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Maharashtra State BoardSSC (English Medium) 10th Standard

The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its total surface area.

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Question

The radii of ends of a frustum are 14 cm and 6 cm respectively and its height is 6 cm. Find its total surface area.

Sum
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Solution

Here, r1 = 14 cm, r2 = 6 cm and h = 6 cm.
Slant height of the frustum,

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

= \[\sqrt{6^2 + \left( 14 - 6 \right)^2}\]

= \[\sqrt{6^2 + 8^2}\]

= \[\sqrt{36 + 64}\]

= \[\sqrt{100}\]

= 10 cm 

Total surface area of frustrum

\[=\pi l(r_{1}+r_{2})+\pi r_{1}{}^{2}+\pi r_{2}{}^{2}\]

\[ = 3 . 14 \times \left( 14 + 6 \right) \times 10 + 3 . 14 \times {14}^2 + 3 . 14 \times 6^2 \]

\[ = 3 . 14 \times 20 \times 10 + 3 . 14 \times 196 + 3 . 14 \times 36\]

\[ = 628 + 615 . 44 + 113 . 04\]

\[ = 1356 . 48 { cm}^2\]

∴ The total surface area of the frustum is 1356.48 cm2.

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Chapter 7: Mensuration - Practice set 7.2 [Page 148]

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Balbharati Geometry Mathematics 2 [English] Standard 10 Maharashtra State Board
Chapter 7 Mensuration
Practice set 7.2 | Q 2.2 | Page 148

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