# A cone has a height of 9 cm and its base has a radius of 4 cm. If the cone is horizontally cut into two segments 3 cm from the base, what would the surface area of the bottom segment be?

May 27, 2018

:.color(purple)(=141.36cm^2 to the nearest 2 decimal places $c {m}^{2}$

#### Explanation:

:.Pythagoras: ${c}^{2} = {9}^{2} + {4}^{2}$

$\therefore c = L = \sqrt{{9}^{2} + {4}^{2}}$

:. c=Lcolor(purple)(=9.849cm

:.9/4=tan theta=2.25=66^@02’15”

:.$\text{color(purple)(S.A} .$$= \pi \cdot r \cdot L$

:.S.A.$= \pi \cdot 4 \cdot 9.849$

:.S.A.$= 123.766$

:.Total S.A.color(purple)(=123.766cm^2

:.Cot 66^@02’15”*6=2.667cm=radius of top part

:.Pythagoras: ${c}^{2} = {6}^{2} + {2.667}^{2}$

$\therefore c = L = \sqrt{{6}^{2} + {2.667}^{2}}$

:. c=Lcolor(purple)(=6.566cm top part

:.S.A. top part$= \pi \cdot r \cdot L$

S.A. top part$\therefore \pi \cdot 2.667 \cdot 6.566$

S.A. top part$\therefore = 55.014$

S.A. top part:.color(purple)(=55.014cm^2

:.S.A. Bottom partcolor(purple)(=123.766-55.014=68.758cm^2

:.S.A. Bottom part$= 68.758 + \pi {r}^{2} + \pi {r}^{2}$

$\therefore 68.758 + 22.340 + 50.265$

:.color(purple)(=141.363cm^2 to the nearest 2 decimal places $c {m}^{2}$