# A cone has a height of 14 cm and its base has a radius of 7 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?

##### 1 Answer
Oct 11, 2017

Total surface area of the bottom segment = 419.6876 sq. cm.

#### Explanation:

$\frac{r 1}{h 1} = \frac{r 2}{h 2}$
$\frac{7}{14} = \frac{r 2}{9}$
$r 2 = \frac{9}{2}$

$l 1 = \sqrt{{\left(r 1\right)}^{2} + {\left(h 1\right)}^{2}} = \sqrt{{7}^{2} + {14}^{2}} = 7 \sqrt{5}$
Lateral surface area of bigger one $= \left(\pi\right) \cdot r 1 \cdot l 1 = \left(\frac{22}{7}\right) \cdot 7 \cdot 7 \sqrt{5}$
$= 154 \sqrt{5} = 344.3545$

$l 2 = \sqrt{{\left(r 2\right)}^{2} + {\left(h 2\right)}^{2}} = \sqrt{{\left(\frac{9}{2}\right)}^{2} + {9}^{2}} = \left(\frac{9}{2}\right) \sqrt{5}$
Lateral surface area of smaller cone
$= \left(\pi\right) \cdot r 2 \cdot l 2$
$= \frac{22 \cdot 9 \cdot 9 \sqrt{5}}{7 \cdot 2 \cdot 2}$
$= \left(\frac{891}{14}\right) \sqrt{5} = 142.3098$

Lateral surface area of the bottom segment
$= 344.3545 - 142.3098 = 202.0447$

Area of larger base $= \frac{22 \cdot 7 \cdot 7}{7} = 154$

Area of smaller base $= \frac{22 \cdot 9 \cdot 9}{7 \cdot 2 \cdot 2} = \frac{891}{14} = 63.6429$

Surface area of the bottom segment
$= 202.0447 + 154 + 63.6429 = \ast 419.6876 \ast$cm^2#