Class 9 Maths Ganita Manjari Chapter 14 Solutions — Math of Space: Surface Area and Volume

Part II • NCERT first edition, August 2026

Class 9 Ganita Manjari Chapter 14 solutions: surface area, volume, cones, cylinders, spheres, all exercises and clearly labelled estimates.

Choose the correct surface and keep units consistent

Surface area is measured in square units; volume in cubic units. Use internal dimensions for capacity. Remove hidden joined faces, and include a lid or base only when the question requires it. Exact answers in π are shown first where useful; numerical answers use $\pi \approx \frac{22}{7}$ unless another approximation is stated.

Solid	Surface area	Volume
Cuboid	$2(lw+lh+wh)$	$lwh$
Cube	$6a^2$	$a^3$
Cylinder, closed	$2\pi rh+2\pi r^2$	$\pi r^2h$
Cone	$\pi rl+\pi r^2$	$\frac13\pi r^2h$
Sphere	$4\pi r^2$	$\frac43\pi r^3$
Hemisphere, including flat base	$3\pi r^2$	$\frac23\pi r^3$
Pyramid	Add the areas of all its faces	$\frac13Bh$

For a cone, $l=\sqrt{r^2+h^2}$ is the slant height, while h is perpendicular height. B denotes a pyramid’s base area. Cylinder curved area is 2πrh; cone curved area is πrl; hemisphere curved area is $2\pi r^{2}$ .

1\text{ m}^3=1000\text{ L},\qquad1\text{ L}=1000\text{ cm}^3

Page 124 reflection: stack layers of base area lw and thickness δh. Each contributes lwδh; adding the layer thicknesses to h gives lwh. For a cube set $l=w=h=a$ , giving $6a^{2}$ and $a^{3}$ . Example 1: the 6 cm cube and $9\times 6\times 4\,\mathrm{cm}$ cuboid both have volume $216\,\mathrm{cm}^{3}$ ; their surface areas are $216\,\mathrm{cm}^{2}$ and $228\,\mathrm{cm}^{2}$ respectively. The cuboid exposes $12\,\mathrm{cm}^{2}$ more area despite equal capacity.

Exercise Set 14.1 — Cubes and cuboids

14.1-1. Cube of volume $64\,\mathrm{cm}^{3}$

a=\sqrt[3]{64}=4\text{ cm},\qquad S=6(4)^2=96\text{ cm}^2

14.1-2. Small cubes in a 2 m box

Convert the box side to 200 cm. Exactly $\frac{200}{20}=10$ small cubes fit along each direction.

N=10^3=1000

This is an exact grid packing, assuming the stated box dimensions are internal.

14.1-3. Boxes in a godown

\begin{aligned}&N\\&=\frac{40}{2}\times\frac{25}{1.25}\times\frac{10}{1}\\&=20\times20\times10\\&=4000\end{aligned}

The dimensions divide exactly, so the volume ratio is achievable by aligned rows. The model ignores aisles, walls and handling space.

14.1-4. Two joined cubes

Each cube has side 5 cm because $5^{3}=125$ . Joining complete faces gives a $10\times 5\times 5\,\mathrm{cm}$ cuboid.

S=2(10\cdot5+10\cdot5+5\cdot5)=250\text{ cm}^2

Equivalently, subtract the two hidden $25\,\mathrm{cm}^{2}$ faces from $2\times 150\,\mathrm{cm}^{2}$ .

14.1-5. Cut a cube into unit cubes

S_{\rm original}=6(4)^2=96\text{ cm}^2

N=4^3=64,\qquad S_{\rm separate}=64\cdot6=384\text{ cm}^2

S_{\rm original}:S_{\rm separate}=1:4

The sum of all small-cube surface areas counts newly exposed cut faces. If they remain tightly assembled, the outer surface is still $96\,\mathrm{cm}^{2}$ .

14.1-6. Three face areas

(lw)(wh)(hl)=(lwh)^2=6\cdot15\cdot10=900

V=\sqrt{900}=30\text{ cm}^3

One consistent set of edge lengths is 2 cm, 3 cm, 5 cm.

14.1-7. Painted unit cubes

Painted faces	Location	Number
Exactly 3	Eight corners	8
Exactly 2	Interior positions on 12 edges	$12\left(5-2\right)=36$
Exactly 1	Interior positions on 6 faces	$6\left(5-2\right)^{2}=54$
None	Interior $3\times 3\times 3$ block	$\left(5-2\right)^{3}=27$

Check: $8+36+54+27=125=5^{3}$ .

14.1-8. All integer-sided cuboids with area $100\,\mathrm{cm}^{2}$

Assume $1\le a\le b\le c$ to avoid counting rotations repeatedly. Then $ab+bc+ca=50$ . Since $3a^{2}\le 50$ , a can only be 1,2,3 or 4. Rearrange:

(a+b)(a+c)=50+a^2

a	Required factor product	Valid ordered dimensions
1	$\left(b+1\right)\left(c+1\right)=51=3\times 17$	(1,2,16)
2	$\left(b+2\right)\left(c+2\right)=54=6\times 9$	(2,4,7)
3	$\left(b+3\right)\left(c+3\right)=59$ , a prime	None with $b\ge 3$
4	$\left(b+4\right)\left(c+4\right)=66$	None with both factors ≥8

Thus there are exactly two cuboids up to permutation: $1\times 2\times 16\,\mathrm{cm}$ and $2\times 4\times 7\,\mathrm{cm}$ . Their pairwise-product sums are $2+32+16=50$ and $8+28+14=50$ . The bound on a and factor checks prove completeness.

Swimming-pool reflection, sample comparison: A holds $\begin{aligned}8\times 50\times 20&=8000\,\mathrm{ft}^{3}\\&\approx 226{,}400\,\mathrm{L}\text{;}\end{aligned}$ B holds $\begin{aligned}6\times 100\times 15&=9000\,\mathrm{ft}^{3}\\&\approx 254{,}700\,\mathrm{L}\text{,}\end{aligned}$ using the supplied $\frac{28.3\,\mathrm{L}}{\mathrm{ft}^{3}}$ . B has 50% more water-surface area and is shallower, but needs 12.5% more water and a longer site. Neither 6 ft nor 8 ft depth should automatically be described as suitable for every school pupil. The choice depends on intended users, supervision, space, cost, accessibility and depth design; dimensions alone do not settle it.

Cylinder explanation, net and Example 2

Cutting the curved wall along a vertical line opens it into a rectangle of width 2πr and height h. Its area is 2πrh. Add two circles for a closed cylinder, or one for an open-topped vessel. A constant circular cross-section of area $\pi r^{2}$ stacked through height h gives volume $\pi r^{2}h$ .

Net of a closed cylinder. The rectangle’s width equals the circumference of each circular end. Fold the rectangle and attach both circles; omit one circle for an open top.

Example 2: the kaleidoscope uses only the curved wall:

A=2\pi rh=2\cdot\frac{22}{7}\cdot3.5\cdot25=550\text{ cm}^2

This excludes any extra overlap needed for joining.

Exercise Set 14.2 — Cylinders

14.2-1. Double the radius, halve the height

\frac{\mathrm{CSA}_B}{\mathrm{CSA}_A}=\frac{(2r)(h/2)}{rh}=1

\frac{V_B}{V_A}=\frac{(2r)^2(h/2)}{r^2h}=2

Therefore CSA $A:B=1:1$ , volume $A:B=1:2$ .

14.2-2. Radii 2:3 and heights 3:2

V_1:V_2=(2^2\cdot3):(3^2\cdot2)=2:3

\mathrm{CSA}_1:\mathrm{CSA}_2=(2\cdot3):(3\cdot2)=1:1

14.2-3. Cylinder cut from a cube

In the intended inscribed arrangement, the circular base is inscribed in a square face. The cube edge r is the cylinder’s diameter and also its height. Thus cylinder radius = $\frac{r}{2}$ .

V=\pi\left(\frac r2\right)^2r=\frac{\pi r^3}{4}\text{ cm}^3

14.2-4. Keep volume unchanged after increasing radius

\begin{aligned}(1.1r)^2h^\prime&=r^2h\\\frac{h^\prime}{h}&=\frac1{1.21}=\frac{100}{121}\\x&=100\left(1-\frac{100}{121}\right)\\&=\frac{2100}{121}\approx17.36\end{aligned}

The height must decrease by approximately 17.36%, not 20%.

14.2-5. Recasting a cube into rods

V_{\rm cube}=12^3=1728\text{ cm}^3

\begin{aligned}&V_{\rm rod}\\&=\pi(2)^2(12)\\&=48\pi\text{ cm}^3\\&\approx150.86\text{ cm}^3\end{aligned}

\frac{V_{\rm cube}}{V_{\rm rod}}=\frac{36}{\pi}\approx11.45

Only 11 complete rods can be formed, with some metal left over. Do not round up to 12. This assumes no material loss in melting.

Cones, pyramids and the volume activities

A right triangle rotated about one leg makes a right circular cone: the rotation axis becomes h, the other leg becomes r, and the hypotenuse becomes l. Unrolling the curved wall produces a circular sector with radius l and arc length 2πr. Therefore:

\begin{aligned}&\frac{\theta}{360^\circ}\\&=\frac rl,\qquad A_{\rm curved}\\&=\frac12(2\pi r)l\\&=\pi rl\end{aligned}

Cone height, radius and slant height. The right triangle has perpendicular legs h and r and hypotenuse l. The perspective cone is not to scale; use the labelled lengths.

Salt and clay activities: make a cone and cylinder with the same internal base radius and perpendicular height. Level each fill consistently. Three cone-fulls should approximately fill the cylinder; the same clay volume can form three such cones. Measurements and packing introduce experimental error, so this supports the formula rather than replacing a proof.

V_{\rm cone}=\frac13\pi r^2h,\qquad V_{\rm pyramid}=\frac13Bh

Example 3: rotate the 5–12–13 triangle about its 12 cm leg. Then $r=5$ , $h=12$ , $l=13$ , and $V=100\pi \,\mathrm{cm}^{3}$ . A pyramid uses its perpendicular height, even when the apex is not directly above the base centre.

Exercise Set 14.3 — Cones

14.3-1. Total area from diameter and slant height

r=\frac{24}{2}=12\text{ m}

\mathrm{TSA}=\pi r(l+r)=\pi(12)(21+12)=396\pi\text{ m}^2

Using $\pi \approx \frac{22}{7}$ gives $1244.57\,\mathrm{m}^{2}$ .

14.3-2. Curved area

\mathrm{CSA}=\pi rl=\frac{22}{7}(7)(10)=220\text{ cm}^2

14.3-3. Height 16 cm, radius 12 cm

l=\sqrt{16^2+12^2}=\sqrt{400}=20\text{ cm}

\mathrm{CSA}=240\pi\text{ cm}^2\approx754.29\text{ cm}^2

\begin{aligned}&\mathrm{TSA}\\&=240\pi+144\pi\\&=384\pi\text{ cm}^2\\&\approx1206.86\text{ cm}^2\end{aligned}

14.3-4. Find the radius from volume

Use $\pi \approx 3.14$ for the given volume $1570\,\mathrm{cm}^{3}$ .

\begin{aligned}1570&=\frac13(3.14)r^2(15)\\r^2&=\frac{1570}{15.7}=100\\r&=10\text{ cm}\end{aligned}

The positive root is used because a radius is positive.

14.3-5. Curved area $308\,\mathrm{cm}^{2}$

r=\frac{308}{\pi(14)}=\frac{308}{44}=7\text{ cm}

\mathrm{TSA}=308+\frac{22}{7}(7)^2=462\text{ cm}^2

14.3-6. Ten conical caps

l=\sqrt{7^2+24^2}=25\text{ cm}

\begin{aligned}&A_{\rm ten}\\&=10\pi rl\\&=10\cdot\frac{22}{7}\cdot7\cdot25\\&=5500\text{ cm}^2\end{aligned}

Caps have no circular base. This area excludes overlaps or additional cutting waste.

14.3-7. Tarpaulin for the tent

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

A=\pi rl=60\pi\text{ m}^2

L=\frac{60\pi}{3}+0.20=20\pi+0.20\text{ m}\approx63.06\text{ m}

The 20 cm allowance is 0.20 m and is added to the length, as the problem states. The model treats sheet area as fully usable apart from that allowance.

14.3-8. Rotate a 6–8–10 triangle

Rotation about the 8 cm leg gives $h=8\,\mathrm{cm}$ , $r=6\,\mathrm{cm}$ and $l=10\,\mathrm{cm}$ .

V=\frac13\pi(6)^2(8)=96\pi\text{ cm}^3\approx301.71\text{ cm}^3

\begin{aligned}&\mathrm{CSA}\\&=\pi(6)(10)\\&=60\pi\text{ cm}^2\\&\approx188.57\text{ cm}^2\end{aligned}

14.3-9. Fill a conical cup to half-depth

For a cup with the vertex at the bottom, the water forms a similar cone of height $\frac{h}{2}$ and radius $\frac{r}{2}$ . All linear dimensions are halved, so:

\frac{V_{\rm water}}{V_{\rm cup}}=\left(\frac12\right)^3=\frac18

Half the depth is only one-eighth of the full volume. The vertex-down orientation is essential.

Half-depth water in a vertex-down conical cup. Axial section. The water surface radius is half the full rim radius; volume scales with the cube of the length ratio.

Sphere and hemisphere activities

A sphere of radius r has surface area $4\pi r^{2}$ and volume $\frac{4\pi r^{3}}{3}$ . A surrounding cylinder of radius r and height 2r has curved area $4\pi r^{2}$ . For the tiny-region explanation, join small surface patches to the centre: the sum of the approximate pyramid volumes approaches one-third of r times the sphere’s surface area. This is a limiting argument, not an exact decomposition into a finite number of ordinary cones.

String activity: use closely spaced windings around a ball and then lay that same string in flat spirals filling circles of the ball’s radius. The ideal result is approximately four circles. Use teacher assistance for fixing the string and keep spacing consistent. Gaps, string thickness and stretching affect the estimate.

Sphere and hemisphere surfaces. Perspective sketches, not to scale. The hemisphere’s flat circular base contributes

\pi r^{2}

when total area is requested.

Example 4: a hemispherical bowl of internal radius 3.5 cm holds:

\begin{aligned}&V\\&=\frac23\cdot\frac{22}{7}\cdot(3.5)^3\\&=\frac{539}{6}\text{ cm}^3\\&\approx89.83\text{ mL}\end{aligned}

Exercise Set 14.4 — Spheres and hemispheres

14.4-1. Ball bearing

S=4\pi(0.7)^2=1.96\pi\text{ cm}^2\approx6.16\text{ cm}^2

14.4-2. Two metal spheres

Same metal means equal density, so mass ratio equals volume ratio.

\begin{aligned}&\frac{V_L}{V_S}\\&=\frac{5920}{740}\\&=8,\qquad\frac{r_L}{r_S}\\&=\sqrt[3]{8}\\&=2\end{aligned}

The smaller radius is $\frac{5}{2}=2.5\,\mathrm{cm}$ . Therefore the larger radius is 5 cm.

14.4-3. Moon-to-Earth surface-area ratio

S_{\rm Moon}:S_{\rm Earth}=\left(\frac14\right)^2:1=1:16

Use the approximate diameter ratio supplied in the question.

14.4-4. Hemisphere of radius 21 cm

\mathrm{CSA}=2\cdot\frac{22}{7}\cdot21^2=2772\text{ cm}^2

\mathrm{TSA}=3\cdot\frac{22}{7}\cdot21^2=4158\text{ cm}^2

14.4-5. Preservative liquid around 16 spheres

V_{\rm box}=16\cdot8\cdot8=1024\text{ cm}^3

\begin{aligned}&V_{\rm spheres}\\&=16\cdot\frac43\pi(2)^3\\&=\frac{512\pi}{3}\text{ cm}^3\end{aligned}

V_{\rm liquid}=1024-\frac{512\pi}{3}\approx488\text{ cm}^3

Rounded to the nearest integer: $488\,\mathrm{cm}^{3}$ . The 4 cm diameters fit as $4\times 2\times 2$ spheres. The liquid fills the gaps left between them.

14.4-6. Increase radius by 10%

\frac{V^\prime}{V}=(1.10)^3=1.331

\text{percentage increase}=(1.331-1)100=33.1\%

The result is exactly 33.1% in the ideal model with an exact 10% radius increase.

14.4-7. Volume increases by 72.8%

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

1+\frac{x}{100}=1.2\quad\Longrightarrow\quad x=20

14.4-8. Painting a hemispherical dome

r=\frac{35.2}{2\pi}=5.6\text{ m}

A=2\pi r^2=197.12\text{ m}^2

The stated price is ₹10 per $100\,\mathrm{cm}^{2}$ . Since $1\,\mathrm{m}^{2}=10{,}000\,\mathrm{cm}^{2}$ , the rate is ₹1000 per $m^{2}$ .

\text{cost}=197.12(1000)=\text{Rs }197120

Paint only the exposed curved dome, not an imaginary flat base. The quoted cost follows the question’s unusually high rate exactly.

Objects, measurements and worked estimates

Sample shape list: a dice is approximately cubical; a cupboard cuboidal; a straight tin cylindrical; a party cap conical; a ball spherical; a square pyramid model pyramidal. Real objects have thickness, rounded edges and fittings, so name the approximation used.

Sample object / assumed internal dimensions	Model	Approximate capacity
Cylindrical vessel, $r=10\,\mathrm{cm}$ , $h=30\,\mathrm{cm}$	$\pi(10)^2(30)$	9.42 L
Water tank, $100\times 80\times 60\,\mathrm{cm}$	$100\times 80\times 60$	480 L
Cupboard interior, $90\times 45\times 180\,\mathrm{cm}$	$90\times 45\times 180$	$0.729\,\mathrm{m}^{3}$ of enclosed space
Tapered bucket, top radius15 cm, bottom10 cm, height30 cm	$\frac{\pi h}{3}(R^2+Rr+r^2)$	14.92 L; frustum model, an extension

These are invented demonstration measurements, not measurements of your belongings. A gas cylinder’s external shape may be modelled as a cylinder plus rounded ends, but external dimensions do not establish its certified gas capacity. For a bucket, treating it as a perfect cylinder should be stated as an approximation.

Example 5, golgappas: the vessel contains $5000\pi \,\mathrm{cm}^{3}$ . With radius 1.5 cm and 60% filling, pani per piece is $0.6\times \left(\frac{4}{3}\right)\pi \left(1.5\right)^{3}=2.7\pi \,\mathrm{cm}^{3}\text{,}$ giving about 1852 pieces. With radius 1.75 cm and 80% filling, the corresponding estimate is about 875 pieces. The book’s approximately 1850 and 870 come from rounding intermediate volumes. Retaining extra digits avoids false precision; real serving sizes and losses vary.

Example 6, tennis-ball packing: use a consistent conversion $1\,\mathrm{ft}=30.48\,\mathrm{cm}$ . The classroom dimensions are $914.4\times 914.4\times 457.2\,\mathrm{cm}$ and its volume is 382,277,428. $992\,\mathrm{cm}^{3}$ . Each ball of radius 3 cm occupies $36\pi \,\mathrm{cm}^{3}$ . Dividing volumes gives an upper bound of about 3.38 million, not an attainable packing count.

\begin{aligned}&N_A\\&=\left\lfloor\frac{914.4}{6}\right\rfloor^2\left\lfloor\frac{457.2}{6}\right\rfloor\\&=152^2\cdot76\\&=1755904\end{aligned}

The book’s 1,687,500 estimate uses the rougher 30 cm per foot, giving $150\times 150\times 75$ . Both are estimates under stated conversion choices; do not mix the 30 cm length approximation with its inconsistent cube $28{,}000\,\mathrm{cm}^{3}$ .

Sample extensions for the staggered arrangements: the small pictures do not fully specify layer coordinates, so state a model. In a triangular layer, each sphere has horizontal cell area $2\sqrt{3}r^{2}$ . Stacking such layers directly above one another gives cell volume $4\sqrt{3}r^{3}$ and roughly 2.04 million balls. Nesting successive layers in triangular gaps makes the vertical separation $2r\sqrt{\frac{2}{3}}$ , giving cell volume $4\sqrt{2}r^{3}$ and roughly 2.50 million balls. These bulk estimates omit boundary losses, furniture and deformation; they are not exact maxima for the finite room.

End-of-Chapter Exercises

E1. Ice-cream scoops

Sample guess: around 50 scoops. Assume each scoop is a full sphere of radius 2.5 cm and ignore packing/serving losses.

V_{\rm tub}=10\cdot15\cdot20=3000\text{ cm}^3

V_{\rm scoop}=\frac43\pi(2.5)^3\approx65.45\text{ cm}^3

N\approx\frac{3000}{65.45}\approx45.8

About 46 scoop-equivalents, or 45 complete scoops plus a partial one. A different assumed scoop size changes the estimate.

E2. Unit cubes needed for the next cube

(a+1)^3-a^3=3a^2+3a+1

This counts the additional unit cubes, not the total in the enlarged cube.

E3. Water in a lifetime and classroom bricks

(i) Sample assumptions: 2 litres per day for 80 years, purely as a mathematical estimate. Total = $\begin{aligned}2\times 365.25\times 80&=58{,}440\,\mathrm{L}\\&=58.44\,\mathrm{m}^{3}\text{.}\end{aligned}$ A $9\times 6\times 3\,\mathrm{m}$ classroom has volume $162\,\mathrm{m}^{3}$ , so this would fill about 36% of it, not the whole room. Different lifetime and room assumptions give different conclusions.

(ii) Sample: an $8\times 6\,\mathrm{m}$ internal classroom, 3 m high, with walls 0.20 m thick, has wall-envelope volume $[\left(8.4\times 6.4\right)-\left(8\times 6\right)]\times 3=17.28\,\mathrm{m}^{3}\text{.}$ Subtract one $1\times 2\,\mathrm{m}$ door and four $1.5\times 1.2\,\mathrm{m}$ windows: opening volume = $9.2\times 0.20=1.84\,\mathrm{m}^{3}$ . Net wall volume = $15.44\,\mathrm{m}^{3}$ . If mortar occupies 10%, brick volume is $13.896\,\mathrm{m}^{3}$ . At $0.20\times 0.10\times 0.06=0.0012\,\mathrm{m}^{3}$ per brick, the estimate is 11,580 bricks, about 11,600 before wastage. These dimensions are examples, not a measured classroom.

E4. Which chocolate option gives more?

V_A=50^3=125000\text{ mm}^3

V_B=100(10^3)=100000\text{ mm}^3

Option A gives $25{,}000\,\mathrm{mm}^{3}$ more chocolate, or 25% more than B, assuming the same material and no hollow spaces.

E5. Area for the world’s people

Sample estimate: take a round hypothetical population of 8 billion and allocate $0.5\,\mathrm{m}^{2}$ per person. This is an assumed planning density, not a current census claim.

A=8\times10^9(0.5)=4\times10^9\text{ m}^2=4000\text{ km}^2

That is a square roughly 63.2 km on a side. Passageways and practical facilities would increase it.

E6. Milk for 1600 students

r=\frac72=3.5\text{ cm}

V_{\rm glass}=\frac{22}{7}(3.5)^2(12)=462\text{ cm}^3

V_{\rm total}=\frac{1600\cdot462}{1000}=739.2\text{ L}

E7. Cone from the sphere-area relation

4\pi(5)^2=5(\pi\cdot4\cdot l)

100\pi=20\pi l\quad\Longrightarrow\quad l=5\text{ cm}

h=\sqrt{5^2-4^2}=3\text{ cm}

V=\frac13\pi(4)^2(3)=16\pi\text{ cm}^3\approx50.29\text{ cm}^3

E8. Planet and Sun volume ratios

The factor $\frac{4\pi }{3}$ cancels when two spherical volumes are compared. Use the radii supplied, in the same units.

\frac{V_J}{V_E}=\left(\frac{69900}{6370}\right)^3\approx1321.34

\begin{aligned}&\frac{V_{\rm Sun}}{V_E}\\&=\left(\frac{695700}{6370}\right)^3\\&\approx1302709.90\end{aligned}

Approximately 1321:1 and 1.303 million:1, respectively. These are volume ratios under the stated spherical model.

E9. Sphere inside its smallest enclosing cylinder

The enclosing cylinder has radius r and height 2r.

\begin{aligned}&\frac{V_{\rm sphere}}{V_{\rm cylinder}}\\&=\frac{\frac43\pi r^3}{\pi r^2(2r)}\\&=\frac23\end{aligned}

E10. A string one metre longer

2\pi(R+\delta)-2\pi R=1\text{ m}

\begin{aligned}&\delta\\&=\frac1{2\pi}\text{ m}\\&\approx0.15915\text{ m}\\&=15.915\text{ cm}\end{aligned}

The answer is the same for Earth, Moon, Jupiter and a volleyball. The original radius cancels; only the added circumference matters. The second string must be held at uniform height all around.

E11. Square pyramid inside a cylinder

The square base has diagonal 2r, so if its side is s, $2s^{2}=\left(2r\right)^{2}$ and its area is $s^{2}=2r^{2}$ . Its perpendicular height is h.

V_P=\frac13(2r^2)h

V_P:V_C=\frac23r^2h:\pi r^2h=2:3\pi

Square base inscribed in the circular cylinder base. Top view. All four square vertices lie on the circle; the pyramid volume uses perpendicular height h.

E12. Changes in volume

(i) Increase length alone:

(l+1)wh-lwh=wh

Correct option: (c), wh cubic units.

(ii) Increase cylinder radius alone:

\pi(r+1)^2h-\pi r^2h=2\pi rh+\pi h

Correct option: (e).

(iii) For $r\ge 1$ , reducing a sphere’s radius by 1 gives new-minus-old change:

\begin{aligned}&\Delta V\\&=\frac43\pi\big((r-1)^3-r^3\big)\\&=-4\pi r^2+4\pi r-\frac43\pi\end{aligned}

The positive amount of volume lost is $4\pi r^2-4\pi r+\frac43\pi$ cubic units.

E13. Surface area as a function of cube volume

a=\sqrt[3]V,\qquad S=6a^2=6V^{2/3}

Use consistent units; for example V in $\mathrm{cm}^{3}$ gives S in $\mathrm{cm}^{2}$ .

E14. Volume as a function of cube surface area

a=\sqrt{\frac S6},\qquad V=a^3=\left(\frac S6\right)^{3/2}

E15. Marbles raise the water level

V_{\rm displaced}=\pi(4)^2(20-16)=64\pi\text{ cm}^3

V_{\rm marble}=\frac43\pi(1)^3

N=\frac{64\pi}{4\pi/3}=48

48 marbles are needed if they are fully submerged, do not absorb water and no water spills. The water remains below the 25 cm rim.

E16. Ink in a pen

Sample measurement model: treat the ink column as a cylinder of internal radius 0.05 cm and length 8 cm. Use the inner bore, not the outside plastic tube.

V=\pi(0.05)^2(8)=0.02\pi\text{ cm}^3\approx0.063\text{ mL}

This is an illustrative estimate. Measure the actual ink-filled length and internal diameter for your pen; a partially filled or tapered refill gives a different value.

E17. Chapatis and edible fruit

(i) Sample guess: around 30 chapatis. The dough sphere has volume $\left(\frac{4}{3}\right)\pi 6^{3}=288\pi \,\mathrm{cm}^{3}$ . Model a chapati as a cylinder of radius 7 cm and thickness 0.2 cm, volume $9.8\pi \,\mathrm{cm}^{3}$ .

N=\frac{288\pi}{9.8\pi}\approx29.39

About 29 complete chapatis under this ideal model; thickness, size and handling loss change the result.

(ii) Sample coconut half: measure outer edible-flesh radius $R=6\,\mathrm{cm}$ and inner cavity radius $r=5\,\mathrm{cm}$ , excluding the hard shell. Model the edible part as a hemispherical shell.

\begin{aligned}&V_{\rm flesh}\\&=\frac23\pi(R^3-r^3)\\&=\frac{182\pi}{3}\text{ cm}^3\\&\approx190.6\text{ cm}^3\end{aligned}

For a muskmelon, similarly subtract the seed cavity and excluded rind from the appropriate half-ellipsoid or hemisphere model. Use your own measurements; the fruit need not be perfectly spherical.

E18. Textbook pages and storeroom packing

(i) Sample dimensions: assume a page is $21\times 28\,\mathrm{cm}$ . The supplied PDFs contain 184 page images including preliminary and blank pages. If printed on both sides, that is 92 physical leaves, excluding the cover. Laying those leaves out once covers $92\times 0.21\times 0.28=5.4096\,\mathrm{m}^{2}$ . Printing each page image on a separate sheet would cover $10.8192\,\mathrm{m}^{2}$ . Either is far below an assumed $9\times 9\,\mathrm{m}$ classroom floor $\left(81\,\mathrm{m}^{2}\right)$ . State whether “pages” means page faces or physical sheets.

(ii) Assume a complete book is $21\times 28\times 1\,\mathrm{cm}$ and use $1\,\mathrm{ft}=30.48\,\mathrm{cm}$ . The storeroom is $457.2\times 609.6\times 914.4\,\mathrm{cm}$ . One achievable uniform orientation places the 28 cm dimension along 457.2 cm, the 21 cm dimension along 609.6 cm and the 1 cm thickness along 914.4 cm:

N=16\cdot29\cdot914=424096

The volume upper bound is:

\begin{aligned}&N\le\left\lfloor\frac{457.2\cdot609.6\cdot914.4}{21\cdot28\cdot1}\right\rfloor\\&=433421\end{aligned}

The achievable count is the largest of the six uniform grid orientations for these assumed dimensions. It is not a proved global packing maximum: mixed orientations, actual book thickness, usable room space and access requirements are unspecified.

E19. One giant cube for people

Sample model: assume 8 billion people and an effective volume of $0.1\,\mathrm{m}^{3}$ per person including an allowance for gaps. The total modelled volume is 800 million $m^{3}$ .

a=\sqrt[3]{8\times10^8}\text{ m}\approx928.3\text{ m}

So the side is roughly 0.93 km. This is a geometric thought experiment, not a practical occupancy design or a current population count. Different space allowances materially change the answer.

E20. A uniform water layer around Earth

Take the supplied Earth radius $R=6371\,\mathrm{km}$ and water volume $W=1.38\times 10^{9}\,\mathrm{km}^{3}$ . Let t be the added outer-layer thickness.

\frac43\pi\big((R+t)^3-R^3\big)=W

t=\sqrt[3]{R^3+\frac{3W}{4\pi}}-R

t\approx2.70439\text{ km}\approx2704\text{ m}

The thin-layer approximation $\frac{W}{4\pi R^{2}}$ gives about 2.70554 km and is close because the layer is thin relative to Earth’s radius. This assumes the water is added outside the reference spherical surface.

E21. Half the volume with twice the surface area

The question does not specify how the dimensions change, so there is no unique answer. A valid pair of cuboids is:

Cuboid	Dimensions	Volume	Total area
Original	$8\times 12\times 15$	1440	$2\left(96+180+120\right)=792$
Changed	$1\times 12\times 60$	720	$2\left(12+720+60\right)=1584$

With all dimensions in cm, volumes are in $\mathrm{cm}^{3}$ and areas in $\mathrm{cm}^{2}$ . The changed volume is half and its surface area is double. If uniform enlargement or reduction had been intended, it would be impossible: a scale factor k gives area factor $k^{2}$ and volume factor $k^{3}$ , which cannot simultaneously be 2 and $\frac{1}{2}$ .

E22. House-volume project: sample report

Question: estimate the enclosed interior volume of a house. Method: split it into non-overlapping rectangular regions, measure internal dimensions in metres, calculate each lwh, and add them. Avoid counting the same room or stairwell twice.

Sample region	Internal dimensions	Volume
Ground-floor enclosed region	$10\times 8\times 3\,\mathrm{m}$	$240\,\mathrm{m}^{3}$
Upper-floor enclosed region	$8\times 6\times 3\,\mathrm{m}$	$144\,\mathrm{m}^{3}$
Total	Two separate levels	$384\,\mathrm{m}^{3}$

Assumptions: flat ceilings, no double-height overlap, dimensions rounded to the nearest convenient value. Exclude open terraces; model a sloping roof separately if included. Conclusion: this example house has approximately $384\,\mathrm{m}^{3}$ of enclosed internal space. Replace the invented measurements with those of your own house and report measurement uncertainty.

Learning-material activities and final checks

The supplied end pages provide graph paper and nets for a cuboid, cube, cylinder, cone, triangular pyramid and square pyramid. Fold a copy along the indicated edges, keeping joining tabs separate from mathematical face areas. For each model, label perpendicular height, base measurements and any slant height, then verify that adjoining edges match. A cone’s sector arc must equal its base circumference. These are construction activities, not additional numbered exercise sets.

All guesstimates above are explicitly labelled samples. Packing counts distinguish a volume upper bound from a feasible arrangement, and from a proved maximum. The textbook’s ball-packing conversion uses different rough approximations; consistent units are used here. Question 21 requires a change of shape, and its missing transformation rule is acknowledged.

Textbook and answer notes

Based on NCERT Ganita Manjari, Grade 9, Part II, first edition August 2026 (ISBN 978-93-5729-236-8). Question numbers follow the supplied textbook. Question cues are shortened; consult the book for the complete wording and original figures. Solutions and diagrams on this page are ClearStudy explanations. Open-ended responses are examples, not the only acceptable answers.