A cone has a height of #18 cm# and its base has a radius of #9 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?

Question

A cone has a height of #18  cm# and its base has a radius of #9  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?

Ella Armstrong · Accepted Answer

Surface area #~~148.833cm^2#

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#R_2# equals #18/3=3/R_2# #18R_2=9# #R_2=9/18=1/2=0.5#

#s# equals #s=sqrt((3-0.5)^2+3^2)# #s=sqrt(6.25+9)# #s=sqrt(15.25)#

Surface area equals #A_s=pi(sqrt15.25(3-0.5)+3^2+0.5^2)# #A_s=pi(38.125+9+0.25)# #A_s=pi(47.375)# #A_s~~148.833cm^2#, Therefore the surface area of the bottom cone segment is roughly 148.833#cm^2#.

I hope I was of assistance.

Sadie Allen · Answer

undefined To find the surface area of the bottom segment of the cone, you can use the formula for the surface area of a cone segment, which is given by:

$$ A = \pi r_1 (r_1 + l) $$

Where:
- $ r_1 $ is the radius of the base of the segment
- $ l $ is the slant height of the cone segment

Given:
- Radius of the base ($ r $) = 9 cm
- Height of the cone ($ h $) = 18 cm

First, we need to find the slant height ($ l $) of the cone segment. Using the Pythagorean theorem:

$$ l = \sqrt{r^2 + h^2} $$

$$ l = \sqrt{9^2 + 18^2} $$
$$ l = \sqrt{81 + 324} $$
$$ l = \sqrt{405} $$
$$ l = 3\sqrt{45} $$

Now, the radius ($ r_1 $) of the base of the cone segment is the radius of the whole cone, which is 9 cm.

Plug the values into the formula:

$$ A = \pi \times 9 \times (9 + 3) $$
$$ A = \pi \times 9 \times 12 $$
$$ A = 108\pi $$

Therefore, the surface area of the bottom segment of the cone is $ 108\pi $ square centimeters.