A cone has a height of $18 cm$ and its base has a radius of $6 cm$ . If the cone is horizontally cut into two segments $3 cm$ from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2016-04-09T14:39:59

200.9cm^2

self drawn
For Original larger cone

The radius of the cone = $R=6cm$

The height of the cone = $H=18cm$

The slant height of the cone

= $L=sqrt(H^2+R^2)=sqrt(18^2+6^2)=6sqrt10cm$

The surface area of larger cone

$A_b=pixxRxxL=pixx6xx6sqrt10=36sqrt10picm^2$

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For the smaller cone (cut off)

The height of the smaller cone = $h=(18-3)cm=15cm$

If the radius of the smaller cone be r ,then

$r/h=R/H=>r=h/HxxR=15/18xx6=5cm$

The slant height of the smaller cone

$=l=sqrt(h^2+r^2)=sqrt(15^2+5^2)=5sqrt10cm$

The Curved surface area of smaller cone

$A_s=pixxrxxl=pixx5xx5sqrt10=25sqrt10picm^2$

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The Curved surface area of the bottom segment

$A_b-A_s=(36sqrt10pi-25sqrt10pi)=11sqrt10picm^2=109.3cm^2$

The total area of two circular planes of the bottom portion is
$A_c=pi(r^2+R^2)=pixx(5^2+6^2)=191.6cm^2$

The Total surface area of the bottom segment
$A_"Total"=(109.3+191.6)cm^2=200.9cm^2$

A cone has a height of 18 cm18 cm and its base has a radius of 6 cm6 cm. If the cone is horizontally cut into two segments 3 cm3 cm from the base, what would the surface area of the bottom segment be?