Table of Contents
Exercise 10.1 Solutions
Class X Mathematics textbook by the State Council of Educational Research and Training, Telangana, Hyderabad
Problem 1
A joker’s cap is in the form of right circular cone whose base radius is 7cm and height is 24 cm. Find the area of the sheet required to make 10 such caps.
Solution:
Given: Radius (r) = 7 cm, Height (h) = 24 cm
First find slant height (l):
\( l = \sqrt{r^2 + h^2} = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \, \text{cm} \)
Curved surface area of one cone = πrl = \(\frac{22}{7} × 7 × 25 = 550 \, \text{cm}^2\)
For 10 caps: \( 10 × 550 = 5500 \, \text{cm}^2 \)
Problem 2
A sports company was ordered to prepare 100 paper cylinders for packing shuttle cocks. The required dimensions of the cylinder are 35 cm length/height and its radius is 7 cm. Find the required area of thick paper sheet needed to make 100 cylinders?
Solution:
Given: Radius (r) = 7 cm, Height (h) = 35 cm
Curved surface area of one cylinder = 2πrh = \( 2 × \frac{22}{7} × 7 × 35 = 1540 \, \text{cm}^2 \)
Total area for 100 cylinders = \( 100 × 1540 = 154000 \, \text{cm}^2 \)
Add base area if needed (not specified in problem):
Base area = πr² = \(\frac{22}{7} × 49 = 154 \, \text{cm}^2\)
Total area with one base = \( 154000 + (100 × 154) = 169400 \, \text{cm}^2 \)
Problem 3
Find the volume of right circular cone with radius 6 cm and height 7 cm.
Solution:
Given: Radius (r) = 6 cm, Height (h) = 7 cm
Volume = \(\frac{1}{3}πr^2h = \frac{1}{3} × \frac{22}{7} × 6^2 × 7 = 264 \, \text{cm}^3 \)
Problem 4
The lateral surface area of a cylinder is equal to the curved surface area of a cone. If their bases are the same, find the ratio of the height of the cylinder to the slant height of the cone.
Solution:
Let common radius = r, cylinder height = h, cone slant height = l
Given: Lateral surface area of cylinder = Curved surface area of cone
\( 2πrh = πrl \) ⇒ \( 2h = l \) ⇒ \( \frac{h}{l} = \frac{1}{2} \)
Thus, ratio is 1:2
Problem 5
A self help group wants to manufacture joker’s caps of 3 cm radius and 4 cm height. If the available paper sheet is 1000 cm², then how many caps can be manufactured from that paper sheet?
Solution:
Given: Radius (r) = 3 cm, Height (h) = 4 cm
Slant height (l) = \( \sqrt{3^2 + 4^2} = 5 \, \text{cm} \)
Curved surface area per cap = πrl = \(\frac{22}{7} × 3 × 5 ≈ 47.14 \, \text{cm}^2 \)
Number of caps = \( \frac{1000}{47.14} ≈ 21.21 \)
Since we can’t make a fraction of a cap, maximum 21 caps can be made.
Problem 6
A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio of 3:1.
Solution:
Let common radius = r, common height = h
Volume of cylinder = πr²h
Volume of cone = \(\frac{1}{3}πr²h\)
Ratio = \( \frac{\text{Volume of cylinder}}{\text{Volume of cone}} = \frac{πr²h}{\frac{1}{3}πr²h} = 3 \)
Thus, the ratio is 3:1
Problem 7
The shape of solid iron rod is cylindrical. Its height is 11 cm and base diameter is 7 cm. Then find the total volume of 50 such rods.
Solution:
Given: Diameter = 7 cm ⇒ Radius (r) = 3.5 cm, Height (h) = 11 cm
Volume of one rod = πr²h = \(\frac{22}{7} × (3.5)^2 × 11 = 423.5 \, \text{cm}^3 \)
Total volume for 50 rods = \( 50 × 423.5 = 21175 \, \text{cm}^3 \)
Problem 8
A heap of rice is in the form of a cone of diameter 12 m and height 8 m. Find its volume. How much canvas cloth is required to cover the heap? (Use π = 3.14)
Solution:
Given: Diameter = 12 m ⇒ Radius (r) = 6 m, Height (h) = 8 m
Volume = \(\frac{1}{3}πr²h = \frac{1}{3} × 3.14 × 6^2 × 8 = 301.44 \, \text{m}^3 \)
For canvas cloth (curved surface area):
Slant height (l) = \( \sqrt{6^2 + 8^2} = 10 \, \text{m} \)
Area = πrl = 3.14 × 6 × 10 = 188.4 m²
Problem 9
The curved surface area of a cone is \(4070 \, \text{cm}^2\) and its diameter is \(70 \, \text{cm}\). What is its slant height?
Solution:
Given: Curved surface area = 4070 cm², Diameter = 70 cm ⇒ Radius (r) = 35 cm
Curved surface area = πrl ⇒ \( \frac{22}{7} × 35 × l = 4070 \)
\( 110 × l = 4070 \) ⇒ \( l = \frac{4070}{110} = 37 \, \text{cm} \)
