A cone has a height of 14 cm and its base has a radius of 7 cm. If the cone is horizontally cut into two segments 5 cm from the base, what would the surface area of the bottom segment be?

1 Answer
Oct 11, 2017

Total surface area of the bottom segment = 419.6876 sq. cm.

Explanation:

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(r1)/(h1)=(r2)/(h2)
7/14=(r2)/9
r2=9/2

l1=sqrt((r1)^2+(h1)^2)=sqrt(7^2+14^2)=7sqrt5
Lateral surface area of bigger one =(pi)*r1*l1=(22/7)*7*7sqrt5
=154sqrt5=344.3545

l2=sqrt((r2)^2+(h2)^2)=sqrt((9/2)^2+9^2)=(9/2)sqrt(5)
Lateral surface area of smaller cone
=(pi)*r2*l2
=(22*9*9sqrt5)/(7*2*2)
=(891/14)sqrt5=142.3098

Lateral surface area of the bottom segment
=344.3545-142.3098=202.0447

Area of larger base =(22*7*7)/7=154

Area of smaller base =(22*9*9)/(7*2*2)=891/14=63.6429

Surface area of the bottom segment
=202.0447+154+63.6429= **419.6876** cm^2#