A cone has a height of $11 cm$ and its base has a radius of $5 cm$ . If the cone is horizontally cut into two segments $2 cm$ from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2018-01-06T12:25:07

Total surface area of bottom segment is $193.86$ sq.cm.

The cone is cut at 2 cm from base, So upper radius of the frustum

of cone is $r_2=(11-2)/11*5~~4.09$ cm . Slant height is

$l=sqrt(2^2+(5-4.09)^2)=sqrt(4+0.83)=sqrt 4.83 ~~2.2$ cm.

Top surface area $A_t=pi*4.09^2~~52.57$ sq.cm

Bottom surface area $A_b=pi*5^2 ~~78.54$ sq.cm

Slant Area $A_s=pi*l*(r_1+r_2)=pi*2.2*(5+4.09)~~62.74$

sq.cm. Total surface area of bottom segment is

$A_t+A_b+A_s=52.57+78.54+62.74 ~~193.86 (2dp)$

sq.cm [Ans]

A cone has a height of 11 cm11 cm and its base has a radius of 5 cm5 cm. If the cone is horizontally cut into two segments 2 cm2 cm from the base, what would the surface area of the bottom segment be?