# 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?

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 $= \left(1\right) + \left(2\right) + \left(3\right)$
$= 1633.6282 + 706.8583 + 78.5398 = 2419.0263 c {m}^{2}$