Mensuration (2D & 3D) – Solved Problems
Mensuration deals with the measurement of geometric figures such as area, perimeter, surface area, and volume. This page covers 2D and 3D mensuration solved problems commonly asked in school exams, competitive exams, and aptitude tests.
Problem 1: Area of a Rectangle
The length and breadth of a rectangle are 12 cm and 8 cm respectively. Find its area and perimeter.
Solution:
Area = 12 × 8 = 96 cm² Perimeter = 2 × (12 + 8) = 40 cm
Answer: Area = 96 cm², Perimeter = 40 cm
Problem 2: Area of a Circle
Find the area of a circle whose radius is 7 cm. (Take π = 22/7)
Solution:
Area = (22/7) × 7 × 7 = 154 cm²
Answer: Area = 154 cm²
Problem 3: Area of a Triangle
The base and height of a triangle are 10 cm and 6 cm respectively. Find its area.
Solution:
Area = ½ × 10 × 6 = 30 cm²
Answer: Area = 30 cm²
Problem 4: Curved Surface Area of a Cylinder
Find the curved surface area of a cylinder with radius 7 cm and height 10 cm.
Solution:
CSA = 2 × (22/7) × 7 × 10 = 440 cm²
Answer: Curved Surface Area = 440 cm²
Problem 5: Volume of a Cube
Find the volume of a cube whose edge length is 5 cm.
Solution:
Volume = 5³ = 125 cm³
Answer: Volume = 125 cm³
Problem 6: Total Surface Area of a Cuboid
Find the total surface area of a cuboid whose dimensions are 8 cm, 5 cm, and 4 cm.
Solution:
TSA = 2(8×5 + 5×4 + 4×8) = 2(40 + 20 + 32) = 184 cm²
Answer: TSA = 184 cm²
Problem 7: Volume of a Cone
Find the volume of a cone with radius 3.5 cm and height 6 cm.
Solution:
Volume = (1/3) × (22/7) × 3.5 × 3.5 × 6 = 77 cm³
Answer: Volume = 77 cm³
Problem 8: Volume of a Sphere
Find the volume of a sphere whose radius is 7 cm.
Solution:
Volume = (4/3) × (22/7) × 7³ = 1437.33 cm³ (approx)
Answer: Volume ≈ 1437.33 cm³
Conclusion: Mensuration problems require a clear understanding of formulas and correct substitution of values. Regular practice of 2D and 3D problems strengthens spatial understanding and exam confidence.