Mensuration Questions for Placement 2026 (with Solutions)

Last Updated: March 2026

Introduction to Mensuration

Mensuration is the branch of mathematics that deals with the measurement of geometric figures and their parameters like length, area, volume, and surface area. This topic is crucial for placement exams as it tests your spatial reasoning, formula application, and calculation skills. Nearly every company includes mensuration questions in their aptitude tests.

Why This Topic is Important

Mensuration questions assess:

Understanding of geometric properties
Ability to visualize 2D and 3D shapes
Formula recall and application
Unit conversion skills
Problem-solving with real-world scenarios

Companies That Ask Mensuration Questions (with Frequency)

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

KEY FORMULAS / CONCEPTS

╔══════════════════════════════════════════════════════════════════╗
║                    COMPLETE MENSURATION FORMULA SHEET            ║
╠══════════════════════════════════════════════════════════════════╣
║                                                                  ║
║  2D SHAPES                                                       ║
║  ────────────────────────────────────────────────────────────   ║
║  Square:                                                         ║
║    Area = a², Perimeter = 4a, Diagonal = a√2                    ║
║                                                                  ║
║  Rectangle:                                                      ║
║    Area = l × b, Perimeter = 2(l+b), Diagonal = √(l²+b²)        ║
║                                                                  ║
║  Triangle:                                                       ║
║    Area = ½ × base × height                                     ║
║    Area (Heron's) = √[s(s-a)(s-b)(s-c)] where s=(a+b+c)/2       ║
║    Area (Equilateral) = (√3/4) × a²                             ║
║                                                                  ║
║  Parallelogram:                                                  ║
║    Area = base × height, Perimeter = 2(a+b)                     ║
║                                                                  ║
║  Rhombus:                                                        ║
║    Area = ½ × d₁ × d₂, Side = ½√(d₁²+d₂²)                       ║
║                                                                  ║
║  Trapezium:                                                      ║
║    Area = ½ × (a+b) × h                                         ║
║                                                                  ║
║  Circle:                                                         ║
║    Area = πr², Circumference = 2πr, Diameter = 2r               ║
║    Arc Length = (θ/360) × 2πr                                   ║
║    Sector Area = (θ/360) × πr²                                  ║
║                                                                  ║
║  3D SHAPES                                                       ║
║  ────────────────────────────────────────────────────────────   ║
║  Cube (side = a):                                                ║
║    Volume = a³, TSA = 6a², LSA = 4a²                            ║
║    Space Diagonal = a√3                                         ║
║                                                                  ║
║  Cuboid (l × b × h):                                             ║
║    Volume = l×b×h, TSA = 2(lb+bh+hl)                            ║
║    LSA = 2h(l+b), Diagonal = √(l²+b²+h²)                        ║
║                                                                  ║
║  Cylinder (radius r, height h):                                  ║
║    Volume = πr²h, CSA = 2πrh, TSA = 2πr(r+h)                    ║
║                                                                  ║
║  Cone (radius r, height h, slant height l):                      ║
║    Volume = ⅓πr²h, CSA = πrl, TSA = πr(r+l)                     ║
║    l = √(r²+h²)                                                 ║
║                                                                  ║
║  Sphere (radius r):                                              ║
║    Volume = (4/3)πr³, Surface Area = 4πr²                       ║
║                                                                  ║
║  Hemisphere (radius r):                                          ║
║    Volume = (2/3)πr³, CSA = 2πr², TSA = 3πr²                    ║
║                                                                  ║
╚══════════════════════════════════════════════════════════════════╝

30 Practice Questions with Step-by-Step Solutions

Question 1

Find the area of a rectangle whose length and breadth are increased by 20% and 10% respectively. The original area was 200 m².

Solution: New length = 1.2L, New breadth = 1.1B New Area = 1.2 × 1.1 × LB = 1.32 × 200 = 264 m²

Question 2

The ratio of length to breadth of a rectangle is 3:2. If the perimeter is 60 cm, find the area.

Solution: Let length = 3x, breadth = 2x Perimeter = 2(3x + 2x) = 10x = 60 x = 6 Length = 18 cm, Breadth = 12 cm Area = 18 × 12 = 216 cm²

Question 3

A square and a rectangle have the same perimeter. The rectangle's sides are in ratio 2:3. Find the ratio of their areas.

Solution: Let square side = a, rectangle sides = 2x, 3x 4a = 2(2x+3x) = 10x → a = 2.5x Area of square = 6.25x² Area of rectangle = 6x² Ratio = 6.25:6 = 25:24

Question 4

The area of an equilateral triangle is 36√3 cm². Find its perimeter.

Solution: Area = (√3/4)a² = 36√3 a² = 144 → a = 12 cm Perimeter = 3 × 12 = 36 cm

Question 5

A right triangle has sides 9 cm, 12 cm, and 15 cm. Find the altitude to the hypotenuse.

Solution: Area = ½ × 9 × 12 = 54 cm² Also, Area = ½ × 15 × h = 54 h = 108/15 = 7.2 cm

Question 6

The diagonals of a rhombus are in ratio 3:4 and its area is 150 cm². Find the side length.

Solution: Let diagonals be 3x and 4x Area = ½ × 3x × 4x = 6x² = 150 x² = 25 → x = 5 Diagonals: 15 cm and 20 cm Side = ½√(15² + 20²) = ½√625 = 12.5 cm

Question 7

A circular ring has outer radius 13 cm and inner radius 12 cm. Find the area of the ring.

Solution: Area = π(R²-r²) = π(169-144) = 25π = 78.57 cm²

Question 8

The perimeter of a semicircular protractor is 36 cm. Find its diameter.

Solution: Perimeter = πr + 2r = r(π+2) = 36 r = 36/(22/7 + 2) = 36/(36/7) = 7 cm Diameter = 14 cm

Question 9

A cube has surface area 294 cm². A cuboid has same surface area with sides in ratio 1:2:3. Find the cuboid's volume.

Solution: For cube: 6a² = 294 → a² = 49 → a = 7 For cuboid: 2(lb+bh+hl) = 294 Let sides be x, 2x, 3x 2(2x² + 6x² + 3x²) = 294 22x² = 294 → x² = 49/11 ≈ Not clean, let's verify: x = 7/√11 Volume = x × 2x × 3x = 6x³ = 6 × (343/11√11) = 343/√11 ≈ 103.4 cm³

(Revised cleaner version): Sides 1:2:3 with TSA 294 2(x·2x + 2x·3x + 3x·x) = 294 2(2x² + 6x² + 3x²) = 294 22x² = 294, x² = 294/22 = 147/11 Volume = 6x³ = 6 × (147/11)^(3/2) = approx 108 cm³

Question 10

A cone, hemisphere, and cylinder stand on equal bases and have the same height. Find the ratio of their volumes.

Solution: Let radius = r, height = r (for cone and cylinder) Cone: ⅓πr³ Hemisphere: ⅔πr³ Cylinder: πr³ Ratio = ⅓ : ⅔ : 1 = 1:2:3

Question 11

Water flows at 5 km/hr through a pipe of diameter 14 cm into a tank 50 m × 44 m. Find time to raise water level by 7 cm.

Solution: Volume needed = 50 × 44 × 0.07 = 154 m³ Flow rate: Area × Speed = π(0.07)² × 5000 m/hr = π × 0.0049 × 5000 m³/hr = 24.5π m³/hr Time = 154/(24.5π) = 154/(24.5 × 22/7) = 154 × 7/(24.5 × 22) = 2 hours = 2 hours

Question 12

A sphere of diameter 12 cm is dropped into a cylindrical vessel of diameter 24 cm. Find the rise in water level.

Solution: Volume of sphere = (4/3)π(6)³ = 288π cm³ Volume displaced = π(12)² × h = 288π 144h = 288 → h = 2 cm

Question 13

The curved surface area of a cone is 308 cm² and its slant height is 14 cm. Find the radius of the base.

Solution: CSA = πrl = 308 (22/7) × r × 14 = 308 44r = 308 → r = 7 cm

Question 14

A room is 8 m long, 6 m broad, and 3.5 m high. Find the cost of papering walls at ₹20/m², allowing 12 m² for doors and windows.

Solution: Wall area = 2 × 3.5 × (8+6) = 98 m² Papering area = 98 - 12 = 86 m² Cost = 86 × 20 = ₹1720

Question 15

Find the volume of the largest sphere that can be carved from a cube of side 14 cm.

Solution: Diameter of sphere = 14 cm, radius = 7 cm Volume = (4/3)π(7)³ = (4/3) × (22/7) × 343 = 4312/3 = 1437.33 cm³

Question 16

A metallic cone of radius 6 cm and height 24 cm is melted and recast into spheres of radius 3 cm. How many spheres are formed?

Solution: Volume of cone = ⅓π(6)²(24) = 288π cm³ Volume of sphere = (4/3)π(3)³ = 36π cm³ Number = 288π/36π = 8 spheres

Question 17

The area of a sector of angle 72° is 77 cm². Find the radius of the circle.

Solution: Sector area = (72/360) × πr² = 77 (1/5) × (22/7) × r² = 77 r² = 77 × 5 × 7/22 = 122.5 r = √122.5 = 7√2.5 ≈ 11.07 cm (Or r = 7√(5/2))

Question 18

A trapezium has parallel sides 60 cm and 40 cm, and non-parallel sides 13 cm each. Find the area.

Solution: Height = √[13² - 10²] = √(169-100) = √69 cm Area = ½ × (60+40) × √69 = 50√69 = 50√69 ≈ 415.2 cm²

Question 19

A cylindrical wire of radius 0.5 mm is drawn from 44 kg of copper (density 8.8 g/cm³). Find wire length.

Solution: Volume = 44000/8.8 = 5000 cm³ Cross-section = π(0.05)² = 0.0025π cm² Length = 5000/(0.0025π) = 2,000,000/π cm = 20000/π m ≈ 6366.2 m

Question 20

The volume of a cube is 512 cm³. Find the length of its diagonal.

Solution: a³ = 512 → a = 8 cm Diagonal = a√3 = 8√3 = 8√3 ≈ 13.86 cm

Question 21

A rectangular tank 20 m × 15 m is filled with water to depth 3 m. Find work done to pump all water to height 5 m above bottom. (Water weight = 1000 kg/m³, g = 9.8 m/s²)

Solution: Volume = 20 × 15 × 3 = 900 m³ Mass = 900 × 1000 = 900,000 kg Center of mass at 1.5 m, lifted to 5 m Height lifted = 5 - 1.5 = 3.5 m Work = 900,000 × 9.8 × 3.5 = 30,870,000 J = 30.87 MJ

Question 22

The base of a prism is a regular hexagon of side 4 cm. If height is 10√3 cm, find volume.

Solution: Area of hexagon = (3√3/2) × a² = (3√3/2) × 16 = 24√3 cm² Volume = 24√3 × 10√3 = 24 × 10 × 3 = 720 cm³

Question 23

A hollow sphere has external diameter 16 cm and internal diameter 12 cm. Find volume of material.

Solution: Volume = (4/3)π(R³-r³) = (4/3)π(512-216) = (4/3)π × 296 = 1184π/3 ≈ 1239.5 cm³

Question 24

A cone and cylinder have same radius and height. If cone volume is 100 cm³, find cylinder volume.

Solution: V_cone = ⅓πr²h = 100 V_cylinder = πr²h = 3 × 100 = 300 cm³

Question 25

The perimeter of a rectangle is 68 cm and its diagonal is 26 cm. Find the area.

Solution: 2(l+b) = 68 → l+b = 34 l²+b² = 676 (l+b)² = l²+b²+2lb = 1156 676 + 2lb = 1156 2lb = 480 → lb = 240 cm²

Question 26

A cylindrical vessel 30 cm diameter contains water. A sphere of diameter 15 cm is immersed. Find water rise.

Solution: Volume displaced = (4/3)π(7.5)³ = 562.5π cm³ π(15)² × h = 562.5π 225h = 562.5 → h = 2.5 cm

Question 27

Find the area of a quadrilateral with diagonal 24 cm and perpendiculars from opposite vertices 8 cm and 12 cm.

Solution: Area = ½ × d × (p₁+p₂) = ½ × 24 × (8+12) = 12 × 20 = 240 cm²

Question 28

A frustum has radii 10 cm and 4 cm, height 8 cm. Find its volume.

Solution: V = (πh/3)(R²+Rr+r²) = (π×8/3)(100+40+16) = (8π/3) × 156 = 416π ≈ 1306.9 cm³

Question 29

The area of a circle inscribed in an equilateral triangle is 154 cm². Find triangle area.

Solution: πr² = 154 → r² = 49 → r = 7 cm For equilateral triangle: r = a/(2√3) → a = 14√3 cm Area = (√3/4) × (14√3)² = (√3/4) × 588 = 147√3 ≈ 254.6 cm²

Question 30

A metal pipe is 21 cm long with external radius 4 cm, internal radius 3 cm. Find weight if metal weighs 8 g/cm³.

Solution: Volume = πh(R²-r²) = (22/7) × 21 × (16-9) = 66 × 7 = 462 cm³ Weight = 462 × 8 = 3696 g = 3.696 kg

SHORTCUTS & TRICKS

Trick 1: Percentage Change in Area

If side increases by x%, area increases by (2x + x²/100)%
If side increases by 20%, area increases by 44%

Trick 2: Equilateral Triangle Relations

Area = (√3/4)a²
Height = (√3/2)a
Inradius = a/(2√3), Circumradius = a/√3

Trick 3: Circle Inscribed in Square

Circle diameter = Square side
Circle area/Square area = π/4 ≈ 0.785

Trick 4: Hemisphere vs Sphere

Hemisphere volume = ½ sphere volume
Hemisphere TSA = 3πr² (includes base)

Trick 5: Cylinder from Rectangle

If rectangle l × b forms cylinder: Volume = πr²h
If rolled along length: r = b/(2π), h = l

Trick 6: Quick Cone Calculations

Slant height l = √(r²+h²)
Common Pythagorean triples: (3,4,5), (6,8,10), (5,12,13)

Trick 7: Sphere in Cube

Largest sphere diameter = cube side
Volume ratio = π/6 ≈ 0.524

Trick 8: Similar Figures

Linear ratio k → Area ratio k² → Volume ratio k³

Common Mistakes to Avoid

Radius vs Diameter: Always check which is given. Many students use diameter as radius.
Slant vs Vertical Height: For cones, CSA uses slant height, volume uses vertical height.
Unit Consistency: Convert all units to the same system before calculating.
TSA vs CSA: TSA includes all faces; CSA excludes bases for cylinders/cones.
Hemisphere Base: Don't forget to include the circular base in TSA calculations.
π Approximation: Use 22/7 for multiples of 7, 3.14 for decimals.
Composite Shapes: Break complex shapes into simple components first.

5 Frequently Asked Questions

Q1: How do I approach complex mensuration problems? A: Break them into basic shapes, calculate each separately, then combine using addition/subtraction. Always draw a diagram.

Q2: Should I memorize all formulas? A: Yes, but understand the derivation. Most 3D formulas extend from 2D: multiply by height for prisms, use ⅓ for pyramids/cones.

Q3: What if I forget a formula during the exam? A: Use basic principles. For example, cone volume is ⅓ of cylinder with same base and height.

Q4: How do I handle unit conversions quickly? A: Memorize: 1 m = 100 cm, 1 m³ = 10⁶ cm³ = 1000 liters. For area: 1 m² = 10,000 cm².

Q5: Which mensuration topics are most important for placements? A: Cubes, cuboids, cylinders, cones, spheres, circles, triangles, and rectangles. Focus 80% of your time on these.

Practice these 30 questions thoroughly to ace mensuration in your placement exams!