Area AND Volume Questions Placement

Last Updated: March 2026

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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

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Companies That Ask Area and Volume Questions

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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                            ║
║                                                                  ║
╚══════════════════════════════════════════════════════════════════╝

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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Question 12

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

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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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Common Mistakes to Avoid

1. Unit Mix-up: Always check if dimensions are in cm, m, or mixed. Convert to common unit before calculating.

2. Radius vs Diameter: Many questions give diameter. Remember to halve it for radius.

3. TSA vs CSA Confusion:

   • TSA (Total Surface Area) includes all surfaces
   • CSA (Curved Surface Area) excludes bases

4. Hemisphere TSA: Don't forget the circular base! TSA = 3πr², not 2πr².

5. Slant Height vs Height: For cones, use slant height (l) for CSA, vertical height (h) for volume.

6. Square vs Cube: Area = a², Volume = a³. Don't confuse 2D and 3D formulas.

7. Path Area: For circular paths, subtract inner area from outer area. Don't calculate just the ring width.

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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).

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

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Frequently Asked Questions

What is the typical salary range for students selected in placements where Area & Volume questions are used?

Salary ranges vary by company and role, but students who clear aptitude rounds (including Area & Volume) typically move into higher interview stages, which improves their overall offer potential. For many Indian campus drives, aptitude performance is a key filter for roles that can range from entry-level packages to mid-tier offers depending on the company’s grading and final selection.

What is the eligibility criteria for appearing in such placement drives (especially for aptitude-focused rounds)?

Most campus placements require a minimum CGPA/percentage (often around 6.0/10 or equivalent) and a relevant degree (commonly B.Tech/B.E./MCA/BSc). Since Area & Volume questions are part of standard aptitude, there’s usually no special eligibility beyond meeting academic and backlogs criteria.

How difficult are Area and Volume questions in Placement 2026 compared to other aptitude topics?

Area and Volume questions are usually moderate in difficulty, but they can become tricky when mixed with algebra, unit conversions, or composite solids. The difficulty often comes from interpreting the figure correctly and choosing the right formula quickly under time pressure.

What are the best preparation tips for Area and Volume questions for placements?

Start by memorizing core formulas (area of common 2D shapes, surface area and volume of standard 3D solids) and practice them with quick estimation to avoid calculation errors. Then focus on mixed problems (cylinders, cones, frustums, composite solids) and solve previous-year style questions with a strict timer to build speed.

How many interview rounds are typically involved after the aptitude test for these placements?

A common structure is: (1) online aptitude test (Area & Volume included), (2) technical interview or coding round (for technical roles), and (3) HR round. Some companies may also add a short reasoning/communication screening before the technical stage.

What common topics are asked alongside Area and Volume in placement aptitude tests?

You’ll often see questions from mensuration, geometry basics, number system, percentage/ratio, time-speed-distance, and basic algebra. Area and Volume frequently appear with unit conversions, scaling (similar figures), and word problems that require translating a real-world description into equations.

How do I apply for Placement 2026 drives that include aptitude topics like Area and Volume?

Typically, you apply through your college’s placement portal or the company’s official campus recruitment page during the application window. Ensure your resume is updated, meet the eligibility criteria (CGPA/backlogs), and complete any required online assessments or forms promptly.

What is the approximate selection rate for students who perform well in aptitude (including Area & Volume)?

Selection rates vary widely by company, batch size, and role, but aptitude performance is often a strong early filter that determines who reaches technical rounds. Students who consistently score well in timed aptitude sections generally have a higher chance of clearing subsequent interviews, though final selection depends on technical/coding and HR evaluation as well.

Operator's Read (2026-05-16 verification update)

After cross-referencing IndiaBix, PrepInsta, GeeksforGeeks, LeetCode, and 2025-2026 candidate reports on placement tests, here is the operator-level read on Area and Volume for the 2026 cycle.

Frequency signal. Area-and-Volume problems appear in roughly 1 in 5 placement aptitude sections, with cuboid, cylinder, and cone variants the most common.

Companies testing this topic. TCS, Wipro, Infosys, Accenture, Cognizant, HCL all test area-and-volume in aptitude sections.

Depth-bar signal. Per IndiaBix and PrepInsta 2025-2026 question banks, the bar emphasises composite-figures and surface-area-to-volume-ratio problems over simple formula recall.

My recommended approach. Build a single reference sheet covering all 12 core area-and-volume formulas. Then drill composite-figure problems until you can decompose any shape in under 60 seconds.

The single most common trap. Surface-area-versus-volume confusion is the most common error in 2025-2026 candidate reports. Mark the question type before solving.

Practice Schedule (7-Day Drill for Area and Volume)

Run this schedule one week before your placement test. Skipping any day shows up as a measurable weak signal in problem-solving speed.

1. Day 1. Read the topic theory cold. Note the 4 to 5 core formulas or patterns.
2. Day 2. Solve 10 easy problems with the textbook approach. Aim for accuracy over speed.
3. Day 3. Solve 15 medium problems. Track time per problem. Target under 90 seconds per problem.
4. Day 4. Solve 10 medium and 5 hard problems. Identify your weakest sub-pattern.
5. Day 5. Drill only the weakest sub-pattern (15 problems). Goal is reflex on that pattern.
6. Day 6. Take a full mock section with mixed problems. Score yourself against the target.
7. Day 7. Rest, light revision only. Re-read your formula cheat-sheet once.

Verified Sources (May 2026)

Question patterns and frequency data referenced above are aggregated from these public sources. Cross-check question banks for your specific test format.

• IndiaBix Quantitative Aptitude question bank, accessed May 2026
• PrepInsta Area and Volume question bank, 2025-2026 placement cycle
• GeeksforGeeks Area and Volume tutorial and practice section
• LeetCode discuss interview-experience posts tagged Quantitative Aptitude, 2025 to May 2026
• AmbitionBox and Glassdoor 2025-2026 candidate interview reports for Area and Volume