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A cone has a height of #9 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

Aurora Ayala

Total surface area of bottom segment is #458.55(2dp)# sq.cm

The cone is cut at 3 cm from base, So upper radius of the frustum of cone is #r_2=(9-3)/9*8=16/3#cm ;

Slant ht #l=sqrt(3^2+(8-16/3)^2)=sqrt(9+64/9)=sqrt (145/9)=4.01#cm

Top surface area #A_t=pi*(16/3)^2=89.36 # sq.cm

Bottom surface area #A_b=pi*8^2=201.06# sq.cm

Slant Area #A_s=pi*l*(r_1+r_2)=pi*4.01*(8+16/3)= 168.13 # sq.cm

Total surface area of bottom segment #=A_t+A_b+A_s=89.36+201.06+168.13=458.55(2dp)#sq.cm [Ans]

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Answer 2

Sadie Adams

The surface area of the bottom segment of the cone would be approximately ( 179.19 , \text{cm}^2 ).

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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.