# A cone has a height of 16 cm and its base has a radius of 5 cm. If the cone is horizontally cut into two segments 12 cm from the base, what would the surface area of the bottom segment be?

Jun 8, 2018

$330.31 c {m}^{2} \text{to the nearest 2 decimal places}$

#### Explanation:

$\frac{16}{5} = \tan \theta = {72}^{\circ} 38 ' 46 ' '$

$\cot {72}^{\circ} 38 ' 46 ' ' \times 4.0 = 1.25 c m = \text{r top area}$

$\therefore \text{F=lateral area} = \pi \left({r}_{1} + {r}_{2}\right) \sqrt{{\left({r}_{1} - {r}_{2}\right)}^{2} + {h}^{2}}$

$\therefore F = \pi \left(5 + 1.25\right) \sqrt{{\left(5 - 1.25\right)}^{2} + {12}^{2}}$

$\therefore F = \pi \left(6.25\right) \sqrt{{\left(3.75\right)}^{2} + {12}^{2}}$

$\therefore F = 19.63495408 \sqrt{14.0625 + 144}$

$\therefore F = 19.63495408 \sqrt{158.0625}$

$\therefore F = 19.63495408 \times 12.57229096$

$\therefore F = 246.8563557$

$\therefore S = F + \pi \left({r}_{1}^{2} + {r}_{2}^{2}\right)$

$\therefore S = 246.8563557 + \pi \left({5}^{2} + {1.25}^{2}\right)$

$\therefore S = 246.8563557 + \pi \left(25 + 1.5625\right)$

$\therefore S = 246.8563557 + \pi \left(26.5625\right)$

$\therefore S = 246.8563557 + 83.44855486$

$\therefore S = 330.3049105$

$\therefore \text{Total surface area o frustum or truncated cone=330.31"cm^2 }$

$\text{to the nearest 2 decimal places}$