A cone has a height of $8 cm$ and its base has a radius of $5 cm$ . If the cone is horizontally cut into two segments $6 cm$ from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2018-04-25T11:33:02

Total surface area of bottom segment is $222.38$ sq.cm

The cone is cut at 6 cm from base, So upper radius of the frustum

of cone is $r_2=(8-6)/8*5=1.25$ cm ; Slant ht

$l=sqrt(6^2+(5-1.25)^2)=sqrt(36+14.0625)$

$=sqrt 50.0625 ~~ 7.08$ cm

Top surface area $A_t=pi*1.25^2 ~~ 4.91$ sq.cm

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

Slant Area $A_s=pi*l*(r_1+r_2)=pi*7.08*(5+1.25)$

$~~138.93$ sq.cm

Total surface area of bottom segment

$=A_t+A_b+A_s=4.91+78.54+138.93$

$~~ 222.38$ sq.cm [Ans]

A cone has a height of 8 cm8 cm and its base has a radius of 5 cm5 cm. If the cone is horizontally cut into two segments 6 cm6 cm from the base, what would the surface area of the bottom segment be?