A cone has a height of $12 c m$ $12 cm$ and its base has a radius of $15 c m$ $15 cm$ . If the cone is horizontally cut into two segments $3 c m$ $3 cm$ from the base, what would the surface area of the bottom segment be?

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A cone has a height of $12 c m$ $12 cm$ and its base has a radius of $15 c m$ $15 cm$ . If the cone is horizontally cut into two segments $3 c m$ $3 cm$ from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2017-01-02T11:24:49

Total surface area of bottom segment is $1500.5 (1 d p)$ $1500.5(1dp)$ sq.cm

Explanation:

The cone is cut at 3 cm from base, So upper radius of the frustum of cone is $r_{2} = \frac{12 - 3}{12} \cdot 15 = 11.25$ $r_2=(12-3)/12*15=11.25$ cm ; slant ht $l = \sqrt{3^{2} + {(15 - 11.25)}^{2}} = \sqrt{9 + 14.0625} = \sqrt{23.0625} = 4.80$ $l=sqrt(3^2+(15-11.25)^2)=sqrt(9+14.0625)=sqrt 23.0625=4.80$ cm

Top surface area $A_{t} = π \cdot {11.25}^{2} = 397.61$ $A_t=pi*11.25^2=397.61$ sq.cm
Bottom surface area $A_{b} = π \cdot 15^{2} = 706.858$ $A_b=pi*15^2=706.858$ sq.cm
Slant Area $A_{s} = π \cdot l \cdot (r_{1} + r_{2}) = π \cdot 4.80 \cdot (15 + 11.25) = 396.03$ $A_s=pi*l*(r_1+r_2)=pi*4.80*(15+11.25)=396.03$ sq.cm

Total surface area of bottom segment $= A_{t} + A_{b} + A_{s} = 397.61 + 706.86 + 396.03 = 1500.5 (1 d p)$ $=A_t+A_b+A_s=397.61+706.86+396.03=1500.5(1dp)$ sq.cm[Ans]

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A cone has a height of $12 c m$ $12 cm$ and its base has a radius of $15 c m$ $15 cm$ . If the cone is horizontally cut into two segments $3 c m$ $3 cm$ from the base, what would the surface area of the bottom segment be?

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Explanation:

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A cone has a height of 12 cm12cm12 cm and its base has a radius of 15 cm15cm15 cm. If the cone is horizontally cut into two segments 3 cm3cm3 cm from the base, what would the surface area of the bottom segment be?

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Explanation:

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A cone has a height of $12 c m$ $12 cm$ and its base has a radius of $15 c m$ $15 cm$ . If the cone is horizontally cut into two segments $3 c m$ $3 cm$ from the base, what would the surface area of the bottom segment be?