Advanced Analytical Skills — II

Unit 3: Surface Area & Volume

From flat shapes to 3-D solids — master every formula, solve painted-cube puzzles, and calculate volumes of real-world objects like water tanks, capsules, and domes.

⏱️ Time to Complete: 10–12 hours | 📝 30 MCQs (Bloom's Mapped) | 15 Worked Examples

Section A

Opening Hook — The Geometry That Built the World

🏛️ How Engineers Calculate the Marble for the Taj Mahal's Dome

The Taj Mahal's main dome is a hemisphere with a diameter of roughly 17.7 m. To estimate how much marble covers it, architects computed the curved surface area — about 2πr² ≈ 493 m². The four smaller domes, the cylindrical minarets (each 40 m tall), and the rectangular plinth all required precise surface-area and volume calculations centuries before calculators existed.

Today, the same formulas drive real decisions worth crores: How many litres does a spherical water tank hold? How much paint covers a cuboidal building? How much ice-cream fills a cone topped with a hemisphere scoop? Every civil engineer, architect, and product designer uses these formulas daily.

What if YOU could solve these instantly? This chapter turns you into a surface-area & volume powerhouse — from 2-D basics all the way to frustums and combined solids.

🏗️ Architecture🚰 Water Tanks🏭 Manufacturing🍦 Packaging💊 Pharma🏟️ Sports Stadiums

An Olympic swimming pool holds exactly 2,500 m³ of water (50 m × 25 m × 2 m — a cuboid!). India's largest water tank at Bhandup, Mumbai, treats 3,750 million litres per day — computing its capacity requires the very formulas you'll learn here.

Section B

Learning Outcomes — Bloom's Taxonomy Mapped

Bloom's Level	Learning Outcome
🔵 Remember	Recall formulas for area, perimeter, surface area, and volume of standard 2-D and 3-D shapes
🔵 Understand	Explain the difference between curved surface area (CSA), lateral surface area (LSA), and total surface area (TSA) with real-world examples
🟢 Apply	Compute the SA and volume of cubes, cuboids, cylinders, cones, spheres, hemispheres, and frustums using correct formulas
🟢 Analyze	Solve painted-cube problems by decomposing a painted cube into corner, edge, face, and interior unit cubes
🟠 Evaluate	Compare shapes to determine which container design maximises volume for minimum surface area (optimisation)
🟠 Create	Design combined-solid objects (capsule, ice-cream cone, silo) and calculate their total SA and volume

Section C

Perimeter & Area of 2-D Figures

1. Triangle & Circle

📐 Triangle — Area Formulas

Basic Formula

Area = ½ × base × height

Heron's Formula (when all 3 sides are known)

Let a, b, c be the sides. Semi-perimeter s = (a + b + c) / 2

Area = √[ s(s − a)(s − b)(s − c) ]

Perimeter

P = a + b + c

ASCII Diagram
        A
       /\
      /  \        height (h)
     /    \       ↕
    /______\
   B   base   C

   Area = ½ × BC × h

⭕ Circle — Area & Circumference

Formulas

Area = πr²

Circumference = 2πr = πd

where r = radius, d = diameter = 2r, π ≈ 22/7 or 3.1416

ASCII Diagram
         ___
       /     \
      |   •   |     •  = centre
      | ←r→  |     r  = radius
       \_____/

   Area = πr²
   Circumference = 2πr

When the question says "Take π = 22/7" use fractions for cleaner answers. When it says "Take π = 3.14" use decimals. Never mix them — it leads to rounding errors.

2. Trapezium & Parallelogram

⬡ Trapezium

A quadrilateral with one pair of parallel sides (called parallel sides a and b).

Area = ½ × (a + b) × h

where h = perpendicular distance between the parallel sides.

ASCII Diagram
      a
   ________
  /        \          h = height
 /          \         ↕
/____________\
      b

  Area = ½ × (a + b) × h

▱ Parallelogram

A quadrilateral with two pairs of parallel sides.

Area = base × height

Perimeter = 2(a + b)

ASCII Diagram
    ___________
   /          /     height (h)
  /          /      ↕
 /_________ /
    base (b)

  Area = b × h

Students confuse the slant side with the height in a parallelogram. The height is always the perpendicular distance from the base to the opposite side — NOT the slant side length.

Every rectangle is a parallelogram, but not every parallelogram is a rectangle. A rectangle is a special parallelogram where all angles are 90°. Similarly, a square is a special case of both a rectangle and a rhombus.

Section D

Cube & Cuboid — Surface Area & Volume

🧊 Cube (all sides equal = a)

Total Surface Area (TSA) = 6a²

Lateral Surface Area (LSA) = 4a²

Volume = a³

Diagonal of cube = a√3

ASCII Diagram
       ________
      /       /|
     /       / |    a = edge length
    /______ /  |
    |      |   |
    |      |   /    All edges = a
    |      |  /
    |______|/

  TSA = 6a²     Volume = a³

📦 Cuboid (length l, breadth b, height h)

TSA = 2(lb + bh + hl)

LSA = 2h(l + b)

Volume = l × b × h

Diagonal = √(l² + b² + h²)

ASCII Diagram
       _____________
      /            /|
     /            / |  h (height)
    /____________/  |
    |           |   |
    |           |   /  b (breadth)
    |           |  /
    |___________|/
         l (length)

  TSA = 2(lb + bh + hl)
  Volume = l × b × h

3. Painted Cube Problems

One of the most popular competitive-exam question types. A cube of side n units is painted on all 6 faces and then cut into unit cubes (1 × 1 × 1). How many small cubes have 3, 2, 1, or 0 painted faces?

🎨 Painted Cube Master Formula

Painted Faces	Position	Count Formula	Example (n = 4)
3 faces painted	Corner cubes	8 (always)	8
2 faces painted	Edge cubes (not corners)	12(n − 2)	12 × 2 = 24
1 face painted	Face-centre cubes	6(n − 2)²	6 × 4 = 24
0 faces painted	Interior cubes	(n − 2)³	2³ = 8
Total cubes	All	n³	4³ = 64

Verification: 8 + 24 + 24 + 8 = 64 = 4³ ✅

Painted Cube n = 3
  Top View (looking down):
  ┌───┬───┬───┐
  │ C │ E │ C │   C = Corner (3 faces painted)
  ├───┼───┼───┤   E = Edge   (2 faces painted)
  │ E │ F │ E │   F = Face   (1 face painted)
  ├───┼───┼───┤   I = Interior (0 faces)
  │ C │ E │ C │
  └───┴───┴───┘
  (n=3 has 0 interior cubes since (3-2)³ = 1³ = 1)

Quick check: The sum of all categories must equal n³. If your numbers don't add up, re-check your calculation. Also remember: for n = 1, the entire cube has all faces painted (but no "cutting" possible). For n = 2, every piece is a corner piece (8 corners, each with 3 painted faces).

Painted cube problems appear in every major Indian exam — CAT, SSC CGL, Bank PO, UPSC CSAT, GATE. Mastering the formula table above gives you instant answers in under 10 seconds.

Section E

Sphere & Hemisphere

🌍 Sphere (radius r)

Surface Area = 4πr²

Volume = (4/3)πr³

ASCII Diagram
         ___
       /     \
      /       \
     |    •    |    • = centre, r = radius
      \       /
       \_____/

  SA = 4πr²     Volume = 4/3 πr³

🥣 Hemisphere (radius r)

Curved Surface Area (CSA) = 2πr²

Total Surface Area (TSA) = 3πr² (curved + flat circular base)

Volume = (2/3)πr³

ASCII Diagram
         ___
       /     \
      /       \
     |    •    |    flat base = πr²
     |_________|

  CSA = 2πr²    TSA = 3πr²
  Volume = 2/3 πr³

Water Tank Problem (Hemispheric)

A hemispherical water tank has internal radius r. Its capacity in litres = Volume in cm³ ÷ 1000. Remember: 1 litre = 1000 cm³ and 1 m³ = 1000 litres.

Overhead water tanks in Indian villages are often hemispherical. If a tank has radius 1.05 m, its volume = (2/3) × (22/7) × (1.05)³ = 2.425 m³ ≈ 2,425 litres — enough for a family of 5 for about 3 days.

Ball Placed in a Cylindrical Vessel

When a sphere of radius r is fully submerged in a cylinder of radius R (R ≥ r), the water level rises. The volume of water displaced equals the volume of the sphere.

🔬 Ball in Cylinder — Water Rise Formula

Rise in water level h = Volume of sphere ÷ Area of cylinder base

h = (4/3)πr³ ÷ πR² = 4r³ / (3R²)

If the ball fits exactly (r = R): h = 4r/3

ASCII Diagram
    ┌──────────┐  ← water rises by h
    │    ___   │
    │  /     \ │
    │ | ball  |│  radius = r
    │  \_____/ │
    │~~~~~~~~~~│  ← original water level
    │          │
    │          │  Cylinder radius = R
    └──────────┘

  Water rise h = 4r³ / (3R²)

Students forget to use the cylinder's radius (not the ball's radius) as the base for computing rise. The rise depends on the cross-sectional area of the container (πR²), not the ball.

Section F

Cone & Cylinder

🥫 Cylinder (radius r, height h)

CSA (Curved Surface Area) = 2πrh

TSA = 2πr(r + h) (curved surface + two circular ends)

Volume = πr²h

ASCII Diagram
      ________
     /        \     ← top circle (πr²)
    |          |
    |          |    h = height
    |          |
    |          |
     \________/     ← bottom circle (πr²)
         r

  CSA = 2πrh    TSA = 2πr(r+h)
  Volume = πr²h

🍦 Cone (radius r, height h, slant height l)

Slant height: l = √(r² + h²)

CSA = πrl

TSA = πr(r + l) (curved surface + circular base)

Volume = (1/3)πr²h

ASCII Diagram
         /\
        /  \
       / l  \       l = slant height
      /   |  \      h = vertical height
     /    |h  \     r = base radius
    /     |    \
   /______|_____\
         r

  l = √(r² + h²)
  CSA = πrl     Volume = ⅓πr²h

A cone's volume is exactly 1/3 of the cylinder with the same base and height. This means 3 cones of water can fill 1 cylinder of the same dimensions. This fact is a common MCQ trap!

The conical shape is used in rocket nose cones (like ISRO's PSLV) because it minimises air resistance. The slant height determines the surface area that faces the airflow — engineers optimise the ratio r:h for maximum aerodynamic efficiency.

Section G

Frustum & Combined Solids

1. Frustum of a Cone

When a cone is cut by a plane parallel to its base, the portion between the base and the cut is called a frustum. Think of a bucket or a lampshade — that's a frustum.

🪣 Frustum Formulas (R = larger radius, r = smaller radius, h = height)

Slant height: l = √[ h² + (R − r)² ]

CSA = π(R + r)l

TSA = π(R + r)l + πR² + πr²

Volume = (πh/3)(R² + r² + Rr)

ASCII Diagram
       ______
      /  r   \        r = top radius (smaller)
     /        \
    /    | h   \      h = height
   /     |      \     l = slant height
  /______|_______\
       R              R = bottom radius (larger)

  l = √[h² + (R-r)²]
  Volume = πh/3 (R² + r² + Rr)

Indian bucket (balti) is a frustum! A typical steel balti has R = 15 cm, r = 10 cm, h = 30 cm. Its capacity = (π × 30 / 3)(225 + 100 + 150) = 10π × 475 ≈ 14,923 cm³ ≈ 14.9 litres.

2. Combined Solids

Many real objects are combinations of basic shapes. The key principle:

Volume of combined solid = Sum of individual volumes
Surface area = Sum of outer surfaces only (subtract the areas where shapes join)

🔗 Common Combinations

1. Cone on Hemisphere (Ice-cream)
         /\
        /  \        ← Cone
       /    \
      /______\
     (  hemi  )     ← Hemisphere
      \______/

  Volume = ⅓πr²h + ⅔πr³
  TSA = πrl + 2πr²
  (No base areas where they join)

2. Cylinder with Hemisphere Ends (Capsule)
      ___  ___________  ___
     (   )|           |(   )
      ‾‾‾  ‾‾‾‾‾‾‾‾‾‾‾  ‾‾‾
     hemi   cylinder    hemi

  Volume = πr²h + 2 × ⅔πr³ = πr²h + 4/3 πr³
  TSA = 2πrh + 2 × 2πr² = 2πrh + 4πr²
  (No flat circles — they are internal joints)

3. Cone on Cylinder (Silo / Rocket)
         /\
        /  \       ← Cone
       /    \
      |      |
      |      |     ← Cylinder
      |      |
      |______|

  Volume = πr²H + ⅓πr²h
  TSA = 2πrH + πrl + πr²
  (Only one base of cylinder is exposed)

The biggest mistake in combined-solid SA problems: Students add ALL surface areas of both shapes, including the joint surfaces that are hidden inside. Always subtract the area where two solids are joined (usually a circle of area πr²).

Section H

Master Formula Comparison Table

Your one-stop reference — keep this table bookmarked for exams.

2-D Figures

Shape	Area	Perimeter
Square (side a)	a²	4a
Rectangle (l × b)	l × b	2(l + b)
Triangle (base b, height h)	½ × b × h	a + b + c
Triangle (Heron's)	√[s(s−a)(s−b)(s−c)]	a + b + c (s = P/2)
Circle (radius r)	πr²	2πr
Trapezium (∥ sides a, b; height h)	½(a + b) × h	Sum of all 4 sides
Parallelogram (base b, height h)	b × h	2(a + b)

3-D Solids

Solid	CSA / LSA	TSA	Volume
Cube (a)	4a²	6a²	a³
Cuboid (l, b, h)	2h(l + b)	2(lb + bh + hl)	l × b × h
Cylinder (r, h)	2πrh	2πr(r + h)	πr²h
Cone (r, h, l)	πrl	πr(r + l)	⅓πr²h
Sphere (r)	4πr²	4πr² (same)	4/3 πr³
Hemisphere (r)	2πr²	3πr²	2/3 πr³
Frustum (R, r, h, l)	π(R+r)l	π(R+r)l + πR² + πr²	πh/3 (R²+r²+Rr)

Memory trick for volumes: Cone = ⅓ Cylinder, Hemisphere = ⅔ Sphere. Also, Sphere SA = 4 × circle area (4πr²). These relationships help you derive forgotten formulas in exams.

Section I

Worked Examples — 15 Step-by-Step Solutions

Ex. 1 — Area of Triangle Using Heron's Formula

BeginnerTopic: 2-D Figures

Question: Find the area of a triangle with sides 13 cm, 14 cm, and 15 cm.

Solution:

a = 13, b = 14, c = 15

s = (13 + 14 + 15) / 2 = 21

Area = √[s(s−a)(s−b)(s−c)]

= √[21 × (21−13) × (21−14) × (21−15)]

= √[21 × 8 × 7 × 6]

= √[7056]

= 84 cm²

Ex. 2 — Area of Trapezium

BeginnerTopic: 2-D Figures

Question: A trapezium has parallel sides of lengths 12 cm and 8 cm, and the perpendicular distance between them is 5 cm. Find its area.

Solution:

a = 12 cm, b = 8 cm, h = 5 cm

Area = ½ × (a + b) × h

= ½ × (12 + 8) × 5

= ½ × 20 × 5

= 50 cm²

Ex. 3 — Circumference and Area of a Circle

BeginnerTopic: 2-D Figures

Question: A circular garden has diameter 28 m. Find its circumference and area. (Take π = 22/7)

Solution:

d = 28 m → r = 14 m

Circumference = 2πr = 2 × (22/7) × 14 = 88 m

Area = πr² = (22/7) × 14² = (22/7) × 196 = 616 m²

Ex. 4 — Total Surface Area of a Cube

BeginnerTopic: Cube

Question: A Rubik's cube has edge 5.7 cm. Find its TSA and the volume of the cube.

Solution:

a = 5.7 cm

TSA = 6a² = 6 × (5.7)² = 6 × 32.49 = 194.94 cm²

Volume = a³ = (5.7)³ = 185.19 cm³

Ex. 5 — Volume of a Cuboid

BeginnerTopic: Cuboid

Question: A water tank is cuboidal with dimensions 2 m × 1.5 m × 1 m. How many litres of water can it hold?

Solution:

Volume = l × b × h = 2 × 1.5 × 1 = 3 m³

Since 1 m³ = 1000 litres:

Capacity = 3 × 1000 = 3000 litres

Ex. 6 — Painted Cube Problem (n = 4)

IntermediateTopic: Painted Cube

Question: A cube of side 4 cm is painted red on all faces and then cut into 1-cm unit cubes. Find the number of unit cubes with (a) 3 painted faces, (b) 2 painted faces, (c) 1 painted face, (d) 0 painted faces.

Solution: n = 4

Painted Cube Breakdown
  (a) 3 faces painted (corners):  8                = 8
  (b) 2 faces painted (edges):   12 × (4−2) = 12×2 = 24
  (c) 1 face painted (faces):    6 × (4−2)² = 6×4  = 24
  (d) 0 faces painted (interior): (4−2)³    = 2³   = 8
                                             ────
                                  Total      = 64 = 4³ ✓

Answers: (a) 8, (b) 24, (c) 24, (d) 8

Ex. 7 — Surface Area of a Sphere

BeginnerTopic: Sphere

Question: A football has a radius of 11 cm. Find the leather required to make it (i.e., its surface area). (π = 3.14)

Solution:

SA = 4πr² = 4 × 3.14 × 11² = 4 × 3.14 × 121

= 1519.76 cm²

Ex. 8 — Volume of a Hemisphere

BeginnerTopic: Hemisphere

Question: A hemispherical bowl has diameter 21 cm. Find the volume of soup it can hold. (π = 22/7)

Solution:

d = 21 cm → r = 10.5 cm

Volume = (2/3)πr³ = (2/3) × (22/7) × (10.5)³

= (2/3) × (22/7) × 1157.625

= (2/3) × 3637.5

= 2425 cm³ ≈ 2.425 litres

Ex. 9 — Hemispherical Water Tank Capacity

IntermediateTopic: Hemisphere — Real World

Question: A village has a hemispherical overhead water tank with internal radius 2.1 m. If the village uses 500 litres/day, how many days will a full tank last? (π = 22/7)

Solution:

Volume = (2/3)πr³ = (2/3) × (22/7) × (2.1)³

= (2/3) × (22/7) × 9.261

= (2/3) × 29.106

= 19.404 m³

Convert: 19.404 m³ × 1000 = 19,404 litres

Days = 19404 / 500 = 38.8 ≈ 38 full days

Ex. 10 — Ball Placed in Cylindrical Vessel

IntermediateTopic: Sphere + Cylinder

Question: A metallic sphere of radius 3 cm is dropped into a cylindrical vessel of radius 6 cm that is partially filled with water. Find the rise in the water level.

Diagram
    ┌──────────┐
    │  ↑ h=?   │
    │   ___    │
    │ / sph \  │  r_sphere = 3 cm
    │ \_____/  │
    │~~~~~~~~~~│  ← original level
    │          │  R_cyl = 6 cm
    └──────────┘

Solution:

Volume of sphere = (4/3)π(3)³ = (4/3)π × 27 = 36π cm³

Volume displaced = πR²h → 36π = π(6)²h → 36π = 36πh

h = 1 cm

The water level rises by 1 cm.

Ex. 11 — Volume of a Cylinder

BeginnerTopic: Cylinder

Question: A cylindrical pillar has diameter 50 cm and height 3.5 m. Find (a) its CSA and (b) volume. (π = 22/7)

Solution:

d = 50 cm → r = 25 cm = 0.25 m, h = 3.5 m

(a) CSA = 2πrh = 2 × (22/7) × 0.25 × 3.5 = 5.5 m²

(b) Volume = πr²h = (22/7) × (0.25)² × 3.5 = (22/7) × 0.0625 × 3.5

= (22/7) × 0.21875 = 0.6875 m³

Ex. 12 — Slant Height and CSA of a Cone

IntermediateTopic: Cone

Question: A conical tent has a base radius of 7 m and height 24 m. Find (a) its slant height, (b) the area of canvas required (CSA). (π = 22/7)

Solution:

(a) l = √(r² + h²) = √(49 + 576) = √625 = 25 m

(b) CSA = πrl = (22/7) × 7 × 25 = 550 m²

Ex. 13 — Volume of a Frustum (Bucket)

IntermediateTopic: Frustum

Question: A bucket is in the shape of a frustum with top radius 20 cm, bottom radius 12 cm, and height 30 cm. Find its capacity in litres. (π = 3.14)

Solution:

R = 20, r = 12, h = 30

Volume = (πh/3)(R² + r² + Rr)

= (3.14 × 30 / 3)(400 + 144 + 240)

= (31.4)(784)

= 24,617.6 cm³

= 24617.6 / 1000 = 24.62 litres

Ex. 14 — Combined Solid: Cone on Hemisphere (Ice-Cream)

AdvancedTopic: Combined Solids

Question: An ice-cream is modelled as a cone of height 12 cm topped with a hemispherical scoop, both having radius 3.5 cm. Find the total volume of ice-cream. (π = 22/7)

Diagram
        ___
      /     \     ← Hemisphere (r = 3.5)
     (       )
      \_____/
       \   /
        \ /       ← Cone (h = 12, r = 3.5)
         V

Solution:

Volume of cone = (1/3)πr²h = (1/3)(22/7)(3.5)²(12) = (1/3)(22/7)(12.25)(12)

= (1/3)(22/7)(147) = (1/3)(462) = 154 cm³

Volume of hemisphere = (2/3)πr³ = (2/3)(22/7)(3.5)³ = (2/3)(22/7)(42.875)

= (2/3)(134.75) = 89.83 cm³

Total volume = 154 + 89.83 = 243.83 cm³

Ex. 15 — Capsule Shape: Cylinder with Hemisphere Ends

AdvancedTopic: Combined Solids

Question: A medicine capsule is 14 mm long. It consists of a cylindrical part in the middle and hemispherical ends, each of radius 3.5 mm. Find the total surface area and volume. (π = 22/7)

Diagram
    ___  ___________  ___
   (   )|           |(   )    Total length = 14 mm
    ‾‾‾  ‾‾‾‾‾‾‾‾‾‾‾  ‾‾‾    r = 3.5 mm
   hemi   cylinder    hemi
   3.5mm   7 mm      3.5mm

Solution:

r = 3.5 mm, total length = 14 mm

Cylinder height h = 14 − 2(3.5) = 14 − 7 = 7 mm

TSA = CSA of cylinder + 2 × CSA of hemisphere

= 2πrh + 2(2πr²) = 2πrh + 4πr²

= 2(22/7)(3.5)(7) + 4(22/7)(3.5)²

= 154 + 154 = 308 mm²

Volume = πr²h + 2 × (2/3)πr³ = πr²h + (4/3)πr³

= (22/7)(3.5)²(7) + (4/3)(22/7)(3.5)³

= (22/7)(12.25)(7) + (4/3)(22/7)(42.875)

= 269.5 + 179.67 = 449.17 mm³

Section J

MCQ Assessment Bank — 30 Questions (Bloom's Mapped)

Remember / Recall (Q1–Q5)

The formula for the volume of a cone is:

πr²h
⅓πr²h
2πr²h
⅔πr³

Remember

✅ Answer: (B) ⅓πr²h — The volume of a cone is one-third of the volume of a cylinder with the same base and height.

The total surface area of a cube with edge 'a' is:

4a²
6a²
a³
8a²

Remember

✅ Answer: (B) 6a² — A cube has 6 faces, each of area a².

The curved surface area of a hemisphere of radius r is:

πr²
4πr²
2πr²
3πr²

Remember

✅ Answer: (C) 2πr² — The CSA is exactly half the SA of a full sphere (4πr²/2 = 2πr²).

Heron's formula uses which of the following?

Base and height
All three sides and semi-perimeter
Two sides and included angle
Diagonal and height

Remember

✅ Answer: (B) — Heron's formula: Area = √[s(s−a)(s−b)(s−c)] where s = (a+b+c)/2.

The slant height of a cone with radius r and height h is:

r + h
√(r² + h²)
√(r² − h²)
r × h

Remember

✅ Answer: (B) √(r² + h²) — By Pythagoras' theorem applied to the right triangle formed by r, h, and l.

Understand / Explain (Q6–Q10)

Why is the volume of a cone exactly ⅓ of the volume of a cylinder with the same base and height?

Because the cone has one less face
Because the cone tapers to a point, enclosing only ⅓ of the space
Because the cone's height is ⅓ of the cylinder's
Because the cone has a smaller radius

Understand

✅ Answer: (B) — The cone tapers uniformly from full base area to zero, and the integral of the cross-sectional area over the height equals exactly ⅓ of the cylinder's volume.

In a painted cube of side n, why do corner cubes always have exactly 3 painted faces?

Because they touch 3 edges
Because they are at the intersection of exactly 3 faces of the original cube
Because they have 3 visible sides
Because they are the smallest cubes

Understand

✅ Answer: (B) — Each corner of a cube is the meeting point of exactly 3 faces, so the unit cube at each corner has 3 of its faces painted.

The TSA of a hemisphere is 3πr² instead of 2πr² because:

It includes the volume
The curved surface (2πr²) plus the flat circular base (πr²) gives 3πr²
Hemispheres have 3 surfaces
The formula is derived differently from spheres

Understand

✅ Answer: (B) — TSA = CSA + base area = 2πr² + πr² = 3πr².

When computing the TSA of a combined solid (cone on cylinder), why do we subtract πr²?

Because we made a calculation error
Because the circular base of the cone sits on the cylinder and is no longer an external surface
Because the cylinder's top is thicker
Because the cone has no base

Understand

✅ Answer: (B) — Where the cone sits on the cylinder, the cone's base circle and the cylinder's top circle are internal joints, not exposed surfaces.

Q10

A frustum is created by:

Cutting a cone along its slant height
Cutting a cone with a plane parallel to its base
Joining two cones base to base
Cutting a sphere in half

Understand

✅ Answer: (B) — A frustum results when a smaller cone is removed from the top by a cut parallel to the base.

Apply / Calculate (Q11–Q15)

Q11

The volume of a cuboid measuring 5 cm × 4 cm × 3 cm is:

12 cm³
60 cm³
94 cm²
47 cm³

Apply

✅ Answer: (B) 60 cm³ — V = l×b×h = 5×4×3 = 60 cm³.

Q12

A cylinder has radius 7 cm and height 10 cm. Its CSA is: (π = 22/7)

440 cm²
220 cm²
880 cm²
154 cm²

Apply

✅ Answer: (A) 440 cm² — CSA = 2πrh = 2 × (22/7) × 7 × 10 = 440 cm².

Q13

A sphere has radius 3 cm. Its volume is: (π = 22/7)

113.14 cm³
36π cm³
84 cm³
38.5 cm³

Apply

✅ Answer: (A) ≈ 113.14 cm³ — V = (4/3)πr³ = (4/3)(22/7)(27) = (4/3)(84.857) ≈ 113.14 cm³. (B is also acceptable as 36π.)

Q14

A cone has radius 6 cm and height 8 cm. Its slant height is:

14 cm
10 cm
√(100) = 10 cm
Both B and C

Apply

✅ Answer: (D) — l = √(36+64) = √100 = 10 cm. Both B and C state the same answer.

Q15

Area of a trapezium with parallel sides 10 cm and 14 cm, and height 6 cm is:

84 cm²
72 cm²
140 cm²
48 cm²

Apply

✅ Answer: (B) 72 cm² — Area = ½(10+14)×6 = ½×24×6 = 72 cm².

Analyze / Compare (Q16–Q20)

Q16

A cube of side 5 is painted and cut into unit cubes. How many cubes have exactly 2 painted faces?

Analyze

✅ Answer: (B) 36 — Using 12(n−2) = 12(5−2) = 12×3 = 36.

Q17

A metallic sphere is melted and recast into a cylinder of the same radius. If the sphere's radius is r, the height of the cylinder is:

4r/3
4r
r/3
2r/3

Analyze

✅ Answer: (A) 4r/3 — Volume of sphere = volume of cylinder → (4/3)πr³ = πr²h → h = 4r/3.

Q18

Which has greater volume: a sphere of radius 3 cm or a cube of side 5 cm?

Sphere
Cube
Both are equal
Cannot be determined

Analyze

✅ Answer: (B) Cube — Sphere volume = (4/3)π(27) ≈ 113.1 cm³. Cube volume = 125 cm³. Cube is larger.

Q19

A cone, a hemisphere, and a cylinder all have the same radius r and height r. The ratio of their volumes is:

1 : 2 : 3
1 : 3 : 2
2 : 1 : 3
1 : 2 : 4

Analyze

✅ Answer: (A) 1 : 2 : 3 — Cone = ⅓πr³, Hemisphere = ⅔πr³, Cylinder = πr³. Ratio = ⅓ : ⅔ : 1 = 1 : 2 : 3.

Q20

A cube of side 6 is painted. How many unit cubes have NO paint on them?

Analyze

✅ Answer: (A) 64 — Interior cubes = (n−2)³ = (6−2)³ = 4³ = 64.

Evaluate / Judge (Q21–Q25)

Q21

A cylindrical can and a cuboidal box have the same height and base area. Which uses less material (surface area)?

Cylinder always uses less
Cuboid always uses less
Both use the same
Depends on dimensions

Evaluate

✅ Answer: (A) — For a given enclosed area, a circle has the minimum perimeter (isoperimetric inequality), so the cylinder's lateral surface area is always less than any prism with the same base area and height.

Q22

A student claims: "If we double the radius of a sphere, its volume becomes 4 times." Is this correct?

Yes, volume scales with r²
No, volume becomes 8 times (scales with r³)
No, volume becomes 6 times
Yes, because SA doubles

Evaluate

✅ Answer: (B) — Volume = (4/3)πr³. If r → 2r, V → (4/3)π(2r)³ = 8 × (4/3)πr³. Volume scales as the cube of the radius.

Q23

Which container shape holds the maximum volume for a given surface area?

Cube
Cylinder
Sphere
Cone

Evaluate

✅ Answer: (C) Sphere — The sphere has the best volume-to-surface-area ratio of any 3-D shape (isoperimetric inequality in 3-D).

Q24

A student calculated TSA of a combined cone-on-cylinder as (2πrH + 2πr² + πrl + πr²). What error did they make?

They forgot the cone's CSA
They included both flat circles of the cylinder — one should be excluded (joint surface)
They used wrong formula for cone
No error — the formula is correct

Evaluate

✅ Answer: (B) — The top circle of the cylinder is the joint with the cone's base and is internal. The correct TSA = 2πrH + πr² + πrl (only the bottom circle of the cylinder).

Q25

A hemispherical dome is to be painted. A painter charges ₹50/m². If the dome has radius 7 m, the cost is: (π = 22/7)

₹15,400
₹30,800
₹7,700
₹23,100

Evaluate

✅ Answer: (A) ₹15,400 — CSA = 2πr² = 2(22/7)(49) = 308 m². Cost = 308 × 50 = ₹15,400. (We use CSA, not TSA, since only the curved part is painted.)

Create / Design (Q26–Q30)

Q26

A toy is shaped as a cone surmounted on a hemisphere, both of radius 3 cm. If the total height of the toy is 15 cm, the height of the cone is:

15 cm
12 cm
9 cm
18 cm

Create

✅ Answer: (B) 12 cm — Total height = height of cone + radius of hemisphere → h = 15 − 3 = 12 cm.

Q27

A capsule of total length 18 mm and radius 3 mm has its volume equal to:

π(3)²(12) + (4/3)π(3)³
π(3)²(18) + (4/3)π(3)³
π(3)²(12) + (2/3)π(3)³
2π(3)(12) + 4π(3)²

Create

✅ Answer: (A) — Cylinder height = 18 − 2(3) = 12 mm. Volume = πr²h + 2×(2/3)πr³ = π(9)(12) + (4/3)π(27).

Q28

A solid is formed by placing a cone on top of a cylinder (same radius r = 7, cylinder height = 10, cone height = 24). Its total volume is: (π = 22/7)

1540 + 1232 = 2772 cm³
1540 + 4312 = 5852 cm³
3080 + 1232 = 4312 cm³
1540 + 616 = 2156 cm³

Create

✅ Answer: (A) — Cylinder V = πr²h = (22/7)(49)(10) = 1540. Cone V = ⅓πr²h = ⅓(22/7)(49)(24) = 1232. Total = 2772 cm³.

Q29

A grain silo consists of a cylinder (r = 5 m, h = 8 m) topped with a hemispherical roof. Its total capacity is: (π = 3.14)

628 + 261.67 = 889.67 m³
628 + 523.33 = 1151.33 m³
314 + 261.67 = 575.67 m³
628 + 130.83 = 758.83 m³

Create

✅ Answer: (A) — Cylinder = π(25)(8) = 628. Hemisphere = (2/3)π(125) ≈ 261.67. Total ≈ 889.67 m³.

Q30

A frustum-shaped glass has R = 5 cm, r = 3 cm, h = 10 cm. Its capacity in ml is approximately: (π = 3.14, 1 cm³ = 1 ml)

513 ml
1026 ml
163 ml
487 ml

Create

✅ Answer: (A) ≈ 513 ml — V = (πh/3)(R²+r²+Rr) = (3.14×10/3)(25+9+15) = (10.467)(49) ≈ 512.87 ≈ 513 ml.

Section K

Short Answer Questions (8)

SA Q1

Q: State the formula for the area of a trapezium and explain each variable.

A: Area = ½ × (a + b) × h, where 'a' and 'b' are the lengths of the two parallel sides, and 'h' is the perpendicular distance (height) between them. The formula averages the two parallel sides and multiplies by the height.

SA Q2

Q: A cube has edge 10 cm. Find its LSA and diagonal.

A: LSA = 4a² = 4 × 100 = 400 cm². Diagonal = a√3 = 10√3 ≈ 17.32 cm.

SA Q3

Q: Differentiate between CSA and TSA of a cylinder.

A: CSA (2πrh) is only the curved part — imagine unwrapping the cylinder into a rectangle. TSA (2πr(r+h)) = CSA + areas of two circular ends (2πr²). When a pipe is open-ended, we use CSA; for a closed tin, we use TSA.

SA Q4

Q: A painted cube of side 3 is cut into unit cubes. How many have exactly 1 painted face?

A: Using the formula 6(n−2)² = 6(3−2)² = 6(1)² = 6. There are 6 unit cubes with exactly 1 painted face (one at the centre of each face of the original cube).

SA Q5

Q: Write the relationship between the volumes of a cone, hemisphere, and cylinder when all have the same radius r and height r.

A: Cone = ⅓πr³, Hemisphere = ⅔πr³, Cylinder = πr³. The ratio is 1 : 2 : 3. Three cones fill one cylinder; one hemisphere is ⅔ of the cylinder.

SA Q6

Q: A sphere of radius 4 cm is melted and recast into small spheres of radius 1 cm. How many small spheres are formed?

A: Volume of large sphere = (4/3)π(4)³ = 256π/3. Volume of small sphere = (4/3)π(1)³ = 4π/3. Number = (256π/3) ÷ (4π/3) = 256/4 = 64 small spheres.

SA Q7

Q: What is a frustum? Give two real-life examples.

A: A frustum is the portion of a cone obtained by cutting it with a plane parallel to the base. It has two circular ends of different radii. Real-life examples: a bucket (balti), a lampshade, a glass tumbler, or a flowerpot.

SA Q8

Q: Find the area of a triangle with sides 5 cm, 12 cm, and 13 cm. Identify the type of triangle.

A: Check: 5² + 12² = 25 + 144 = 169 = 13². It's a right triangle. Area = ½ × 5 × 12 = 30 cm². (Heron's formula also gives the same: s = 15, Area = √(15×10×3×2) = √900 = 30.)

Section L

Long Answer Questions (3)

LA Q1 — Painted Cube Analysis

AdvancedMarks: 10

Question: A wooden cube of side 6 cm is painted blue on all faces and then cut into unit cubes (1 cm × 1 cm × 1 cm).

(a) How many unit cubes are formed in total?

(b) How many have exactly 3, 2, 1, and 0 painted faces respectively?

(d) If the cube were of side 'n', derive a general formula for each category.

Answer:

(a) Total = n³ = 6³ = 216 unit cubes

(b)

3 faces painted (corners): 8
2 faces painted (edges): 12(n−2) = 12(4) = 48
1 face painted (face centres): 6(n−2)² = 6(16) = 96
0 faces painted (interior): (n−2)³ = 4³ = 64

(c) Verification: 8 + 48 + 96 + 64 = 216 = 6³ ✓

(d) General formulas for a cube of side n:

Painted Faces	Position	Formula
3	Corners	8 (constant for n ≥ 2)
2	Edges	12(n − 2) for n ≥ 3
1	Face centres	6(n − 2)² for n ≥ 3
0	Interior	(n − 2)³ for n ≥ 3

LA Q2 — Combined Solid with Multiple Parts

AdvancedMarks: 10

Question: A decorative pillar consists of a cylinder of height 2.8 m and radius 0.35 m, surmounted by a cone of height 0.7 m (same base radius). The pillar stands on a cuboid pedestal of dimensions 1 m × 1 m × 0.5 m. Find: (π = 22/7)

(a) Volume of the cylinder

(b) Volume of the cone

(d) Total volume of the pillar (cylinder + cone)

(e) Total outer surface area to be painted (exclude the base of pedestal and joint surfaces)

Answer:

(a) V_cyl = πr²h = (22/7)(0.35)²(2.8) = (22/7)(0.1225)(2.8) = (22/7)(0.343) = 1.078 m³

(b) V_cone = ⅓πr²h = ⅓(22/7)(0.1225)(0.7) = ⅓(22/7)(0.08575) = ⅓(0.2695) = 0.0898 m³

(c) l = √(r² + h²) = √(0.1225 + 0.49) = √(0.6125) = 0.7826 m

(d) Total V = 1.078 + 0.0898 = 1.168 m³

(e) Painted area = LSA of pedestal (4 sides) + top of pedestal (minus cylinder base circle) + CSA of cylinder + CSA of cone

LSA pedestal = 2 × 0.5 × (1 + 1) = 2 m²

Top of pedestal (exposed) = 1 × 1 − π(0.35)² = 1 − 0.385 = 0.615 m²

CSA cylinder = 2πrh = 2(22/7)(0.35)(2.8) = 6.16 m²

CSA cone = πrl = (22/7)(0.35)(0.7826) = 0.8609 m²

Total painted area ≈ 2 + 0.615 + 6.16 + 0.861 = 9.636 m²

LA Q3 — Real-World Water Tank Problem

AdvancedMarks: 10

Question: A village water storage system consists of a cylindrical tank (radius 3 m, height 4 m) with a hemispherical dome on top (same radius). (π = 22/7)

(a) Find the total volume of water the tank can hold (in litres).

(b) Find the total outer surface area (the tank sits on the ground — exclude the bottom circle).

(d) If the village uses 2000 litres/day, how many days will a full tank last?

Answer:

(a)

V_cylinder = πr²h = (22/7)(9)(4) = 113.14 m³

V_hemisphere = (2/3)πr³ = (2/3)(22/7)(27) = 56.57 m³

Total volume = 113.14 + 56.57 = 169.71 m³

In litres = 169.71 × 1000 = 1,69,714 litres

(b)

CSA of cylinder = 2πrh = 2(22/7)(3)(4) = 75.43 m²

CSA of hemisphere = 2πr² = 2(22/7)(9) = 56.57 m²

Total outer SA = 75.43 + 56.57 = 132 m²

(No bottom circle since it sits on the ground; no top circle of cylinder since hemisphere covers it.)

(c) Cost = 132 × 25 = ₹3,300

(d) Days = 1,69,714 / 2000 = 84.86 ≈ 84 full days

Section M

Formula Sheet & Chapter Summary

📋 Quick-Reference Formula Sheet

2-D Shapes

▸ Triangle: A = ½bh | Heron's: A = √[s(s−a)(s−b)(s−c)], s = (a+b+c)/2

▸ Circle: A = πr², C = 2πr

▸ Trapezium: A = ½(a+b)h

▸ Parallelogram: A = bh

3-D Solids

▸ Cube (a): TSA = 6a², LSA = 4a², V = a³

▸ Cuboid (l,b,h): TSA = 2(lb+bh+hl), LSA = 2h(l+b), V = lbh

▸ Cylinder (r,h): CSA = 2πrh, TSA = 2πr(r+h), V = πr²h

▸ Cone (r,h,l): l = √(r²+h²), CSA = πrl, TSA = πr(r+l), V = ⅓πr²h

▸ Sphere (r): SA = 4πr², V = 4/3 πr³

▸ Hemisphere (r): CSA = 2πr², TSA = 3πr², V = 2/3 πr³

▸ Frustum (R,r,h): l = √[h²+(R−r)²], CSA = π(R+r)l, V = πh/3(R²+r²+Rr)

Painted Cube (side n, cut into unit cubes)

▸ 3 painted faces: 8 | 2 painted: 12(n−2) | 1 painted: 6(n−2)² | 0 painted: (n−2)³

Conversions

▸ 1 m³ = 1000 litres | 1 litre = 1000 cm³ | 1 cm³ = 1 ml

Exam strategy: Always write the formula first, substitute values, then simplify. This earns partial marks even if your final answer has an arithmetic error. Also, draw a rough diagram for every 3-D problem — it prevents sign and formula errors.

Self-check: Can you write all 3-D formulas from memory? Cover the formula sheet above and try writing them on a blank paper. If you can recall all formulas within 3 minutes, you're exam-ready!

✅ Unit 3 complete. You've mastered Surface Area & Volume!

[QR: Link to EduArtha video tutorial — Surface Area & Volume]