Area and Volume Questions for Placement 2026 (with Solutions)

Last Updated: March 2026

Introduction to Area and Volume

Area and Volume problems are fundamental to aptitude tests in campus placements. These questions test your understanding of geometric shapes, their properties, and your ability to calculate space occupied by 2D and 3D objects. This topic is extensively covered in technical and non-technical company exams alike.

Why This Topic is Important

Area and Volume questions assess:

Spatial visualization ability
Formula application skills
Unit conversion competency
Problem-solving with composite shapes
Real-world application of mathematics

Companies That Ask Area and Volume Questions

Company	Frequency	Difficulty Level
TCS	High	Easy to Moderate
Infosys	High	Easy to Moderate
Wipro	Moderate	Easy
Cognizant	Moderate	Easy
Accenture	Moderate	Easy
Capgemini	Moderate	Easy
IBM	Low	Moderate
Tech Mahindra	Moderate	Easy
HCL	Moderate	Easy
LTI Mindtree	Low	Easy

KEY FORMULAS / CONCEPTS

╔══════════════════════════════════════════════════════════════════╗
║                  AREA & VOLUME FORMULA SHEET                     ║
╠══════════════════════════════════════════════════════════════════╣
║                                                                  ║
║  2D SHAPES - AREA & PERIMETER                                   ║
║  ────────────────────────────────────────────────────────────   ║
║  Square:        Area = a², Perimeter = 4a                       ║
║  Rectangle:     Area = l × b, Perimeter = 2(l+b)                ║
║  Triangle:      Area = ½ × b × h                               ║
║                 Area = √[s(s-a)(s-b)(s-c)] (Heron's)            ║
║  Circle:        Area = πr², Circumference = 2πr                 ║
║  Semi-circle:   Area = ½πr², Perimeter = πr + 2r               ║
║  Parallelogram: Area = b × h                                   ║
║  Rhombus:       Area = ½ × d₁ × d₂                             ║
║  Trapezium:     Area = ½ × (a+b) × h                           ║
║                                                                  ║
║  3D SHAPES - VOLUME & SURFACE AREA                              ║
║  ────────────────────────────────────────────────────────────   ║
║  Cube:          Volume = a³, TSA = 6a², LSA = 4a²              ║
║  Cuboid:        Volume = l×b×h, TSA = 2(lb+bh+hl), LSA=2h(l+b) ║
║  Cylinder:      Volume = πr²h, CSA = 2πrh, TSA = 2πr(r+h)      ║
║  Cone:          Volume = ⅓πr²h, CSA = πrl, TSA = πr(r+l)       ║
║  Sphere:        Volume = 4/3 πr³, TSA = 4πr²                   ║
║  Hemisphere:    Volume = ⅔πr³, CSA = 2πr², TSA = 3πr²          ║
║                                                                  ║
║  IMPORTANT RELATIONSHIPS                                        ║
║  ────────────────────────────────────────────────────────────   ║
║  Diagonal of square = a√2                                      ║
║  Diagonal of cuboid = √(l²+b²+h²)                              ║
║  Slant height of cone: l = √(r²+h²)                            ║
║  1 m³ = 1000 liters = 1,000,000 cm³                            ║
║  1 hectare = 10,000 m² = 2.47 acres                            ║
║                                                                  ║
╚══════════════════════════════════════════════════════════════════╝

30 Practice Questions with Step-by-Step Solutions

Question 1

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

Solution: Using Heron's formula: s = (13+14+15)/2 = 21 Area = √[21(21-13)(21-14)(21-15)] = √[21 × 8 × 7 × 6] = √[7056] = 84 cm²

Question 2

The diagonal of a square is 10√2 cm. Find its area.

Solution: Diagonal = a√2 = 10√2 So, a = 10 cm Area = a² = 100 cm²

Question 3

A rectangle has length 24 m and diagonal 25 m. Find its area.

Solution: Using Pythagoras: b = √(25² - 24²) = √(625 - 576) = √49 = 7 m Area = 24 × 7 = 168 m²

Question 4

The circumference of a circle is 44 cm. Find its area.

Solution: 2πr = 44 → r = 44/(2π) = 44 × 7/(2 × 22) = 7 cm Area = πr² = (22/7) × 49 = 154 cm²

Question 5

Find the volume of a cylinder with radius 7 cm and height 10 cm.

Solution: Volume = πr²h = (22/7) × 49 × 10 = 1540 cm³

Question 6

A cone has radius 6 cm and height 8 cm. Find its volume and curved surface area.

Solution: Volume = ⅓πr²h = ⅓ × π × 36 × 8 = 96π = 301.71 cm³ l = √(r²+h²) = √(36+64) = 10 cm CSA = πrl = π × 6 × 10 = 60π = 188.57 cm²

Question 7

The surface area of a sphere is 616 cm². Find its volume.

Solution: 4πr² = 616 → r² = 616 × 7/(4 × 22) = 49 → r = 7 cm Volume = (4/3)πr³ = (4/3) × (22/7) × 343 = 4312/3 = 1437.33 cm³

Question 8

A cuboid has dimensions 12 cm × 8 cm × 6 cm. Find its diagonal and total surface area.

Solution: Diagonal = √(12²+8²+6²) = √(144+64+36) = √244 = 2√61 cm TSA = 2(12×8 + 8×6 + 6×12) = 2(96 + 48 + 72) = 2 × 216 = 432 cm²

Question 9

The perimeter of a rhombus is 40 cm and one diagonal is 12 cm. Find the area.

Solution: Side = 40/4 = 10 cm Half of d₁ = 6 cm Half of d₂ = √(10² - 6²) = √64 = 8 cm d₂ = 16 cm Area = ½ × 12 × 16 = 96 cm²

Question 10

A trapezium has parallel sides 20 cm and 30 cm, height 12 cm. Find its area.

Solution: Area = ½ × (20+30) × 12 = ½ × 50 × 12 = 300 cm²

Question 11

Water flows through a cylindrical pipe of radius 1 cm at 10 cm/s. How much water flows in 1 minute?

Solution: In 1 minute, length of water column = 10 × 60 = 600 cm Volume = πr²h = π × 1 × 600 = 600π = 1885.71 cm³ = 1.886 liters

Question 12

A cube has surface area 150 cm². Find its volume.

Solution: 6a² = 150 → a² = 25 → a = 5 cm Volume = a³ = 125 cm³

Question 13

The base of a pyramid is a square of side 10 cm and height is 12 cm. Find its volume.

Solution: Volume = ⅓ × Base Area × Height = ⅓ × 100 × 12 = 400 cm³

Question 14

Find the area of a sector with radius 14 cm and angle 45°.

Solution: Area = (θ/360) × πr² = (45/360) × (22/7) × 196 = 77 cm²

Question 15

A hemispherical bowl has radius 21 cm. Find its capacity in liters.

Solution: Volume = ⅔πr³ = ⅔ × (22/7) × 9261 = 19404 cm³ = 19.404 liters

Question 16

A metallic sphere of radius 9 cm is melted and recast into a cylinder of radius 6 cm. Find the height of the cylinder.

Solution: Volume of sphere = Volume of cylinder (4/3)π(9)³ = π(6)²h (4/3) × 729 = 36h 972 = 36h h = 27 cm

Question 17

The perimeter of a semicircle is 36 cm. Find its area.

Solution: Perimeter = πr + 2r = r(π + 2) = 36 r(22/7 + 2) = 36 r × 36/7 = 36 → r = 7 cm Area = ½πr² = ½ × (22/7) × 49 = 77 cm²

Question 18

A room is 12 m long, 8 m wide, and 5 m high. Find the cost of whitewashing walls at ₹20 per m².

Solution: Area of 4 walls = 2h(l+b) = 2 × 5 × (12+8) = 200 m² Cost = 200 × 20 = ₹4000

Question 19

A right circular cone has slant height 25 cm and base radius 7 cm. Find its curved surface area.

Solution: CSA = πrl = (22/7) × 7 × 25 = 550 cm²

Question 20

The area of a circle is increased by 44 cm² when its radius is increased by 2 cm. Find the original radius.

Solution: π(r+2)² - πr² = 44 π(r² + 4r + 4 - r²) = 44 π(4r + 4) = 44 4r + 4 = 44 × 7/22 = 14 4r = 10 r = 2.5 cm

Question 21

A rectangular field has area 720 m² and length 30 m. Find the cost of fencing at ₹15 per meter.

Solution: Width = 720/30 = 24 m Perimeter = 2(30+24) = 108 m Cost = 108 × 15 = ₹1620

Question 22

A cylindrical tank of radius 7 m and height 10 m is filled with water. How many small cylinders of radius 7 cm and height 10 cm can be filled?

Solution: Ratio of radii = 700/7 = 100 Ratio of heights = 1000/10 = 100 Number = 100² × 100 = 1,000,000

Question 23

The diagonal of a cube is 6√3 cm. Find its surface area.

Solution: Diagonal = a√3 = 6√3 → a = 6 cm Surface area = 6a² = 6 × 36 = 216 cm²

Question 24

A hollow cylinder has external radius 8 cm, internal radius 6 cm, and height 14 cm. Find the volume of material.

Solution: Volume = πh(R²-r²) = (22/7) × 14 × (64-36) = 44 × 28 = 1232 cm³

Question 25

A circular park of radius 35 m has a path 7 m wide around it. Find the area of the path.

Solution: Outer radius = 42 m Area of path = π(42² - 35²) = (22/7) × (1764 - 1225) = (22/7) × 539 = 1694 m²

Question 26

Three cubes of side 4 cm are joined end to end. Find the surface area of the resulting cuboid.

Solution: Dimensions: 12 cm × 4 cm × 4 cm TSA = 2(12×4 + 4×4 + 4×12) = 2(48 + 16 + 48) = 2 × 112 = 224 cm²

Question 27

A sphere and a cube have the same surface area. Find the ratio of their volumes.

Solution: 4πr² = 6a² → r²/a² = 6/(4π) = 3/(2π) Volume ratio = (4/3)πr³ : a³ = (4/3)π(3/2π)^(3/2) : 1 After simplification: √6 : √π or approximately 1.38 : 1

Question 28

A right prism has triangular base with sides 5 cm, 12 cm, 13 cm and height 20 cm. Find its volume.

Solution: Base is right-angled (5² + 12² = 13²) Base area = ½ × 5 × 12 = 30 cm² Volume = 30 × 20 = 600 cm³

Question 29

The volume of a hemisphere is 19404 cm³. Find its total surface area.

Solution: ⅔πr³ = 19404 r³ = 19404 × 3 × 7/(2 × 22) = 9261 r = 21 cm TSA = 3πr² = 3 × (22/7) × 441 = 4158 cm²

Question 30

A cylindrical wire of radius 1 mm is drawn from a solid sphere of radius 9 cm. Find the length of wire.

Solution: Volume of sphere = (4/3)π(9)³ = 972π cm³ Cross-section of wire = π(0.1)² = 0.01π cm² Length = 972π/0.01π = 97200 cm = 972 m

SHORTCUTS & TRICKS

Trick 1: Quick π Calculations

Use π = 22/7 when radius is multiple of 7 Use π = 3.14 for decimal answers

Trick 2: Pythagorean Triples

Remember common triples: (3,4,5), (5,12,13), (8,15,17), (7,24,25), (9,40,41)

Trick 3: Heron's Formula Shortcut

For triangle with sides a, b, c:

If s is integer, area will likely be integer
For (13,14,15), s=21, and factors align perfectly

Trick 4: Sphere to Cylinder/Cone Conversions

When a sphere is recast:

Volume remains constant
Equate (4/3)πr³ to new shape's volume formula

Trick 5: Similar Figures Area/Volume Ratio

If linear dimensions ratio is k:

Area ratio = k²
Volume ratio = k³

Trick 6: Sector Area Quick Formula

Sector Area = (θ/360) × πr² = (θ/360) × Circle Area For θ = 60°: Sector = 1/6 of circle For θ = 90°: Sector = 1/4 of circle

Trick 7: Hemisphere Volume Relationship

Volume of hemisphere = ½ × Volume of sphere with same radius

Common Mistakes to Avoid

Unit Mix-up: Always check if dimensions are in cm, m, or mixed. Convert to common unit before calculating.
Radius vs Diameter: Many questions give diameter. Remember to halve it for radius.
TSA vs CSA Confusion:
- TSA (Total Surface Area) includes all surfaces
- CSA (Curved Surface Area) excludes bases
Hemisphere TSA: Don't forget the circular base! TSA = 3πr², not 2πr².
Slant Height vs Height: For cones, use slant height (l) for CSA, vertical height (h) for volume.
Square vs Cube: Area = a², Volume = a³. Don't confuse 2D and 3D formulas.
Path Area: For circular paths, subtract inner area from outer area. Don't calculate just the ring width.

5 Frequently Asked Questions

Q1: How do I remember all the formulas for area and volume? A: Focus on understanding derivations. Most 3D formulas build from 2D: Cylinder = Circle × Height, Cone = ⅓ × Cylinder, Sphere relates to cylinder (Archimedes' discovery).

Q2: When should I use π = 22/7 vs 3.14? A: Use 22/7 when radius is multiple of 7 or when answer choices are fractions. Use 3.14 for decimal precision or when specified.

Q3: How do I approach composite shape problems? A: Break into basic shapes, calculate individually, then add/subtract as needed. Always visualize or sketch the figure.

Q4: What are the most commonly tested shapes? A: Cubes, cuboids, cylinders, cones, spheres, triangles, circles, and combinations thereof. Focus on these for maximum ROI.

Q5: How can I quickly check if my answer is reasonable? A: Do a sanity check: Volume should be positive, area should be larger than individual sides, and units should match (cm² for area, cm³ for volume).

Master these 30 questions and you'll be well-prepared for any Area and Volume question in your placement exams!