NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 12: Surface Areas & Volumes - Exercise 12.2
Question 3

Question. 3

A solid cone of radius \(r\) and height \(h\) is placed over a solid cylinder having same base radius and height as that of the cone. The total surface area of the combined solid is \(\pi r[\sqrt{r^2+h^2}+3r+2h]\).

Answer:

false

Detailed Answer with Explanation:

Step 1: Write down the dimensions in SI units.

  • Radius of both cone and cylinder = \(r\) metres (m).
  • Height of cone and cylinder = \(h\) metres (m).

Step 2: Break the solid into parts.

  • One cone (on top).
  • One cylinder (at bottom).

Step 3: Formula for curved surface area (CSA).

  • CSA of cone = \(\pi r l\), where \(l = \sqrt{r^2+h^2}\) (slant height in metres).
  • CSA of cylinder = \(2\pi r h\).

Step 4: Consider the bases.

  • The top base of cylinder is covered by the cone, so it is not visible.
  • We only need the bottom base of cylinder (circle of radius \(r\)).
  • Area of bottom base = \(\pi r^2\).

Step 5: Add them up to get total surface area (TSA).

TSA = CSA of cone + CSA of cylinder + base area of cylinder

= \(\pi r l + 2\pi r h + \pi r^2\) (square metres).

Step 6: Compare with given expression.

Given = \(\pi r[\sqrt{r^2+h^2}+3r+2h]\).

Our result = \(\pi r[\sqrt{r^2+h^2} + 2h + r]\).

These are not the same. The given expression has extra terms.

Final Answer: The statement is false.

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