A cone has a height of #6 cm# and its base has a radius of #4 cm#. If the cone is horizontally cut into two segments #5 cm# from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2017-02-23T11:20:35

Surface area of bottom segment#=137cm^2#

In triangle ABG: #(AG)/(BG)=t/a=tan angle BrArr6/4=tan angleB=1.5#

#angleB=56°18’36’’#

In triangle ADF:#(AF)/(DF)=t/a=tan angle D#----DE parallel to BC

#angleD=56°18’36’’#

#:.a/t=cot56°18’36’’ xx 1#

multiply both sides by#t/1#

#a=0.666664964cmrArr# radius top cone

Surface area of a cone#=pirl#

Area of circle#=pir^2#

Side of cone=l #:.s/t=cosec56°18’36’’#

multiply both sides by#t/1#

#s=6 xx cosec56°18’36’’#

#s=7.211=l# side of cone

Side of top cone#=s/t=cosec56°18’36’’#

multiply both sides by#t/1#

#s=1 xx cosec56°18’36’’#

#s=1.201850646cm=l# slope side of top cone

Total SA#=pir^2+pirl#

SA#=pi xx 4^2+pi xx 4 xx 7.211#

Total SA#=140.882cm^2#

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SA ot top cone: #=pir^2+pirl#

#=pi xx .666664964^2+pi xx 0.666664964 xx 1.202#

Top SA#=3.913cm^2#

SA of bottom part#=140.882-3.913=136.968=137cm^2#