NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.2
Question 7

Question. 7

The curved surface area of a frustum of a cone is \(\pi l(r_1+r_2)\), where \(l=\sqrt{h^2+(r_1+r_2)^2}\), \(r_1,r_2\) are radii and \(h\) is height.

Answer:

false

Detailed Answer with Explanation:

Step 1: A frustum of a cone is formed when a small cone is cut off from a bigger cone by a plane parallel to its base.

Step 2: The formula for the curved surface area (CSA) of a frustum is:

\[ CSA = \pi l (r_1 + r_2) \]

where:

  • \(r_1\) = radius of the larger circular base,
  • \(r_2\) = radius of the smaller circular base,
  • \(l\) = slant height of the frustum.

Step 3: The slant height \(l\) is found using Pythagoras theorem in the right-angled triangle formed by height \(h\), difference of radii \((r_1 - r_2)\), and slant height \(l\).

So,

\[ l = \sqrt{h^2 + (r_1 - r_2)^2} \]

Step 4: But in the given statement, \(l\) was written as:

\[ l = \sqrt{h^2 + (r_1 + r_2)^2} \]

This is wrong, because we must use the difference of radii \((r_1 - r_2)\), not the sum.

Step 5: Therefore, the statement is False.

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