NCERT Class 10 Maths · Chapter 12
NCERT Solutions Class 10 Maths Chapter 12 — Surface Areas & Volumes
Step-by-step solutions for all exercises in NCERT Class 10 Maths Surface Areas & Volumes.
Chapter Overview
Find surface area and volume of combinations of solids and conversion between solids.
This chapter is part of the NCERT Mathematics textbook for Class 10 and is important for CBSE school examinations. The concepts covered here build the foundation for more advanced topics in higher classes.
Below you will find solved examples from this chapter. Each solution includes detailed step-by-step working so you can understand the method, not just the answer.
Solved Examples from Surface Areas & Volumes
1A toy is in the shape of a cone mounted on a hemisphere of the same base radius. Which expression correctly represents the total surface area of the toy?
Answer: CSA of cone + CSA of hemisphere
Solution:
Step 1: The total surface area of a combined solid is the sum of the curved (or exposed) surface areas of its individual components.
Step 2: When a cone is mounted on a hemisphere, the circular base of the cone and the circular top of the hemisphere are joined together, becoming internal surfaces.
Step 3: Therefore, the exposed surface area consists only of the curved surface area of the cone and the curved surface area of the hemisphere.
2A solid is formed by placing a cylinder on top of another larger cylinder. To find the total volume of this combined solid, what approach should be used?
Answer: Add the Volumes of both cylinders.
Solution:
Step 1: Volume is a measure of the three-dimensional space occupied by an object.
Step 2: When two or more solids are combined to form a new solid, the total volume of the resulting solid is the sum of the volumes of the individual solids, assuming there are no hollow spaces or overlaps beyond their joining surfaces.
Step 3: Surface area, on the other hand, measures the area of the exterior surfaces and is not directly additive in the same way.
3Ravi is calculating the total surface area of a cubical block surmounted by a hemisphere. He uses the formula: Total Surface Area = (Total Surface Area of Cube) + (Curved Surface Area of Hemisphere). What mistake, if any, did Ravi make?
Answer: He should have subtracted the area of the base of the hemisphere from the surface area of the cube.
Solution:
Step 1: The cubical block has 6 faces. When a hemisphere is placed on top of one face, a circular area on that face is covered by the base of the hemisphere.
Step 2: This covered area is no longer part of the exposed surface and must be excluded from the cube's total surface area.
Step 3: Therefore, the correct approach is: (TSA of Cube) - (Area of circular base of hemisphere) + (CSA of Hemisphere).
4A solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is 6 cm and the diameter of the base is 4 cm. What is the volume of the toy? (Use π = 22/7)
Answer: 41.90 cm³
Solution:
Step 1: Given diameter = 4 cm, so radius (r) = 4/2 = 2 cm. Height of cone (h) = 6 cm.
Step 2: Volume of cone = (1/3) × π × r² × h = (1/3) × π × (2)² × 6 = (1/3) × π × 4 × 6 = 8π cm³.
Step 3: Volume of hemisphere = (2/3) × π × r³ = (2/3) × π × (2)³ = (2/3) × π × 8 = (16/3)π cm³.
Step 4: Total Volume = Volume of cone + Volume of hemisphere = 8π + (16/3)π = (24/3)π + (16/3)π = (40/3)π cm³ ≈ (40/3) × (22/7) = 880/21 ≈ 41.90 cm³.
5A storage tank is in the shape of a cylinder with two hemispherical ends. The total length of the tank is 13.5 m and the diameter is 7 m. Find the total surface area of the tank. (Use π = 22/7)
Answer: 297 m²
Solution:
Step 1: Diameter = 7 m, so radius (r) = 7/2 m.
Step 2: Total length of the tank = 13.5 m. The height of the cylindrical part (h) = Total length - 2 × r = 13.5 - 2 × (7/2) = 13.5 - 7 = 6.5 m.
Step 3: Total Surface Area = CSA of cylinder + 2 × CSA of hemisphere.
Step 4: TSA = 2πrh + 2 × (2πr²) = 2πr(h + 2r) = 2 × (22/7) × (7/2) × (6.5 + 2 × 7/2) = 22 × (6.5 + 7) = 22 × 13.5 = 297 m².
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