A cone has a height of $8 c m$ $8 cm$ and its base has a radius of $5 c m$ $5 cm$ . If the cone is horizontally cut into two segments $6 c m$ $6 cm$ from the base, what would the surface area of the bottom segment be?

Question

1477 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-04-25T11:33:02

Total surface area of bottom segment is $222.38$ $222.38$ sq.cm

The cone is cut at 6 cm from base, So upper radius of the frustum

of cone is $r_{2} = \frac{8 - 6}{8} \cdot 5 = 1.25$ $r_2=(8-6)/8*5=1.25$ cm ; Slant ht

$l = \sqrt{6^{2} + {(5 - 1.25)}^{2}} = \sqrt{36 + 14.0625}$ $l=sqrt(6^2+(5-1.25)^2)=sqrt(36+14.0625)$

$= \sqrt{50.0625} \approx 7.08$ $=sqrt 50.0625 ~~ 7.08$ cm

Top surface area $A_{t} = π \cdot {1.25}^{2} \approx 4.91$ $A_t=pi*1.25^2 ~~ 4.91$ sq.cm

Bottom surface area $A_{b} = π \cdot 5^{2} \approx 78.54$ $A_b=pi*5^2 ~~ 78.54$ sq.cm

Slant Area $A_{s} = π \cdot l \cdot (r_{1} + r_{2}) = π \cdot 7.08 \cdot (5 + 1.25)$ $A_s=pi*l*(r_1+r_2)=pi*7.08*(5+1.25)$

$\approx 138.93$ $~~138.93$ sq.cm

Total surface area of bottom segment

$= A_{t} + A_{b} + A_{s} = 4.91 + 78.54 + 138.93$ $=A_t+A_b+A_s=4.91+78.54+138.93$

$\approx 222.38$ $~~ 222.38$ sq.cm [Ans]

A cone has a height of 8cm8 cm and its base has a radius of 5cm5 cm. If the cone is horizontally cut into two segments 6cm6 cm from the base, what would the surface area of the bottom segment be?