Surface Areas and Volumes

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CLASS IX Mathematics ~8–10 marks/year Ch 11 of 12
Surface Areas and Volumes

Class 9 · Mathematics · NCERT chapter notes · Akanksha Classes

Snapshot
  • This chapter covers six solid shapes: cuboid, cube, right circular cylinder, right circular cone, sphere, and hemisphere.
  • For each solid you need three distinct measures: Lateral (Curved) Surface Area (LSA/CSA), Total Surface Area (TSA), and Volume.
  • LSA = only the side walls (no top/bottom bases). TSA = every exposed face. Volume = space inside the solid.
  • Real-world application questions (painting a tank, canvas for a tent, water in a pipe) dominate the board exam — learn to identify which formula fits.
  • Board weightage: approx. 8-10 marks/year, spread across 1-2 short-answer (3 marks each) and 1 long-answer (4 marks). Composite-solid questions appear frequently.
  • Key constant: use $\pi = \dfrac{22}{7}$ unless stated otherwise. Use $\pi = 3.14$ when the radius has decimal digits.
Detailed Notes

S1. Cuboid

A cuboid (rectangular box) has length $l$, breadth $b$, and height $h$. It has 6 rectangular faces, 12 edges, and 8 vertices. Every face is a rectangle.

$$\text{LSA (four walls)} = 2h(l + b)$$ $$\text{TSA (all six faces)} = 2(lb + bh + hl)$$ $$\text{Volume} = l \times b \times h$$ $$\text{Body diagonal} = \sqrt{l^2 + b^2 + h^2}$$

Each variable explained:

  • $l$ = length of the base; $b$ = breadth of the base; $h$ = height of the cuboid.
  • LSA counts only the four vertical walls. If you are painting the four walls of a room (floor and ceiling excluded), you use LSA $= 2h(l+b)$. Perimeter of base $\times$ height is an easy way to remember this.
  • TSA adds the two horizontal faces (top and bottom), each of area $lb$. TSA $=$ LSA $+ 2lb = 2h(l+b) + 2lb = 2(lb + bh + hl)$.
  • Volume counts every cubic unit inside. Important conversion: $1\ \text{m}^3 = 1000\ \text{litres}$.
NCERT Example — Cuboid: whitewashing a room

Problem: A room is 5 m long, 4 m wide, and 3 m high. Find the cost of whitewashing the four walls and the ceiling at Rs 7.50 per m2.

Step 1 — Area of four walls (LSA): $2h(l+b) = 2 \times 3 \times (5+4) = 6 \times 9 = 54\ \text{m}^2$.

Step 2 — Area of ceiling: $l \times b = 5 \times 4 = 20\ \text{m}^2$.

Step 3 — Total area to whitewash: $54 + 20 = 74\ \text{m}^2$.

Step 4 — Cost: $74 \times 7.50 = \textbf{Rs 555}.$

NCERT Example — Cuboid: decorating a Christmas box

Problem: Mary wants to cover a wooden box (dimensions 80 cm by 40 cm by 20 cm) with square sheets of paper of size 40 cm by 40 cm. How many sheets are required?

Solution: TSA of box $= 2(lb + bh + hl) = 2(80 \times 40 + 40 \times 20 + 20 \times 80) = 2(3200 + 800 + 1600) = 2 \times 5600 = 11200\ \text{cm}^2$.

Area of one sheet $= 40 \times 40 = 1600\ \text{cm}^2$.

Sheets required $= \dfrac{11200}{1600} = \mathbf{7\ \text{sheets}}.$

NCERT Example — Cuboid: painting bricks

Problem: Paint sufficient to cover $9.375\ \text{m}^2$ is available. How many bricks of dimensions $22.5\ \text{cm} \times 10\ \text{cm} \times 7.5\ \text{cm}$ can be painted?

Solution: TSA of one brick $= 2(22.5 \times 10 + 10 \times 7.5 + 7.5 \times 22.5) = 2(225 + 75 + 168.75) = 937.5\ \text{cm}^2$.

Total paint area $= 9.375\ \text{m}^2 = 93750\ \text{cm}^2$.

Number of bricks $= \dfrac{93750}{937.5} = \mathbf{100\ \text{bricks}}.$

S2. Cube

A cube is a special cuboid where all three edges are equal: $l = b = h = a$. All 6 faces are identical squares.

$$\text{LSA} = 4a^2$$ $$\text{TSA} = 6a^2$$ $$\text{Volume} = a^3$$ $$\text{Body diagonal} = a\sqrt{3}$$

Each variable explained:

  • $a$ = edge (side) length of the cube.
  • LSA = 4 faces (the four side faces, excluding top and bottom) $\times\ a^2 = 4a^2$.
  • TSA = 6 equal square faces $\times\ a^2 = 6a^2$. Verify: substitute $l = b = h = a$ into cuboid TSA formula $2(lb+bh+hl) = 2 \times 3a^2 = 6a^2$. Consistent.
  • A cube of side 10 cm has volume $10^3 = 1000\ \text{cm}^3 = 1\ \text{litre}$. Very handy fact.
NCERT Example — Cube: edge from volume

Problem: A cubical box has a volume of $\dfrac{125}{8}\ \text{m}^3$. Find its TSA.

Solution: $a^3 = \dfrac{125}{8} \Rightarrow a = \dfrac{5}{2}\ \text{m} = 2.5\ \text{m}$.

TSA $= 6a^2 = 6 \times (2.5)^2 = 6 \times 6.25 = \mathbf{37.5\ \text{m}^2}.$

S3. Right Circular Cylinder

A right circular cylinder has a circular base of radius $r$ and a height $h$, with the axis perpendicular to the base. Examples: a water pipe, a tin can, a roller, a well.

$$\text{LSA (Curved Surface Area, CSA)} = 2\pi r h$$ $$\text{TSA} = 2\pi r(r + h)$$ $$\text{Volume} = \pi r^2 h$$ $$\text{Hollow cylinder: Volume of material} = \pi(R^2 - r^2)h$$

Each variable explained:

  • $r$ = radius of the circular base; $h$ = height (length) of the cylinder.
  • CSA = circumference of base $\times$ height $= 2\pi r \times h = 2\pi rh$. This is the area of the curved side wall (like the label on a tin can — if you cut and unroll it, you get a rectangle of width $2\pi r$ and height $h$).
  • TSA = CSA + 2 circular bases $= 2\pi rh + 2\pi r^2 = 2\pi r(r+h)$.
  • Volume = area of base $\times$ height $= \pi r^2 h$.
  • For hollow cylinders (pipes), only the material (between outer radius $R$ and inner radius $r$) is solid. Volume of material $= \pi(R^2 - r^2)h$.
NCERT Example — Cylinder: cost of painting a pillar

Problem: A cylindrical pillar is 50 cm in diameter and 3.5 m in height. Find the cost of painting its curved surface at Rs 12.50 per m2.

Solution: $r = 25\ \text{cm} = 0.25\ \text{m}$; $h = 3.5\ \text{m}$.

CSA $= 2\pi rh = 2 \times \dfrac{22}{7} \times 0.25 \times 3.5 = 2 \times \dfrac{22}{7} \times \dfrac{7}{8} = \dfrac{11}{2} = 5.5\ \text{m}^2$.

Cost $= 5.5 \times 12.50 = \textbf{Rs 68.75}.$

NCERT Example — Cylinder: plastering a well

Problem: The inner diameter of a circular well is 3.5 m and it is 10 m deep. Find (i) its inner curved surface area, (ii) the cost of plastering at Rs 40 per m2.

Solution: $r = 1.75\ \text{m}$, $h = 10\ \text{m}$.

(i) CSA $= 2\pi rh = 2 \times \dfrac{22}{7} \times 1.75 \times 10 = 110\ \text{m}^2$.

(ii) Cost $= 110 \times 40 = \textbf{Rs 4400}.$

NCERT Example — Cylinder: capacity of a vessel in litres

Problem: A cylindrical vessel is 14 cm in diameter and 24 cm tall. How many litres of oil can it hold?

Solution: $r = 7\ \text{cm}$, $h = 24\ \text{cm}$.

Volume $= \pi r^2 h = \dfrac{22}{7} \times 49 \times 24 = 22 \times 7 \times 24 = 3696\ \text{cm}^3$.

Capacity $= \dfrac{3696}{1000} = \mathbf{3.696\ \text{litres}}.$

NCERT Example — Cylinder: mass of a hollow metallic pipe

Problem: A metallic pipe has inner diameter 4 cm, outer diameter 4.4 cm, and length 72 cm. If density of metal is $8\ \text{g/cm}^3$, find its mass.

Solution: $R = 2.2\ \text{cm}$, $r = 2\ \text{cm}$, $h = 72\ \text{cm}$.

Volume of metal $= \pi(R^2 - r^2)h = \dfrac{22}{7} \times (4.84 - 4) \times 72 = \dfrac{22}{7} \times 0.84 \times 72 = 190.08\ \text{cm}^3$.

Mass $= 190.08 \times 8 = \mathbf{1520.64\ \text{g}} \approx 1.52\ \text{kg}.$

S4. Right Circular Cone

A right circular cone has a circular base of radius $r$, a perpendicular height $h$, and a slant height $l$ connecting the apex to the rim of the base. Examples: an ice-cream cone, a funnel, a conical tent, a heap of sand.

$$\text{Slant height:}\quad l = \sqrt{r^2 + h^2}$$ $$\text{LSA (Curved Surface Area)} = \pi r l$$ $$\text{TSA} = \pi r(r + l)$$ $$\text{Volume} = \dfrac{1}{3}\pi r^2 h$$

Each variable explained:

  • $r$ = radius of base circle; $h$ = perpendicular (vertical) height from centre of base to apex; $l$ = slant height (the sloping edge from apex to base rim).
  • The relation $l = \sqrt{r^2 + h^2}$ comes from the right triangle in the cross-section: hypotenuse $l$, legs $r$ and $h$. Always compute $l$ first when given $r$ and $h$ and asked for CSA or TSA.
  • CSA $= \pi r l$ — the area of the curved lateral surface, like the fabric of a conical tent (base circle excluded).
  • TSA $= \pi r l + \pi r^2 = \pi r(r+l)$ — adds the base circle.
  • Volume of cone $= \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \text{Volume of enclosing cylinder}$. A cone holds exactly one-third as much as the cylinder with the same base and height. This is a standard experimental result shown in NCERT.
NCERT Example — Cone: CSA when slant height given

Problem: Find the curved surface area of a cone with slant height 60 cm and base radius 21 cm.

Solution: CSA $= \pi rl = \dfrac{22}{7} \times 21 \times 60 = 22 \times 3 \times 60 = \mathbf{3960\ \text{cm}^2}.$

NCERT Example — Cone: cost of whitewashing a conical tomb

Problem: The slant height and base diameter of a conical tomb are 25 m and 14 m. Find the cost of white-washing its curved surface at Rs 210 per 100 m2.

Solution: $r = 7\ \text{m}$, $l = 25\ \text{m}$.

CSA $= \pi rl = \dfrac{22}{7} \times 7 \times 25 = 22 \times 25 = 550\ \text{m}^2$.

Cost $= \dfrac{550 \times 210}{100} = \textbf{Rs 1155}.$

NCERT Example — Cone: slant height then canvas area

Problem: A conical tent is 10 m high and has base radius 24 m. Find (i) slant height, (ii) cost of canvas at Rs 70 per m2.

Solution: $h = 10\ \text{m}$, $r = 24\ \text{m}$.

(i) $l = \sqrt{r^2 + h^2} = \sqrt{576 + 100} = \sqrt{676} = \mathbf{26\ \text{m}}.$

(ii) Canvas area = CSA $= \dfrac{22}{7} \times 24 \times 26 = \dfrac{13728}{7} \approx 1961.14\ \text{m}^2$.

Cost $\approx 1961.14 \times 70 \approx \textbf{Rs 1,37,280}.$

NCERT Example — Cone: CSA and TSA from radius and height

Problem: Height of a cone = 16 cm, base radius = 12 cm. Find CSA and TSA. (Use $\pi = 3.14$)

Solution: $l = \sqrt{12^2 + 16^2} = \sqrt{144 + 256} = \sqrt{400} = 20\ \text{cm}$.

CSA $= 3.14 \times 12 \times 20 = \mathbf{753.6\ \text{cm}^2}.$

TSA $= \pi r(r+l) = 3.14 \times 12 \times (12+20) = 3.14 \times 12 \times 32 = \mathbf{1205.76\ \text{cm}^2}.$

NCERT Example — Cone: volume of a heap of wheat and canvas to cover it

Problem: A heap of wheat is conical with diameter 10.5 m and height 3 m. Find (i) its volume, (ii) canvas area to cover it from rain.

Solution: $r = 5.25\ \text{m}$, $h = 3\ \text{m}$.

$l = \sqrt{5.25^2 + 3^2} = \sqrt{27.5625 + 9} = \sqrt{36.5625} \approx 6.05\ \text{m}$.

(i) Volume $= \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 5.25^2 \times 3 = \dfrac{22}{7} \times 27.5625 = \mathbf{86.625\ \text{m}^3}.$

(ii) Canvas area = CSA $= \pi rl = \dfrac{22}{7} \times 5.25 \times 6.05 \approx \mathbf{99.55\ \text{m}^2}.$

S5. Sphere

A sphere is a perfectly round solid. Every point on its surface is at the same distance (radius $r$) from the centre. Examples: a ball, a marble, a shot-put, a globe. For a sphere there is no distinction between LSA and TSA — the entire surface is one continuous curved face.

$$\text{Surface Area} = 4\pi r^2$$ $$\text{Volume} = \dfrac{4}{3}\pi r^3$$

Each variable explained:

  • $r$ = radius. If diameter $d$ is given, use $r = d/2$ first.
  • Surface area $= 4\pi r^2$: equals exactly four times the area of the sphere's great circle ($\pi r^2$). A geometric fact worth remembering: if you wrap a sphere perfectly, the paper has area equal to four of the circular cross-sections.
  • Volume $= \dfrac{4}{3}\pi r^3$: a sphere's volume is $\dfrac{2}{3}$ of the volume of its circumscribing cylinder (Archimedes).
NCERT Example — Sphere: surface area

Problem: Find the surface area of a sphere of radius 14 cm.

Solution: SA $= 4\pi r^2 = 4 \times \dfrac{22}{7} \times 196 = 4 \times 22 \times 28 = \mathbf{2464\ \text{cm}^2}.$

NCERT Example — Sphere: surface area from diameter

Problem: Find the surface area of a sphere of diameter 14 cm.

Solution: $r = 7\ \text{cm}$. SA $= 4 \times \dfrac{22}{7} \times 49 = 4 \times 22 \times 7 = \mathbf{616\ \text{cm}^2}.$

NCERT Example — Sphere: mass of a shot-put

Problem: A shot-put is a metallic sphere of radius 4.9 cm. Density of metal $= 7.8\ \text{g/cm}^3$. Find its mass.

Solution: Volume $= \dfrac{4}{3} \times \dfrac{22}{7} \times (4.9)^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times 117.649 \approx 493.1\ \text{cm}^3$.

Mass $= 493.1 \times 7.8 \approx \mathbf{3846.32\ \text{g}} \approx 3.85\ \text{kg}.$

NCERT Example — Sphere: volume from surface area

Problem: Find the volume of a sphere whose surface area is $154\ \text{cm}^2$.

Solution: $4\pi r^2 = 154 \Rightarrow r^2 = \dfrac{154 \times 7}{4 \times 22} = \dfrac{1078}{88} = \dfrac{49}{4} \Rightarrow r = 3.5\ \text{cm}$.

Volume $= \dfrac{4}{3} \times \dfrac{22}{7} \times (3.5)^3 = \dfrac{4}{3} \times \dfrac{22}{7} \times 42.875 = \dfrac{4 \times 22 \times 42.875}{21} \approx \mathbf{179.67\ \text{cm}^3}.$

S6. Hemisphere

A hemisphere is half a sphere, cut along a great circle. It has one curved surface and one flat circular base. Examples: a bowl, a dome, a half-melon.

$$\text{LSA (Curved Surface Area)} = 2\pi r^2$$ $$\text{TSA (solid hemisphere)} = 3\pi r^2$$ $$\text{Volume} = \dfrac{2}{3}\pi r^3$$

Each variable explained:

  • $r$ = radius of the hemisphere.
  • CSA = half the sphere's surface $= \dfrac{1}{2} \times 4\pi r^2 = 2\pi r^2$ (the dome part only).
  • TSA = curved part + flat circular base $= 2\pi r^2 + \pi r^2 = 3\pi r^2$.
  • Volume = half the sphere's volume $= \dfrac{1}{2} \times \dfrac{4}{3}\pi r^3 = \dfrac{2}{3}\pi r^3$.
  • Common exam trap: Use TSA $= 3\pi r^2$ for a solid hemisphere (like a marble cut in half). Use CSA $= 2\pi r^2$ for an open bowl (where only the inner curved surface is painted).
NCERT Example — Hemisphere: TSA of a bowl

Problem: A hemispherical bowl has diameter 14 cm. Find its total surface area.

Solution: $r = 7\ \text{cm}$. TSA $= 3\pi r^2 = 3 \times \dfrac{22}{7} \times 49 = 3 \times 22 \times 7 = \mathbf{462\ \text{cm}^2}.$

NCERT Example — Hemisphere: capacity in litres

Problem: How many litres of milk can a hemispherical bowl of diameter 10.5 cm hold?

Solution: $r = 5.25\ \text{cm}$.

Volume $= \dfrac{2}{3}\pi r^3 = \dfrac{2}{3} \times \dfrac{22}{7} \times (5.25)^3 = \dfrac{2}{3} \times \dfrac{22}{7} \times 144.703 \approx 303.19\ \text{cm}^3$.

Capacity $= \dfrac{303.19}{1000} \approx \mathbf{0.303\ \text{litres}}.$

S7. Composite (Combined) Solids

Many real-world objects are combinations of two or more basic solids. To find their surface area or volume, break them into known parts, compute each separately, then combine carefully.

$$\text{Total Volume} = V_1 + V_2 + \cdots$$ $$\text{Total SA} = \text{sum of all exposed (visible) surfaces only}$$

Key principle: When two solids are joined, the common face is interior — do NOT count it in the total surface area. Subtract any face that disappears into the interior of the composite shape.

Common composite types in NCERT/board exams:

  • Cylinder topped with a cone (rocket shape): TSA = CSA of cylinder + base of cylinder + CSA of cone (no circular face between them counted).
  • Cylinder topped with a hemisphere (capsule): TSA = CSA of cylinder + base of cylinder + CSA of hemisphere.
  • Hemisphere placed inside a cone (ice-cream scoop in cone).
  • Cuboid with a hemispherical cavity scooped out.
NCERT Example — Composite: hollow hemisphere mounted on cylinder (vessel)

Problem: A vessel is a hollow hemisphere mounted on a hollow cylinder. Diameter of hemisphere = 14 cm. Total height = 13 cm. Find the inner surface area of the vessel.

Solution: $r = 7\ \text{cm}$. Height of cylinder $= 13 - 7 = 6\ \text{cm}$.

Inner surface = CSA of cylinder + CSA of hemisphere $= 2\pi rh + 2\pi r^2 = 2\pi r(h + r)$

$= 2 \times \dfrac{22}{7} \times 7 \times (6 + 7) = 44 \times 13 = \mathbf{572\ \text{cm}^2}.$

NCERT Example — Composite: hemisphere on a cylinder (toy)

Problem: A toy is in the form of a hemisphere surmounted on a right circular cylinder. The diameter of the base of the cylinder is 7 cm, height of cylinder is 13 cm. Find the surface area of the toy. (Use $\pi = 22/7$)

Solution: $r = 3.5\ \text{cm}$, $h = 13\ \text{cm}$.

TSA = CSA of cylinder + base of cylinder + CSA of hemisphere

$= 2\pi rh + \pi r^2 + 2\pi r^2 = 2\pi r(h + r) + \pi r^2$

$= 2 \times \dfrac{22}{7} \times 3.5 \times 16.5 + \dfrac{22}{7} \times 12.25$

$= 22 \times 33 + 38.5 = 726 + 38.5 = \mathbf{764.5\ \text{cm}^2}.$

NCERT Example — Composite: volume of a gulab jamun (cylinder + 2 hemispheres)

Problem: A gulab jamun has cylindrical portion of length 5 cm and diameter 2.8 cm, with hemispherical ends. Find its volume.

Solution: $r = 1.4\ \text{cm}$, length of cylinder $= 5\ \text{cm}$.

Total length $= 5 + 1.4 + 1.4 = 7.8\ \text{cm}$ (two hemispheres add $2r$ to length).

Volume $= \pi r^2 h_{\text{cyl}} + 2 \times \dfrac{2}{3}\pi r^3 = \pi r^2 \left(h + \dfrac{4r}{3}\right)$

$= \dfrac{22}{7} \times (1.4)^2 \times \left(5 + \dfrac{4 \times 1.4}{3}\right) = \dfrac{22}{7} \times 1.96 \times 6.867 \approx \mathbf{25.05\ \text{cm}^3}.$

S8. Master Formula Table

Pin this table — everything your board exam tests from this chapter:

Solid LSA / CSA TSA Volume
Cuboid ($l, b, h$) $2h(l+b)$ $2(lb+bh+hl)$ $lbh$
Cube (side $a$) $4a^2$ $6a^2$ $a^3$
Cylinder ($r, h$) $2\pi r h$ $2\pi r(r+h)$ $\pi r^2 h$
Cone ($r, h, l$) $\pi r l$ $\pi r(r+l)$ $\dfrac{1}{3}\pi r^2 h$
Sphere ($r$) $4\pi r^2$ $4\pi r^2$ $\dfrac{4}{3}\pi r^3$
Hemisphere ($r$) $2\pi r^2$ $3\pi r^2$ $\dfrac{2}{3}\pi r^3$

Cone slant height: $l = \sqrt{r^2 + h^2}$. Compute this first whenever $r$ and $h$ are given.

Hollow cylinder (pipe): Volume of material $= \pi(R^2 - r^2)h$, where $R$ = outer radius, $r$ = inner radius.

Unit conversions: $1\ \text{m}^3 = 10^6\ \text{cm}^3 = 1000\ \text{litres}$; $1\ \text{dm}^3 = 1\ \text{litre}$; $1\ \text{cm}^3 = 1\ \text{mL}$.

S9. Common Mistakes and Exam Tips

  • Diameter vs radius: The single most common error. When a problem says "diameter = 14 cm", you must use $r = 7\ \text{cm}$ in every formula. Never plug in the diameter directly.
  • Slant height vs vertical height: CSA of a cone uses slant height $l$, not vertical height $h$. If the problem gives $r$ and $h$, compute $l = \sqrt{r^2+h^2}$ as the very first step.
  • Sphere: no separate LSA and TSA. The whole surface is $4\pi r^2$. Do not add any flat base.
  • Hemisphere TSA trap: For an open bowl (inner surface only), use $2\pi r^2$. For a solid hemisphere (like a rubber dome), use $3\pi r^2$. Read the question carefully.
  • Composite solid SA: The touching face disappears. Two solids joined at a circle of radius $r$ contribute $-2\pi r^2$ to the combined TSA (both sides of that face are interior).
  • The 1/3 in cone volume: Forgetting this triples your answer — a very expensive mistake in a 3-mark question. Cone $= \dfrac{1}{3}$ cylinder.
  • Consistent units: Convert all lengths to the same unit before computing. A common trap: radius in cm but height in m.
  • Choice of pi: Use $\dfrac{22}{7}$ when the radius is a multiple of 7 (gives clean answers). Use $3.14$ when decimals appear. The problem usually indicates which.
  • Volume and capacity: Volume gives cm3; to convert to litres divide by 1000. A cylinder of radius 10 cm and height 14 cm holds $\pi \times 100 \times 14 = 1400\pi \approx 4400\ \text{cm}^3 = 4.4$ litres.
Practice MCQs
1. The lateral surface area of a cube of side 4 cm is:
  1. $16\ \text{cm}^2$
  2. $64\ \text{cm}^2$
  3. $96\ \text{cm}^2$
  4. $48\ \text{cm}^2$
Answer: (B) LSA of cube $= 4a^2 = 4 \times 4^2 = 4 \times 16 = 64\ \text{cm}^2$.
2. The diameter of a sphere is 6 cm. Its volume is:
  1. $\dfrac{4}{3}\pi \times 6^3\ \text{cm}^3$
  2. $\dfrac{4}{3}\pi \times 3^3\ \text{cm}^3$
  3. $4\pi \times 3^2\ \text{cm}^3$
  4. $\pi \times 3^3\ \text{cm}^3$
Answer: (B) $r = 3\ \text{cm}$. Volume $= \dfrac{4}{3}\pi r^3 = \dfrac{4}{3}\pi \times 27 = 36\pi\ \text{cm}^3$. Only option B gives the correct expression.
3. A cone has base radius 7 cm and slant height 25 cm. Its curved surface area is:
  1. $550\ \text{cm}^2$
  2. $1100\ \text{cm}^2$
  3. $175\ \text{cm}^2$
  4. $704\ \text{cm}^2$
Answer: (A) CSA $= \pi rl = \dfrac{22}{7} \times 7 \times 25 = 22 \times 25 = 550\ \text{cm}^2$.
4. The total surface area of a right circular cylinder of radius 7 cm and height 10 cm is:
  1. $440\ \text{cm}^2$
  2. $660\ \text{cm}^2$
  3. $748\ \text{cm}^2$
  4. $374\ \text{cm}^2$
Answer: (C) TSA $= 2\pi r(r+h) = 2 \times \dfrac{22}{7} \times 7 \times (7+10) = 44 \times 17 = 748\ \text{cm}^2$.
5. A cone and a cylinder have the same base radius 6 cm and the same height 9 cm. The ratio of their volumes is:
  1. $1 : 3$
  2. $3 : 1$
  3. $1 : 9$
  4. $9 : 1$
Answer: (A) Volume of cone $= \dfrac{1}{3}\pi r^2 h$; Volume of cylinder $= \pi r^2 h$. Ratio $= \dfrac{1}{3} : 1 = 1 : 3$.
6. The slant height of a cone whose base radius is 8 cm and vertical height is 15 cm is:
  1. $23\ \text{cm}$
  2. $17\ \text{cm}$
  3. $\sqrt{161}\ \text{cm}$
  4. $\sqrt{481}\ \text{cm}$
Answer: (B) $l = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\ \text{cm}$.
7. The total surface area of a solid hemisphere of radius 7 cm is:
  1. $308\ \text{cm}^2$
  2. $154\ \text{cm}^2$
  3. $462\ \text{cm}^2$
  4. $616\ \text{cm}^2$
Answer: (C) TSA $= 3\pi r^2 = 3 \times \dfrac{22}{7} \times 49 = 3 \times 22 \times 7 = 462\ \text{cm}^2$.
8. If the radius of a sphere is doubled, its volume becomes how many times the original?
  1. 2 times
  2. 4 times
  3. 6 times
  4. 8 times
Answer: (D) Volume $\propto r^3$. New volume $= \dfrac{4}{3}\pi (2r)^3 = 8 \times \dfrac{4}{3}\pi r^3 = 8$ times the original.
9. A cuboid has dimensions $5\ \text{cm} \times 4\ \text{cm} \times 3\ \text{cm}$. The length of its space diagonal is:
  1. $\sqrt{25}\ \text{cm}$
  2. $\sqrt{41}\ \text{cm}$
  3. $5\sqrt{2}\ \text{cm}$
  4. $\sqrt{34}\ \text{cm}$
Answer: (C) Diagonal $= \sqrt{l^2+b^2+h^2} = \sqrt{25+16+9} = \sqrt{50} = 5\sqrt{2}\ \text{cm}$.
10. The curved surface area of a hemisphere equals the area of a circle of radius $R$. If the hemisphere has radius $r$, then $R$ is:
  1. $r$
  2. $r\sqrt{2}$
  3. $2r$
  4. $r/\sqrt{2}$
Answer: (B) CSA of hemisphere $= 2\pi r^2 = \pi R^2 \Rightarrow R^2 = 2r^2 \Rightarrow R = r\sqrt{2}$.
Assertion-Reason Questions
A: The total surface area of a solid hemisphere of radius 7 cm is $462\ \text{cm}^2$. R: TSA of a solid hemisphere $= 3\pi r^2$.
Answer: Both A and R are true, and R is the correct explanation of A. $3 \times \dfrac{22}{7} \times 49 = 462\ \text{cm}^2$. Verified.
A: The volume of a cone equals one-third the volume of a cylinder having the same base and height. R: This can be verified by filling a cone and emptying it into the cylinder exactly three times.
Answer: Both A and R are true, and R is the correct explanation of A. This is the standard NCERT activity demonstrating $V_{\text{cone}} = \dfrac{1}{3}V_{\text{cylinder}}$.
Previous-Year Questions
PYQ 1. The radii of two cylinders are in the ratio $2:3$ and their heights are in the ratio $5:3$. Find the ratio of their curved surface areas. (CBSE 2020, 2 marks)
Solution: CSA $= 2\pi rh$. Ratio $= \dfrac{r_1 h_1}{r_2 h_2} = \dfrac{2}{3} \times \dfrac{5}{3} = \dfrac{10}{9}$. Answer: $10 : 9$.
PYQ 2. A cone has height 24 cm and base radius 6 cm. A child reshapes it into a sphere. Find the radius of the sphere. (CBSE 2018, 3 marks)
Solution: Volume of cone $= \dfrac{1}{3}\pi \times 36 \times 24 = 288\pi\ \text{cm}^3$.
Set equal to sphere: $\dfrac{4}{3}\pi R^3 = 288\pi \Rightarrow R^3 = 216 \Rightarrow \mathbf{R = 6\ \text{cm}}$.
PYQ 3. The surface area of a sphere is $616\ \text{cm}^2$. Find its volume. (CBSE Board, 3 marks)
Solution: $4\pi r^2 = 616 \Rightarrow r^2 = \dfrac{616 \times 7}{4 \times 22} = 49 \Rightarrow r = 7\ \text{cm}$.
Volume $= \dfrac{4}{3} \times \dfrac{22}{7} \times 343 = \dfrac{4 \times 22 \times 49}{3} = \dfrac{4312}{3} \approx \mathbf{1437.33\ \text{cm}^3}$.
PYQ 4. A conical tent is to accommodate 11 persons. Each person needs $4\ \text{m}^2$ of floor space and $20\ \text{m}^3$ of air. Find the height of the tent. (CBSE Board)
Solution: Total floor area $= 44\ \text{m}^2 = \pi r^2 \Rightarrow r^2 = \dfrac{44 \times 7}{22} = 14\ \text{m}^2$.
Total volume $= 220\ \text{m}^3 = \dfrac{1}{3}\pi r^2 h = \dfrac{1}{3} \times \dfrac{22}{7} \times 14 \times h = \dfrac{44h}{3}$.
$h = \dfrac{220 \times 3}{44} = \mathbf{15\ \text{m}}$.
PYQ 5. A hemispherical tank is made of iron sheet 1 cm thick. Inner radius = 1 m. Find the volume of iron used. (CBSE Board, 3 marks)
Solution: $r = 100\ \text{cm}$ (inner), $R = 101\ \text{cm}$ (outer).
Volume of iron $= \dfrac{2}{3}\pi(R^3 - r^3) = \dfrac{2}{3} \times \dfrac{22}{7} \times (101^3 - 100^3)$.
$101^3 - 100^3 = (101-100)(101^2 + 101 \times 100 + 100^2) = 30301$.
Volume $= \dfrac{2 \times 22 \times 30301}{21} = \dfrac{1333244}{21} \approx \mathbf{63487\ \text{cm}^3} \approx 0.0635\ \text{m}^3$.
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More Class 9 Mathematics chapters
Number Systems · Polynomials · Coordinate Geometry · Linear Equations in Two Variables · Introduction to Euclid's Geometry · Lines and Angles · Triangles · Quadrilaterals