A cone has a height of #16 cm# and its base has a radius of #5 cm#. If the cone is horizontally cut into two segments #12 cm# from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2018-06-08T17:39:52

#330.31cm^2 "to the nearest 2 decimal places"#

#16/5=tan theta=72^@38'46''#

#cot72^@38'46'' xx4.0=1.25cm="r top area"#

#:."F=lateral area"=pi(r_1+r_2) sqrt ((r_1-r_2)^2+h^2)#

#:.F=pi(5+1.25) sqrt( (5-1.25)^2+12^2)#

#:.F=pi(6.25) sqrt( (3.75)^2+12^2)#

#:.F=19.63495408 sqrt( 14.0625+144)#

#:.F=19.63495408 sqrt( 158.0625)#

#:.F=19.63495408 xx 12.57229096#

#:.F=246.8563557#

#:.S=F+pi(r_1^2+r_2^2)#

#:.S=246.8563557+pi(5^2+1.25^2)#

#:.S=246.8563557+pi(25+1.5625)#

#:.S=246.8563557+pi(26.5625)#

#:.S=246.8563557+83.44855486#

#:.S=330.3049105#

#:."Total surface area o frustum or truncated cone=330.31"cm^2 "#

#"to the nearest 2 decimal places"#