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A solid cone of radius r and height h is placed over a solid cylinder having same base radius and height as that of a cone. The total surface area of the combined solid is πr[r2+h2+3r+2h].

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Question

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

Options

  • True

  • False

MCQ
True or False
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Solution

This statement is False.

Explanation:

When a solid cone is placed over a solid cylinder of same base radius, the base of cone and top of the cylinder will not be covered in total surface area.

Since the height of cone and cylinder is same,

We get,

Total surface area of cone = πrl + πr2, where r = base radius and l = slant height

Total surface area of shape formed = Total surface area of cone + Total surface area of cylinder – 2(Area of base)

Total surface area of cylinder = 2πrh + 2πr2h, where r = base radius and h = height

= πr(r + l) + (2πrh + 2πr2) – 2(πr2)

= πr2 + πrl + 2πrh + 2πr2 – 2πr2

= πr(r + l + h)

= `pi"r"("r" + sqrt("r"^2 + "h"^2) + 2"h")`

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Chapter 12: Surface Areas and Volumes - Exercise 12.2 [Page 142]

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NCERT Exemplar Mathematics Exemplar [English] Class 10
Chapter 12 Surface Areas and Volumes
Exercise 12.2 | Q 3 | Page 142
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