# A cone has a height of 8 cm and its base has a radius of 6 cm. If the cone is horizontally cut into two segments 4 cm from the base, what would the surface area of the bottom segment be?

Jun 14, 2018

$282.74 \text{cm"^2"to the nearest 2 decimal places}$

#### Explanation:

$\tan \theta = \frac{8}{6} = 1.333333333 = {53}^{\circ} 7 ' 48 ' '$

$\text{top radius} = \cot {53}^{\circ} 7 ' 48 ' ' = 0.75 \times 4 = 3.0 c m$

$L a t e r a l a r e a = F = \pi \left({r}_{1} + {r}_{2}\right) \sqrt{{\left({r}_{1} - {r}_{2}\right)}^{2} + {h}^{2}}$

$F = \pi \left(6 + 3\right) \sqrt{{\left(6 - 3\right)}^{2} + {4}^{2}}$

F=pi(9)sqrt((9+16)

$F = 9 \pi \sqrt{25}$

$F = 28.27433388 \times 5 = 141.3716694 {\text{cm}}^{2}$

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

$S = 141.3716694 + \pi \left({6}^{2} + {3}^{2}\right)$

$S = 141.3716694 + \pi \left(36 + 9\right)$

$S = 141.3716694 + \pi \left(45\right)$

$S = 141.3716694 + 141.3716694$

$S = 282.7433388 {\text{cm}}^{2}$

$S = \text{surface area of bottom segment}$

$= 282.74 \text{cm"^2"to the nearest 2 decimal places}$