NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.2
Question 5

Question. 5

The volume of the frustum of a cone is \(\dfrac{1}{3}\pi h[r_1^2+r_2^2-r_1r_2]\), where \(h\) is vertical height of the frustum and \(r_1, r_2\) are the radii of the ends.

Answer:

false

Detailed Answer with Explanation:

Step 1: Recall the correct formula for the volume of a frustum of a cone.

It is: \(V = \dfrac{1}{3}\pi h (r_1^2 + r_2^2 + r_1r_2)\)

  • \(V\) = volume in cubic metres (m3)
  • \(h\) = vertical height in metres (m)
  • \(r_1, r_2\) = radii of the circular ends in metres (m)

Step 2: Compare the given formula with the correct one.

Given: \(\dfrac{1}{3}\pi h(r_1^2 + r_2^2 - r_1r_2)\)

Correct: \(\dfrac{1}{3}\pi h(r_1^2 + r_2^2 + r_1r_2)\)

Step 3: Notice the difference.

The given formula has a minus sign (\(-r_1r_2\)) instead of a plus sign (\(+r_1r_2\)).

Step 4: Therefore, the statement is wrong.

Final Answer: False

NCERT Exemplar Solutions Class 10 – Mathematics – CHAPTER 12: Surface Areas & Volumes – Exercise 12.2 | Detailed Answers