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

Question 3

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.