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A cone has a height of #16 cm# and its base has a radius of #8 cm#. If the cone is horizontally cut into two segments #3 cm# from the base, what would the surface area of the bottom segment be?

Answer 1

Total surface area of bottom segment is #486.58(2dp)# sq,cm

The cone is cut at 3 cm from base, So upper radius of the frustum of cone is #r_2=(16-3)/16*8=6.5#cm ; slant ht #l=sqrt(3^2+(8-6.5)^2)=sqrt(9+2.25)=sqrt 11.25=3.354 cm#

Top surface area #A_t=pi*6.5^2=132.73# cm^2 Bottom surface area #A_b=pi*8^2=201.06#cm^2 Slant Area #A_s=pi*l*(r_1+r_2)=pi*3.354*(8+6.5)=152.79#cm^2 Total surface area of bottom segment #=A_t+A_b+A_s=132.73+201.06+153.79=486.58(2dp)#sq,cm[Ans]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.