Question

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?

Answer 1

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

Explanation:

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`

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`