A cone has a height of $9 c m$ $9 cm$ and its base has a radius of $4 c m$ $4 cm$ . If the cone is horizontally cut into two segments $7 c m$ $7 cm$ from the base, what would the surface area of the bottom segment be?

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A cone has a height of $9 c m$ $9 cm$ and its base has a radius of $4 c m$ $4 cm$ . If the cone is horizontally cut into two segments $7 c m$ $7 cm$ from the base, what would the surface area of the bottom segment be?

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Answer 1 · 2016-10-04T11:01:12

Total surface area of bottom segment is $170.4 (1 d p)$ $170.4(1dp)$ sq,cm

Explanation:

The cone is cut at 7 cm from base, So upper radius of the frustum of cone is $r_{2} = \frac{9 - 7}{9} \cdot 4 = 0.889$ $r_2=(9-7)/9*4=0.889$ cm ; slant ht $l = \sqrt{7^{2} + {(4 - 0.889)}^{2}} = \sqrt{49 + 9.678} = \sqrt{58.678} = 7.66 c m$ $l=sqrt(7^2+(4-0.889)^2)=sqrt(49+9.678)=sqrt 58.678=7.66 cm$
Top surface area $A_{t} = π \cdot {0.889}^{2} = 2.48 c m^{2}$ $A_t=pi*0.889^2=2.48 cm^2$
Bottom surface area $A_{b} = π \cdot 4^{2} = 50.265 c m^{2}$ $A_b=pi*4^2=50.265 cm^2$
Slant Area $A_{s} = π \cdot l \cdot (r_{1} + r_{2}) = π \cdot 7.66 \cdot (4 + 0.889) = 117.65 c m^{2}$ $A_s=pi*l*(r_1+r_2)=pi*7.66*(4+0.889)=117.65 cm^2$

Total surface area of bottom segment $= A_{t} + A_{b} + A_{s} = 2.48 + 50.265 + 117.65 = 170.4 (1 d p)$ $=A_t+A_b+A_s=2.48+50.265+117.65=170.4(1dp)$ sq,cm[Ans]

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A cone has a height of $9 c m$ $9 cm$ and its base has a radius of $4 c m$ $4 cm$ . If the cone is horizontally cut into two segments $7 c m$ $7 cm$ 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 9cm9 cm and its base has a radius of 4cm4 cm. If the cone is horizontally cut into two segments 7cm7 cm from the base, what would the surface area of the bottom segment be?

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A cone has a height of $9 c m$ $9 cm$ and its base has a radius of $4 c m$ $4 cm$ . If the cone is horizontally cut into two segments $7 c m$ $7 cm$ from the base, what would the surface area of the bottom segment be?