A cone has a height of $4 c m$ $4 cm$ and its base has a radius of $6 c m$ $6 cm$ . If the cone is horizontally cut into two segments $2 c m$ $2 cm$ from the base, what would the surface area of the bottom segment be?

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A cone has a height of $4 c m$ $4 cm$ and its base has a radius of $6 c m$ $6 cm$ . If the cone is horizontally cut into two segments $2 c m$ $2 cm$ from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2017-10-18T20:14:04

Total surface area of the bottom segment = 69.1382

Explanation:

Refer diagram below :
$r = 3, r_{2} = \frac{3}{2}, h = 4, h_{1} = 2,$ $r = 3, r_2 = 3/2, h = 4, h_1 = 2,$
$l = \sqrt{h^{2} + r^{2}} = \sqrt{4^{2} + 3^{2}} = 5$ $l = sqrt(h^2+r^2) = sqrt(4^2+3^2) = 5$
$l_{1} = \sqrt{h_{1}^{2} + r_{2}^{2}} = \sqrt{2^{2} + 2^{2}} = 2 \sqrt{2}$ $l_1 = sqrt(h_1^2 + r_2^2) = sqrt(2^2+2^2) = 2sqrt2$

Total surface area of the bottom segment
$= π r^{2} + π r_{2}^{2} + π r l - π r_{2} l_{1}$ $= pir^2 + pir_2^2 + pirl - pir_2l_1$
$= π 3^{2} + π {(\frac{3}{2})}^{2} + π \cdot 3 \cdot 5 - π \cdot (\frac{3}{2}) \cdot 2 \sqrt{2}$ $=pi3^2 + pi(3/2)^2 + pi*3*5 -pi*(3/2)*2sqrt2$
$= 9 π + (\frac{9}{4}) π + 15 π - 3 \sqrt{2} \cdot π$ $= 9pi + (9/4)pi + 15pi - 3sqrt2*pi$
$69.1382 c m^{2}$ $69.1382 cm^2$

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A cone has a height of $4 c m$ $4 cm$ and its base has a radius of $6 c m$ $6 cm$ . If the cone is horizontally cut into two segments $2 c m$ $2 cm$ from the base, what would the surface area of the bottom segment be?

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Explanation:

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A cone has a height of $4 c m$ $4 cm$ and its base has a radius of $6 c m$ $6 cm$ . If the cone is horizontally cut into two segments $2 c m$ $2 cm$ from the base, what would the surface area of the bottom segment be?