NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.1
Question 13

Question.  13

The radii of the top and bottom of a bucket (frustum) are 28 cm and 7 cm, and its slant height is 45 cm. The curved surface area is

(A)

4950 cm²

(B)

4951 cm²

(C)

4952 cm²

(D)

4953 cm²

Detailed Answer with Explanation:

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.

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