# A cone has a height of 36 cm and its base has a radius of 15 cm. If the cone is horizontally cut into two segments 24 cm from the base, what would the surface area of the bottom segment be?

##### 1 Answer
Oct 17, 2017

Total surface area of bottom segment $= 2419.0263$ #### Explanation:

Slanting length (${l}_{1}$) of uncut cone  = sqrt(36^2 + 15^2
${l}_{1} = \sqrt{1521} = 39 c m$

Lateral surface area of uncut cone $= \pi r l$
$= \pi \cdot 15 \cdot 39 = 1837.8317 c {m}^{2}$

Slanting length of cut cone with height 12 cm is
${l}_{2}$, radius ${r}_{2}$
${l}_{2} / {l}_{1} = {h}_{2} / {h}_{1} = {r}_{2} / {r}_{1}$
${l}_{2} / 39 = \frac{12}{36} = {r}_{2} / 15$
${l}_{2} = \frac{39 \cdot 12}{36} = 13 c m .$
${r}_{2} = \frac{15 \cdot 12}{36} = 5 c m$
Lateral surface area of cut cone $= \pi \cdot {r}_{2} \cdot {l}_{2}$
$= \pi \cdot 5 \cdot 13 = 204.2035 c {m}^{2}$

Lateral surface area of cut base $= \left(\pi \cdot {r}_{1} \cdot {l}_{1}\right) - \left(\pi \cdot {r}_{2} \cdot {l}_{2}\right)$
$= 1837.8317 - 204.2035 = 1633.6282 c {m}^{2}$, (1)

Area of uncut cone base $= \pi \cdot {r}_{1}^{2} = \pi \cdot {15}^{2} = 706.8583 c {m}^{2}$, (2)

Area of cut cone base$= \pi \cdot {r}_{2}^{2} = \pi \cdot {5}^{2} = 78.5398 c {m}^{2}$, (3)

Total surface area of cut base $= \left(1\right) + \left(2\right) + \left(3\right)$
$= 1633.6282 + 706.8583 + 78.5398 = 2419.0263 c {m}^{2}$