\pi$ cancels, so the ratio of volumes is the cube of the ratio of radii: $\dfrac{V_1}{V_2}$ $= \dfrac{\frac{4}{3}\pi R_1^3}{\frac{4}{3}\pi R_2^3}$ $= \left(\dfrac{R_1}{R_2}\right)^3$ (i) Jupiter : Earth $\dfrac{69900}{6370} \approx 10.973$ $(10.973)^3 \approx 1321$ (ii) Sun : Earth $\dfrac{695700}{6370} \approx 109.215$ $(109.215)^3 \approx 1302710$ Jupiter’s volume is about 1320 times the Earth’s, and the Sun’s is about 13 lakh (1.3 million) times the Earth’s.Jupiter is only about 11 times as wide as the Earth, but 11 × 11 × 11 is over 1300. Keep the ratio of radii to several decimal places before cubing — rounding 10.973 to 11 changes the answer by almost 1%.

Question 39

Show that the volume of a sphere is equal to $\dfrac{2}{3}$ of the volume of the smallest cylinder which encloses it.

Accepted Answer

Let the sphere have radius $r$. The smallest cylinder that encloses it fits tightly around it: its base must hold the sphere’s widest circle, so its radius is $r$, and its height must reach from the bottom of the sphere to the top, so its height is $2r$. $V_{\text{cylinder}}$ $= \pi r^2 \times 2r$ $= 2\pi r^3$ $V_{\text{sphere}} = \dfrac{4}{3}\pi r^3$ $\dfrac{V_{\text{sphere}}}{V_{\text{cylinder}}}$ $= \dfrac{\frac{4}{3}\pi r^3}{2\pi r^3}$ $= \dfrac{4}{6}$ $= \dfrac{2}{3}$ So the volume of the sphere is $\dfrac{2}{3}$ of the volume of the smallest enclosing cylinder.The same cylinder appeared in Fig. 14.12: its curved surface $2\pi r \times 2r = 4\pi r^2$ equals the sphere’s surface. Archimedes was so proud of these two results that he asked for a sphere inside a cylinder to be carved on his tomb.

Question 40

* 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?

Accepted Answer

Let the Earth’s radius be $R$ metres and let the second string be $x$ metres above the ground. The second string is a circle of radius $R + x$, and it is 1 m longer than the first: $2\pi (R + x) = 2\pi R + 1$ $2\pi R + 2\pi x = 2\pi R + 1$ $2\pi x = 1$ $x = \dfrac{1}{2\pi} \approx \dfrac{1}{6.283} \approx 0.159 \text{ m}$ The second string is about 16 cm above the ground.The radius $R$ cancelled out completely, so the answer is the same 16 cm for (i) the Moon, (ii) Jupiter and (iii) a volleyball — or for anything round.Most people are surprised: 1 metre seems like nothing compared with the Earth’s 40,000 km equator, so we expect a microscopic gap. But the circumference grows by $2\pi$ for every unit the radius grows, whatever the starting size; so an extra metre of string always buys the same $\dfrac{1}{2\pi}$ m of height. The size of the ball only feels as if it should matter.

Question 41

* 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

Accepted Answer

The four corners of the square lie on a circle of radius $r$, so the square’s diagonal passes through the centre and is a diameter: diagonal $= 2r$. The apex is on the top of the cylinder, so the pyramid’s height is $h$.

Left: the square base sits on the cylinder’s base with its four corners on the circle; the apex is the centre of the top. Right: seen from above, the square’s diagonal is a diameter, 2r.

If the side of the square is $s$, its diagonal is $s\sqrt{2}$ (Baudhāyana–Pythagoras):

$s\sqrt{2} = 2r$

$s = \sqrt{2}\,r$

$\text{area of base} = s^2 = 2r^2$

(Or directly: the square is two triangles with base $2r$ and height $r$, area $2 \times \dfrac{1}{2} \times 2r \times r = 2r^2$.)

$V_{\text{pyramid}}$

$= \dfrac{1}{3} \times 2r^2 \times h$

$= \dfrac{2r^2h}{3}$

$V_{\text{cylinder}} = \pi r^2 h$

$\dfrac{V_{\text{pyramid}}}{V_{\text{cylinder}}}$

$= \dfrac{2r^2h/3}{\pi r^2 h}$

$= \dfrac{2}{3\pi}$

The ratio is $2 : 3\pi$, about 0.21 — the pyramid fills roughly a fifth of the cylinder.

Question 42

What 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

Accepted Answer

(i) $(l + 1)wh - lwh$ $= lwh + wh - lwh$ $= wh$ The volume increases by $wh$ cubic units — option (c). You add one slab of thickness 1 on the $w \times h$ face.(ii) $\pi (r + 1)^2 h - \pi r^2 h$ $= \pi h\left[(r + 1)^2 - r^2\right]$ $= \pi h(2r + 1) = 2\pi rh + \pi h$ The volume increases by $2\pi rh + \pi h$ cubic units — option (e). (Option (d) forgets that $(r+1)^2 - r^2$ is $2r + 1$, not $2r + 2$.)(iii) (for $r \ge 1$) $\dfrac{4}{3}\pi r^3 - \dfrac{4}{3}\pi (r - 1)^3$ $= \dfrac{4}{3}\pi\left[r^3 - (r - 1)^3\right]$ $ = \dfrac{4}{3}\pi\left[r^3 - (r^3 - 3r^2 + 3r - 1)\right]$ $= \dfrac{4}{3}\pi (3r^2 - 3r + 1)$ The volume decreases by $\dfrac{4}{3}\pi(3r^2 - 3r + 1) \text{ cubic units}$ Check with $r = 1$: the sphere vanishes completely, and the formula gives $\dfrac{4}{3}\pi(3 - 3 + 1) = \dfrac{4}{3}\pi \ \checkmark$

Question 43

* Given a cube with volume $V$, express its total surface area $S$ in terms of $V$.

Accepted Answer

If the side is $a$, then $V = a^3$, so $a = \sqrt[3]{V} = V^{1/3}$. $S = 6a^2 = 6\left(V^{1/3}\right)^2$ $= 6V^{2/3} = 6\sqrt[3]{V^2}$ $S = 6\sqrt[3]{V^2}$. Check: $V = 64$ gives $a = 4$ and $S = 6 \times 16 = 96$; also $6\sqrt[3]{64^2} = 6\sqrt[3]{4096} = 6 \times 16 = 96$ ✓.

Question 44

* Given a cube with total surface area $S$, express its volume $V$ in terms of $S$.

Accepted Answer

If the side is $a$, then $S = 6a^2$, so $a^2 = \dfrac{S}{6}$ and $a = \sqrt{\dfrac{S}{6}}$. $V = a^3 = \left(\sqrt{\dfrac{S}{6}}\right)^3$ $= \left(\dfrac{S}{6}\right)^{3/2} = \dfrac{S}{6}\sqrt{\dfrac{S}{6}}$ $V = \left(\dfrac{S}{6}\right)^{3/2}$. Check with $S = 96$: $\left(\dfrac{96}{6}\right)^{3/2}$ $= 16^{3/2}$ $= 16 \times 4$ $= 64 \ \checkmark$

Question 45

A 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 glas

Accepted Answer

Each marble sinks and pushes up water equal to its own volume. The water must rise $20 - 16 = 4 \text{ cm}$. The extra volume needed is a cylinder of radius 4 cm and height 4 cm:

$\pi \times 4^2 \times 4 = 64\pi \text{ cm}^3$

$\text{volume of one marble} = \dfrac{4}{3}\pi \times 1^3$

$= \dfrac{4}{3}\pi \text{ cm}^3$

$\text{number of marbles} = \dfrac{64\pi}{\frac{4}{3}\pi}$

$= 64 \times \dfrac{3}{4}$

$= 48$

The crow should drop 48 marbles. (The $\pi$ cancels, so no approximation is needed. The glass is 25 cm tall, so the water at 20 cm does not overflow.)

Question 46

Find the volume of ink in a new ball point pen. Take the necessary measurements and make approximations, as needed.

Accepted Answer

Model: the ink fills a thin cylinder inside the refill tube. Measure (with a ruler, on a new refill):length of the ink column $\approx 9 \text{ cm}$; inner diameter of the tube $\approx 2 \text{ mm}$, so $r \approx 0.1 \text{ cm}$ (the outside of a refill is about 3 mm across; its plastic wall takes up the rest). $V = \pi r^2 h \approx 3.14 \times 0.1^2 \times 9$ $= 3.14 \times 0.09 \approx 0.28 \text{ cm}^3$ A new pen holds roughly $0.3 \text{ cm}^3$ of ink — about a third of a millilitre.This is an estimate, not a fixed answer. Your own measurements will differ. Notice that the inner radius matters most: it is squared, so a tube 3 mm across inside instead of 2 mm would hold more than twice as much ink.

Question 47

Solve: (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.

Accepted Answer

(i) Volume of the dough: $\dfrac{4}{3}\pi \times 6^3$ $= 288\pi \text{ cm}^3 \approx 904 \text{ cm}^3$ Assumption: a rolled chapati is a thin cylinder about 18 cm across (radius 9 cm) and 2 mm (0.2 cm) thick. $\text{one chapati} = \pi \times 9^2 \times 0.2$ $= 16.2\pi \text{ cm}^3$ $\text{number} = \dfrac{288\pi}{16.2\pi}$ $= \dfrac{2880}{162} \approx 17.8$ The dough makes about 17 chapatis. (The $\pi$ cancels, so no approximation of $\pi$ is needed.)(ii) Model: the edible part is a thick spherical shell — a sphere of radius $R$ (the outside of the flesh) with a hollow sphere of radius $r$ removed from the middle. Cut the fruit in half and measure both radii across the cut face. For example, for a coconut with $R \approx 5.5 \text{ cm}$ and flesh about 1 cm thick ($r \approx 4.5 \text{ cm}$), one half holds $\dfrac{2}{3}\pi\left(R^3 - r^3\right)$ $= \dfrac{2}{3} \times 3.14 \times (166.375 - 91.125)$ $ = \dfrac{2}{3} \times 3.14 \times 75.25 \approx 157.5 \text{ cm}^3$ so the whole coconut has about $315 \text{ cm}^3$ of flesh. For a muskmelon with $R \approx 7 \text{ cm}$ (inside the rind) and a seed hollow of radius about 3 cm, the fruit is \(\df

Question 48

Looking 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 \text

Accepted Answer

Measurements: a page of the book is about $17 \text{ cm} \times 24 \text{ cm}$, and Part 2 has about 164 numbered pages (82 sheets). The closed book is about 1 cm thick. (Measure your own copy.)(i) Laying every page (each printed side) flat: $164 \times 17 \times 24$ $= 66912 \text{ cm}^2 \approx 6.7 \text{ m}^2$ A classroom floor of $8 \text{ m} \times 6 \text{ m} = 48 \text{ m}^2$ is about seven times larger. (If you lay out the 82 sheets instead, they cover only about $3.3 \text{ m}^2$.) No — the pages cover only about a seventh of the floor.(ii) Volume of one book: $17 \times 24 \times 1 = 408 \text{ cm}^3$. Using $1 \text{ ft}^3 \approx 28317 \text{ cm}^3$ (the conversion used in the chapter): $\text{storeroom} = 15 \times 20 \times 30$ $= 9000 \text{ ft}^3$ $\approx 9000 \times 28317$ $= 254853000 \text{ cm}^3$ $\dfrac{254853000}{408} \approx 624600$ Books are cuboids, so they stack almost without gaps, but whole books must fit along each wall. In cm the room is $457.2 \times 609.6 \times 914.4$. Stacking the books flat with the 24 cm side along the 457.2 cm wall: \(\left\lfloor \dfrac{457.2}{24} \right\rfloor \times \left\lfloor \dfrac{609.6}{17} \r

Question 49

* If the entire human population decided to climb into one giant cube, how long would the side have to be?

Accepted Answer

Assumptions: about $8.2 \times 10^9$ people. Arrange them standing, floor upon floor (as in the book’s picture): each person gets $0.5 \text{ m} \times 0.5 \text{ m}$ of floor and each floor is 2 m high, so each person uses $0.5 \times 0.5 \times 2 = 0.5 \text{ m}^3$. $\text{total volume} = 8.2 \times 10^9 \times 0.5$ $= 4.1 \times 10^9 \text{ m}^3$ $\text{side} = \sqrt[3]{4.1 \times 10^9} \approx 1600 \text{ m}$ (Check: $1600^3 = 4.096 \times 10^9$.) The cube would be only about 1.6 km on each side.This is an estimate, not a fixed answer. If people were packed with no space at all (a body is about $0.065 \text{ m}^3$), the side would shrink to about 0.8 km. Either way, all of humanity fits in a cube smaller than many cities.

Question 50

* The Earth’s surface has an estimated volume of 1.38 billion $\text{km}^3$ 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

Accepted Answer

(i) Let $R = 6371 \text{ km}$, $V = 1.38 \times 10^9 \text{ km}^3$ and the thickness be $t$. The water is the gap between a sphere of radius $R + t$ and the Earth: $\dfrac{4}{3}\pi (R + t)^3 - \dfrac{4}{3}\pi R^3 = V$ $(R + t)^3 = R^3 + \dfrac{3V}{4\pi}$ $t = \sqrt[3]{R^3 + \dfrac{3V}{4\pi}} - R$ Because the layer is very thin compared with $R$, it is almost a flat sheet spread over the surface, so a simpler expression is $t \approx \dfrac{V}{4\pi R^2}$ (ii) Surface of the Earth: $4\pi R^2$ $= 4 \times 3.1416 \times 6371^2 \approx 5.10 \times 10^8 \text{ km}^2$ $t \approx \dfrac{1.38 \times 10^9}{5.10 \times 10^8} \approx 2.7 \text{ km}$ The exact expression gives $t \approx 2.70 \text{ km}$ as well. The water layer would be about 2.7 km thick.

Question 51

* Give the dimension of a cuboid whose volume is halved when its surface area is doubled.

Accepted Answer

This asks for a cuboid that can be reshaped into another cuboid with half the volume but twice the surface area. Flattening is the idea: a thin, wide slab has a lot of surface for very little volume.

Take the cuboid $8 \text{ cm} \times 14 \text{ cm} \times 14 \text{ cm}$:

$V = 8 \times 14 \times 14 = 1568 \text{ cm}^3$

$S = 2(8 \times 14 + 14 \times 14 + 14 \times 8)$

$= 2(112 + 196 + 112)$

$= 840 \text{ cm}^2$

Change it to a flat slab $1 \text{ cm} \times 28 \text{ cm} \times 28 \text{ cm}$ (thickness divided by 8, the other two edges doubled):

$V = 1 \times 28 \times 28$

$= 784 \text{ cm}^3$

$= \dfrac{1568}{2}$

$S = 2(28 + 784 + 28)$

$= 1680 \text{ cm}^2$

$= 2 \times 840$

So the cuboid $8 \text{ cm} \times 14 \text{ cm} \times 14 \text{ cm}$ works: as the $1 \times 28 \times 28$ slab its volume is halved and its surface area is doubled.

The answer is not unique — this is the only such pair with whole-number edges up to 29 cm, found by checking every possibility, but many pairs with fractional edges exist. Any pair you find should be checked the same way: compute both volumes and both surface areas.

Question 52

Project: Find the volume of your house making necessary approximations. Present how you solved it.

Accepted Answer

Method. Break the house into simple solids and add up their volumes.1. Draw a rough plan of each floor. Treat every room, the kitchen, bathrooms and passages as cuboids. 2. Measure length and width of each (a measuring tape, or count floor tiles and multiply by the tile size) and the ceiling height (count the steps of a staircase and multiply by the height of one step if you cannot reach). 3. Find each volume with $l \times w \times h$ and add. 4. Add any other shapes: a sloping roof is a triangular prism, a water tank a cylinder or cuboid. 5. Say clearly whether you measured the inside (air space) or the outside (including walls).A worked example (one floor, ceiling 3 m): two bedrooms $4 \times 3.5$, a hall $5 \times 4$, a kitchen $3 \times 2.5$, a bathroom $2 \times 1.5$ (all in metres). $\text{floor area} = 2(14) + 20 + 7.5 + 3$ $= 58.5 \text{ m}^2$ $\text{volume} \approx 58.5 \times 3 = 175.5 \text{ m}^3$ This is an estimate, not a fixed answer. Every house is different, so there is no single answer; your presentation should show the plan, the measurements, the shapes you chose and the arithmetic, as above.

Math of Space: Surface Area and Volume

Exercise Set 14.1

Exercise Set 14.2

Exercise Set 14.3

Exercise Set 14.4

End-of-Chapter Exercises