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NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.3
(This page features Exercise 12.3 (surface Areas and Volumes) from NCERT Class 10 Maths, which has been removed from the current syllabus [2025-26] but remains a useful reference.)
The NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.3 will guide you beyond basic three-dimensional objects such as cones, cubes, etc. While learning in exercise 12.3, you will be learning the basic concepts of a frustum of a cone, how it is formed, and the formula involved in the frustum of a cone. This is an important topic learned not just in mathematics but also in prominent practical disciplines such as architecture. So, let's learn this vital subject of chapter 12.
1.0Download NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.3 : Free PDF
2.0Introduction to the Frustum of a Cone
The frustum of a cone is formed when a right circular cone is sliced parallel to its base, slicing off a smaller cone from the top. The rest of the shape is the frustum, which consists of two circular ends with varying radii. You could see this shape in most day-to-day objects, like drinking tumblers, funnels, and some types of vases. The prominent features of the frustum are two circular ends, a slant height, and a certain height.
Radii of the conical ends of a frustum of a cone is represented using r1 and r2 such that r1 is the larger radius (lower part) and r2 is the smaller one (upper part). Similar to ordinary cones, the height and slant height of the frustum of a cone are represented using h and l, respectively.
3.0Key Concepts of Exercise 12.3: Overview
Exercise 12.3 is primarily concerned with the calculation of the volume, curved surface area, and the total surface area of the frustum of a cone. Let us discuss these significant formulas of a frustum, with r1 and r2 being the radius of the bottom and top and "h" being the height of the frustum, to compute these questions:
Volume of the Frustum of a Cone:
The volume is the space bounded by the boundaries of a solid. In the same way, the volume of the frustum of a cone is the area bounded by the two bases and the surface of the frustum. The Volume V can be calculated using the given formula, which is written as:
Curved Surface Area of the Frustum of a Cone
The Curved Surface Area of the Frustum of a Cone is the area bounded under the lateral sides of a frustum. The formula is denoted by ACurved and can be expressed as:
Here,
l is the frustum's slant height, i.e., the distance between two radii of the frustum in a sloping way. It is given by the formula:
Total Surface Area of the Frustum of a Cone:
The total Surface Area is the area of the solid, which is found on all the sides of the figure. Total Surface Area of the Frustum of the Cone includes the area as well as its lateral side and top and bottom of the figure. It can be represented by the use of Atotal, and it can be expressed as:
4.0Practical Application of Frustum of a Cone
The frustum of a cone is a highly prevalent idea in everyday use, such as the building of cones, cups, and mechanical components. The determination of the volume and surface areas of a frustum allows engineers and designers to determine the amount of material required and other specifications for these objects. The idea is also useful in solving most of the geometry questions in the exam.
For a better grasp and problem-solving, stress on practising NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes - Exercise 12.3 to enhance your understanding of the concepts of the frustum of a cone.
5.0NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.3: Detailed Solutions
1. A drinking glass is in the shape of a frustum of a cone of height 14 cm. The diameters of its two circular ends are 4 cm and 2 cm. Find the capacity of the glass.
Solution:
R = 2 cm, r = 1 cm, h = 14 cm
Capacity of the glass = volume of the frustum with radii of ends as 2 cm and 1 cm and height 14 cm
= (1/3) * π * h * {R² + r² + Rr}
= (1/3) * π * 14 * {(2)² + (1)² + 2(1)} cm³
= (1/3) * (22/7) * 14 * 7 cm³
= 308/3 = 102 2/3 cm³
2. The slant height of a frustum of a cone is 4 cm and the perimeters (circumference) of its circular ends are 18 cm and 6 cm. Find the curved surface area of the frustum.
Sol. We have, Slant height
2πr₁ = 18 cm and 2πr₂ = 6 cm
=> πr₁ = 18/2 = 9 cm and πr₂ = 6/2 = 3 cm
∴ Curved surface area of the frustum of the cone = π(r₁ + r₂)l = (πr₁ + πr₂)l
= (9 + 3) × 4 cm²
= 12 × 4 cm² = 48 cm².
3. A fez, the cap used by the Turks, is shaped like the frustum of a cone (see fig.). If its radius on the open side is 10 cm, radius at the upper base is 4 cm and its slant height is 15 cm, find the area of material used for making it.
Solution:
Given: R = 10 cm, r = 4 cm, l = 15 cm
Curved surface area = π × l × {R + r}
= π × 15 × {10 + 4} cm²
= (22/7) × 15 × 14 cm²
= 660 cm²
Area of the closed side:
Closed
=𝜋 r² = (22/7) × (4)² = 352/7 = 50 2/7 cm²
Total area of the material used = curved surface area of cap + area of base
= (660 + 50 2/7) cm² = 710 2/7 cm²
4. A container, opened from the top and made up of a metal sheet, is in the form of a frustum of a cone of height 16 cm with radii of its lower and upper ends as 8 cm and 20 cm, respectively. Find the cost of the milk which can completely fill the container, at the rate of Rs. 20 per litre. Also find the cost of the metal sheet used to make the container, if it costs Rs. 8 per 100 cm^2. (Take π = 3.14).
Sol.
We have: r₁ = 20 cm, r₂ = 8 cm and h = 16 cm.
Therefore, Volume of the frustum = (1/3)πh[r₁² + r₂² + r₁r₂]
= (1/3) × (314/100) × 16 [20² + 8² + 20 × 8] cm³
= (1/3) × (314/100) × 16 [400 + 64 + 160] cm³
= (1/3) × (314/100) × 16 × 624 cm³
= (314/100) × 16 × 208 cm³
= (314 × 16 × 208) / 100000 litres
Therefore, Cost of milk = ₹ 20 × (314 × 16 × 208) / 100000 litres
= ₹ 208.998 ≈ ₹ 209
Now, slant height of the given frustum l = √(h² + (r₁ - r₂)²)
= √(16² + (20 - 8)²)
= √(16² + 12²)
= √(256 + 144)
= √400 = 20 cm.
Therefore, Curved surface area = π(r₁ + r₂)l
= 3.14 × (20 + 8) × 20 cm²
= (314/100) × 28 × 20 cm²
= 1758.4 cm²
Area of the bottom = πr₂²
= (314/100) × 8 × 8 cm²
= 200.96 cm²
Therefore, Total area of metal required = CSA + area of bottom
= 1758.4 cm² + 200.96 cm²
= 1959.36 cm²
Cost of metal required = ₹ (8/100) × 1959.36 cm²
= ₹ 156.75
5. A metallic right circular cone 20 cm high and whose vertical angle is 60° is cut into two parts at the middle of its height by a plane parallel to its base. If the frustum is drawn into a wire of diameter (1/16) cm, find the length of the wire.
Solution:
R/20 = tan 30° = 1/√3, i.e., R = 20/√3 cm
r/10 = tan 30° = 1/√3, i.e., r = 10/√3 cm
h = 10 cm is the height of the frustum.
Volume of the material in the frustum ACDB
= (1/3) x π x h x (R² + r² + Rr)
= (1/3) x π x 10 x ((400/3) + (100/3) + (200/3)) cm³
= (7000/9) x π cm³
Now let us suppose a wire of diameter 1/16 cm is made of length x cm.
Then, π x (1/32)² x x = (7000/9) x π
x = (7000 x 32 x 32) / 9 cm
x = 71680/9 m
x = 7964.4 m
Therefore, required length of wire = 7964 m.
6.0Benefits of Studying NCERT Solutions Class 10 Maths Chapter 12 Surface Areas and Volumes Exercise 12.3
- Explains the surface area of a frustum of a cone clearly with proper derivations and logic.
- Detailed methods help students follow the correct approach to solve each problem.
- Increases familiarity with important questions likely to appear in board exams.
- Builds confidence by practicing a variety of problems based on the same concept.
- Lays groundwork for advanced geometry in exams like NTSE and Olympiads