NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.3
Question 7

Question. 7

Two cones with same base radius 8 cm and height 15 cm are joined together along their bases. Find the surface area of the shape so formed.

Answer:

\(272\pi\;\text{cm}^2\)

Detailed Answer with Explanation:

Step 1: Write down the known values.

  • Radius of base, \(r = 8\;\text{cm}\)
  • Height of cone, \(h = 15\;\text{cm}\)

Step 2: Find the slant height \(l\) of the cone using Pythagoras theorem:

\[ l = \sqrt{r^2 + h^2} = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17\;\text{cm} \]

Step 3: Since the two cones are joined along their bases, the base areas will not be visible. So, the surface area of the new solid = sum of curved surface areas of the two cones.

Step 4: Formula for curved surface area (CSA) of one cone:

\[ \text{CSA} = \pi r l \]

Substitute values for one cone:

\[ \pi r l = \pi \times 8 \times 17 = 136\pi\;\text{cm}^2 \]

Step 5: For two cones, total surface area:

\[ 2 \times 136\pi = 272\pi\;\text{cm}^2 \]

Final Answer: The surface area of the solid is \(272\pi\;\text{cm}^2\).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 12: Surface Areas & Volumes – Exercise 12.3 | Detailed Answers