NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.1
Question 15

Question.  15

Two solid hemispheres of the same base radius \(r\) are joined along their bases. The curved surface area of the new solid is

(A)

\(4\pi r^2\)

(B)

\(6\pi r^2\)

(C)

\(3\pi r^2\)

(D)

\(8\pi r^2\)

Detailed Answer with Explanation:

Step 1: Recall that a hemisphere is half of a sphere.

Step 2: Each hemisphere has a curved surface area (CSA) given by:

\( \text{CSA of one hemisphere} = 2\pi r^2 \).

Step 3: When two hemispheres of radius \(r\) are joined together along their flat bases, the flat parts disappear, and they form a complete sphere.

Step 4: The surface area of a sphere is entirely curved surface (no flat faces).

Formula: \( \text{CSA of a sphere} = 4\pi r^2 \).

Step 5: Therefore, the new solid (a sphere) has curved surface area:

\( 4\pi r^2 \).

Final Answer: Option A (\(4\pi r^2\)).

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 12: Surface Areas & Volumes – Exercise 12.1 | Detailed Answers