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

Question 13

Step 1: Recall the formula for the curved surface area (CSA) of a frustum of a cone:

\[ CSA = \pi (R + r) l \]

where:

• \(R\) = radius of the bigger circular end (top)
• \(r\) = radius of the smaller circular end (bottom)
• \(l\) = slant height of the frustum

Step 2: Substitute the given values:

\(R = 28\,\text{cm},\; r = 7\,\text{cm},\; l = 45\,\text{cm}\)

Step 3: Add the radii:

\(R + r = 28 + 7 = 35\)

Step 4: Multiply with slant height:

\((R + r) \times l = 35 \times 45 = 1575\)

Step 5: Multiply with \(\pi\) (take \(\pi = 3.1416\)):

\(CSA = 1575 \times 3.1416 \approx 4949.98\,\text{cm}^2\)

Step 6: Round off to the nearest whole number:

\(CSA \approx 4950\,\text{cm}^2\)

Final Answer: The curved surface area is 4950 cm².