A cone has a height of 14 cm14cm and its base has a radius of 7 cm7cm. If the cone is horizontally cut into two segments 5 cm5cm 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)r1h1=r2h2
7/14=(r2)/9714=r29
r2=9/2r2=92

l1=sqrt((r1)^2+(h1)^2)=sqrt(7^2+14^2)=7sqrt5l1=(r1)2+(h1)2=72+142=75
Lateral surface area of bigger one =(pi)*r1*l1=(22/7)*7*7sqrt5=(π)r1l1=(227)775
=154sqrt5=344.3545=1545=344.3545

l2=sqrt((r2)^2+(h2)^2)=sqrt((9/2)^2+9^2)=(9/2)sqrt(5)l2=(r2)2+(h2)2=(92)2+92=(92)5
Lateral surface area of smaller cone
=(pi)*r2*l2=(π)r2l2
=(22*9*9sqrt5)/(7*2*2)=22995722
=(891/14)sqrt5=142.3098=(89114)5=142.3098

Lateral surface area of the bottom segment
=344.3545-142.3098=202.0447=344.3545142.3098=202.0447

Area of larger base =(22*7*7)/7=154=22777=154

Area of smaller base =(22*9*9)/(7*2*2)=891/14=63.6429=2299722=89114=63.6429

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