# 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 - Geometry

Sum

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

#### 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

$= \pi\left( r_1 + r_2 \right)l + \pi r_1^2 + \pi r_2^2$
$= 3 . 14 \times \left( 14 + 6 \right) \times 10 + 3 . 14 \times {14}^2 + 3 . 14 \times 6^2$
$= 3 . 14 \times 20 \times 10 + 3 . 14 \times 196 + 3 . 14 \times 36$
$= 628 + 615 . 44 + 113 . 04$
$= 1356 . 48 { cm}^2$

Concept: Frustum of a Cone
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#### APPEARS IN

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