A cone has a height of $24 c m$ $24 cm$ and its base has a radius of $12 c m$ $12 cm$ . If the cone is horizontally cut into two segments $16 c m$ $16 cm$ from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2017-12-20T14:02:02

T S A = 1430.716

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$O A = h = 24 c m, O B r = 12 c m,, A F = h_{1} = 24 - 16 = 8 c m$ $OA = h = 24 cm, OB r = 12 cm, , AF = h_1 = 24-16 = 8 cm$

$F D = r_{2} = (\frac{h_{1}}{h}) \cdot r = (\frac{8}{24}) \cdot 6 = 2 c m$ $FD = r_2 =( h_1 /h)*r = (8/ 24) * 6 = 2 cm$

$A C = l = \sqrt{h^{2} + r^{2}} = \sqrt{24^{2} + 12^{2}} = 26.8328 c m$ $AC = l = sqrt(h^2 + r^2) = sqrt(24^2 + 12^2) = 26.8328 cm$

$A E = l_{1} = \sqrt{h_{1}^{2} + r_{2}^{2}} = \sqrt{8^{2} + 2^{2}} = 8.2462 c m$ $AE = l_1 = sqrt(h_1^2 + r_2^2) = sqrt(8^2 + 2^2) = 8.2462 cm$

$π r^{2} = π \cdot 12^{2} = 452.3893 c m^{2}$ $pir^2 = pi*12^2 = 452.3893 cm^2$

$π r_{2}^{2} = π \cdot 2^{2} = 12.5664 c m^{2}$ $pir_2^2 = pi*2^2 = 12.5664 cm^2$

#pirl= pi* 12 * 26.8328 = 1011.5727 cm^2

#pir_2l_1 = pi* 2 * 8.2462 = 51.8124 cm^2

Total surface area = $(π r^{2} + π r_{2}^{2} + π . r . l - π . r_{2} . l_{1})$ $(pir^2 + pir_2^2 + pi.r.l - pi.r_2.l_1)$

$T S A = 458.3893 + 12.5664 + 1011.5727 - 51.8124 = * 1430.716 c m^{2} *$ $T S A = 458.3893 + 12.5664 + 1011.5727 - 51.8124 = **1430.716 cm^2**$

A cone has a height of 24cm24 cm and its base has a radius of 12cm12 cm. If the cone is horizontally cut into two segments 16cm16 cm from the base, what would the surface area of the bottom segment be?