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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.
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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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