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NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)

Page No. 257

[Use π = 22/7, unless stated otherwise]
Q 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.
Sol. Given, upper diameter - 4 cm
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
Height of glass = 14 cm
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)

Q 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.
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
Sol. We have:
Slant height (l) = 4 cm
2πr1 = 18 cm
and 2πr2 = 6 cm
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
∴ 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.

Q 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. 
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
Sol. Here, the radius of the open side (r1) = 10 cm
The radius of the upper base (r2) = 4 cm
Slant height (l) = 15 cm
∴ Area of the material required
= [Curved surface area of the frustum] + [Area of the top end]
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)

Q 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 cm2. (Take π = 3.14)
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
Sol. We have:
r1 = 20 cm, r2 = 8 cm
and h = 16 cm
∴ Volume of the frustum
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
Now, the slant height of the given frustum
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
∴ Curved surface area
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
Area of the bottom
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
∴ Total area of metal required = 1758.4 cm2 + 200.96 cm2 = 1959.36 cm2
Cost of metal required NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)

Q 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 so obtained be drawn into a wire of diameter 1/16 cm, find the length of the wire.
Sol. Let us consider the frustum DECB of the metallic cone ABC
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)

NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
Now, the volume of the frustum DBCE
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
Let l be the length and D be the diameter of the wire drawn from the frustum. 

Since the wire is in the form of a cylinder,
∴ Volume of the wire = πr2l
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
∵ [Volume of the frustum] = [Volume of the wire]
NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)

NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)
Thus, the required length of the wire = 7964.44 m.

The document NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4) is a part of the Bank Exams Course NCERT Mathematics for Competitive Exams.
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FAQs on NCERT Solutions for Class 10 Maths Chapter 12 - Surface Areas and Volumes (Exercise 12.4)

1. How do I calculate the surface area of a cylinder?
Ans. To calculate the surface area of a cylinder, you need to find the sum of the areas of its curved surface and its two circular bases. The formula for the surface area of a cylinder is 2πrh + 2πr^2, where r is the radius of the base and h is the height of the cylinder.
2. What is the formula for finding the volume of a cone?
Ans. The formula for finding the volume of a cone is (1/3)πr^2h, where r is the radius of the circular base and h is the height of the cone. This formula represents the volume of a cone as one-third of the volume of a cylinder with the same base and height.
3. How can I determine the surface area of a sphere?
Ans. To determine the surface area of a sphere, you can use the formula 4πr^2, where r is the radius of the sphere. This formula represents the total area of the curved surface of the sphere.
4. Is there a formula to find the volume of a cuboid?
Ans. Yes, there is a formula to find the volume of a cuboid. The formula is lwh, where l represents the length, w represents the width, and h represents the height of the cuboid. By multiplying these three dimensions, you can calculate the volume of a cuboid.
5. How do I find the surface area of a cone?
Ans. To find the surface area of a cone, you need to calculate the sum of the area of the curved surface and the area of the base. The formula for the surface area of a cone is πr(r + l), where r is the radius of the base and l is the slant height of the cone. The slant height can be found using the Pythagorean theorem in a right-angled triangle formed by the height, radius, and slant height.
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