# A cone has a height of 12 cm and its base has a radius of 8 cm. If the cone is horizontally cut into two segments 8 cm from the base, what would the surface area of the bottom segment be?

Jun 29, 2018

$A = \left(\frac{640}{9} + \frac{256}{9} \sqrt{13}\right) \pi \approx 173.7 \pi \text{ "cm^2 ~~549.6 " } c {m}^{2}$

#### Explanation:

Given: cone of $h = 12 c m , b = 8 c m$. Cut $8 c m$ from the base. Find the surface area of the bottom segment.

The new shape is called a conical frustum . It has the surface area formula :

$A = {A}_{\text{lateral" + A_"bases}}$

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

${A}_{\text{bases}} = \pi \left({\left({r}_{1}\right)}^{2} + {\left({r}_{2}\right)}^{2}\right)$, where

r_1 = "base radius = 8 cm", " "r_2 = "top radius" = ?,

${h}_{f} = \text{the height of the frustum} = 8 c m$

${h}_{\text{top cone"/h_"bottom come}} = \frac{12 - 8}{12} = \frac{4}{12}$

$\textcolor{b l u e}{\text{Find the radius of the top}}$ using proportions of the two cones:

${h}_{\text{top cone"/h_"bottom come" = r_2/r_1; " }} \frac{4}{12} = {r}_{2} / 8$

Use the cross-product: $12 {r}_{2} = 4 \cdot 8 = 32$

${r}_{2} = \frac{32}{12} = \frac{8}{3} c m$

${A}_{\text{lateral}} = \pi \left(8 + \frac{8}{3}\right) \sqrt{{\left(8 - \frac{8}{3}\right)}^{2} + {8}^{2}} = \frac{32}{3} \pi \sqrt{{\left(\frac{16}{3}\right)}^{2} + 64}$

${A}_{\text{lateral}} = \frac{32}{3} \pi \sqrt{\frac{256}{9} + \frac{576}{9}} = \frac{32}{3} \pi \frac{\sqrt{832}}{3}$

${A}_{\text{lateral}} = \frac{32}{9} \pi \sqrt{16} \sqrt{4} \sqrt{13} = \frac{32}{9} \pi \cdot 8 \sqrt{13}$

${A}_{\text{lateral" = 256/9 sqrt(13) pi " }} c {m}^{2}$

${A}_{\text{bases" = pi(8^2 + (8/3)^2) = 640/9 pi " }} c {m}^{2}$

$A = \left(\frac{640}{9} + \frac{256}{9} \sqrt{13}\right) \pi \approx 173.7 \pi \text{ "cm^2 ~~549.6 " } c {m}^{2}$

Jun 30, 2018

$\approx 545.60 \text{cm"^2"to the nearest 2 decimal places}$

#### Explanation:

$\tan \theta = \frac{12}{8} = 1.5 = {56}^{\circ} 18 ' 36 ' '$

$\text{top radius} = \cot {56}^{\circ} 18 ' 36 ' ' = 0.666666666 \times 4 = 2.666666666 c m$

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

$F = \pi \left(8 + 2.666666666\right) \sqrt{{\left(8 - 2.666666666\right)}^{2} + {8}^{2}}$

F=pi(10.667)sqrt((5.333^2+8^2)

$F = 33.511 \sqrt{28.441} + 64$

$F = 33.511 \sqrt{92.441}$

$F = 33.511 \times 9.615$

$F = 322.196 {\text{cm}}^{2}$

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

$S = 322.196 + \pi \left({8}^{2} + {2.666666666}^{2}\right)$

$S = 322.196 + \pi \left(64 + 7.111\right)$

$S = 322.196 + \pi \left(71.113\right)$

$S = 322.196 + 223.402$

$S = 545.598 {\text{cm}}^{2}$

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

$= 545.60 \text{cm"^2"to the nearest 2 decimal places}$