A cone has a height of 24 cm and its base has a radius of 6 cm. If the cone is horizontally cut into two segments 5 cm from the base, what would the surface area of the bottom segment be?

Sep 19, 2017

Total surface area of bottom segment is 358.04 (2dp)# sq.cm.

Explanation:

The cone is cut at 5 cm from base, So upper radius of the frustum of

cone is ${r}_{2} = \frac{24 - 5}{24} \cdot 6 = 4.75$cm ; slant ht:

$l = \sqrt{{5}^{2} + {\left(6 - 4.75\right)}^{2}} = \sqrt{25 + 1.5625} = \sqrt{26.5625} = 5.15$

Top surface area ${A}_{t} = \pi \cdot {4.75}^{2} = 70.88$ sq.cm

Bottom surface area ${A}_{b} = \pi \cdot {6}^{2} = 113.1$ sq.cm

Slant Area ${A}_{s} = \pi \cdot l \cdot \left({r}_{1} + {r}_{2}\right) = \pi \cdot 5.15 \cdot \left(6 + 4.75\right) = 174.06$ sq.cm

Total surface area of bottom segment

$= {A}_{t} + {A}_{b} + {A}_{s} = 70.88 + 113.1 + 174.06 = 358.04 \left(2 \mathrm{dp}\right)$

sq.cm [Ans]