A cone has a height of $36 c m$ $36 cm$ and its base has a radius of $9 c m$ $9 cm$ . If the cone is horizontally cut into two segments $12 c m$ $12 cm$ from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2017-10-20T11:40:35

Total surface area of bottom segment is $950.46$ $950.46$ sq.cm

The cone is cut at 12 cm from base, So upper radius of the frustum

of cone is $r_{2} = \frac{36 - 12}{36} \cdot 9 = 6$ $r_2=(36-12)/36*9=6$ cm ; slant height of the frustum

is $l = \sqrt{12^{2} + {(9 - 6)}^{2}} = \sqrt{144 + 9} = \sqrt{153} = 12.37$ $l=sqrt(12^2+(9-6)^2)=sqrt(144+9)=sqrt 153=12.37$ cm

Top surface area $A_{t} = π \cdot 6^{2} = 113.1$ $A_t=pi*6^2=113.1$ sq.cm

Bottom surface area $A_{b} = π \cdot 9^{2} = 254.47$ $A_b=pi*9^2=254.47$ sq.cm

Slant area $A_{s} = π \cdot l \cdot (r_{1} + r_{2}) = π \cdot 12.37 \cdot (9 + 6) = 582.89$ $A_s=pi*l*(r_1+r_2)=pi*12.37*(9+6)=582.89$ sq.cm

Total surface area of bottom segment

$= A_{t} + A_{b} + A_{s} = 113.1 + 254.47 + 582.89 = 950.46 (2 d p)$ $=A_t+A_b+A_s=113.1+254.47+582.89=950.46 (2dp)$ sq.cm [Ans]

A cone has a height of 36 cm36cm36 cm and its base has a radius of 9 cm9cm9 cm. If the cone is horizontally cut into two segments 12 cm12cm12 cm from the base, what would the surface area of the bottom segment be?