A cone has a height of #24 cm# and its base has a radius of #12 cm#. If the cone is horizontally cut into two segments #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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#OA = h = 24 cm, OB r = 12 cm, , AF = h_1 = 24-16 = 8 cm#

#FD = r_2 =( h_1 /h)*r = (8/ 24) * 6 = 2 cm#

#AC = l = sqrt(h^2 + r^2) = sqrt(24^2 + 12^2) = 26.8328 cm #

#AE = l_1 = sqrt(h_1^2 + r_2^2) = sqrt(8^2 + 2^2) = 8.2462 cm#

#pir^2 = pi*12^2 = 452.3893 cm^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 = #(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 cm^2**#