NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.4

Question 14

Step 1: Write down the given dimensions

• Radius of both cylinder and cone, \(r = 3\,\text{cm}\)
• Height of cylinder, \(h_{\rm cyl} = 12\,\text{cm}\)
• Slant height of cone, \(l = 5\,\text{cm}\)

Step 2: Find the vertical height of the cone

The cone’s vertical height \(h_{\rm cone}\) can be found using Pythagoras theorem:

\[ h_{\rm cone} = \sqrt{l^2 - r^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4\,\text{cm}. \]

Step 3: Total Surface Area (TSA)

TSA includes:

• Curved surface area (CSA) of cylinder = \(2\pi r h_{\rm cyl}\)
• Curved surface area of cone = \(\pi r l\)
• Base of cylinder (only one base, since top is attached to cone) = \(\pi r^2\)

So,

\[ TSA = 2\pi r h_{\rm cyl} + \pi r^2 + \pi r l \]

Substitute values:

\[ TSA = 2 \times 3.14 \times 3 \times 12 + 3.14 \times 3^2 + 3.14 \times 3 \times 5 \]

\[ = 226.08 + 28.26 + 47.1 = 301.44\,\text{cm}^2 \]

Step 4: Volume

Total volume = Volume of cylinder + Volume of cone

\[ V = \pi r^2 h_{\rm cyl} + \tfrac{1}{3}\pi r^2 h_{\rm cone} \]

Substitute values:

\[ V = 3.14 \times 3^2 \times 12 + \tfrac{1}{3} \times 3.14 \times 3^2 \times 4 \]

\[ = 339.12 + 37.68 = 376.8\,\text{cm}^3 \]

Final Answer:

Total Surface Area = 301.44 cm²

Volume = 376.8 cm³