A cone has a height of 7 cm7cm and its base has a radius of 5 cm5cm. If the cone is horizontally cut into two segments 1 cm1cm from the base, what would the surface area of the bottom segment be?

1 Answer
Binayaka C.
Jun 12, 2017

Total surface area of bottom segment is 172.09(2dp)172.09(2dp)sq.cm

Explanation:

The cone is cut at 1 cm from base, So upper radius of the frustum of cone is r_2=(7-1)/7*5=30/7~~4.286r2=7−17⋅5=307≈4.286cm ; slant ht l~~sqrt(1^2+(5-4.29)^2)~~1.23l≈√12+(5−4.29)2≈1.23

Top surface area A_t=pi*4.286^2~~57.70At=π⋅4.2862≈57.70 sq.cm
Bottom surface area A_b=pi*5^2~~78.54Ab=π⋅52≈78.54 sq.cm
Slant Area A_s=pi*l*(r_1+r_2)=pi*1.23*(5+4.285)=35.85As=π⋅l⋅(r1+r2)=π⋅1.23⋅(5+4.285)=35.85 sq.cm

Total surface area of bottom segment =A_t+A_b+A_s~~57.70+78.54+35.85 ~~172.09(2dp)=At+Ab+As≈57.70+78.54+35.85≈172.09(2dp)sq.cm[Ans]

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