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?

Question

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?

Sadie Adams · Accepted Answer

#~~545.60"cm"^2"to the nearest 2 decimal places"# #tan theta=12/8=1.5=56^@18'36''# #"top radius"=cot 56^@18'36''=0.666666666 xx4=2.666666666cm # #Lateral area= F=pi(r_1+r_2)sqrt((r_1-r_2)^2+h^2)# #F=pi(8+2.666666666)sqrt((8-2.666666666)^2+8^2)# #F=pi(10.667)sqrt((5.333^2+8^2)# #F=33.511sqrt28.441+64# #F=33.511sqrt92.441# #F=33.511xx9.615# #F=322.196"cm"^2# #S=F+ pi (r_1^2+r_2^2)# #S=322.196+pi(8^2+2.666666666^2)# #S=322.196+pi(64+7.111)# #S=322.196+pi(71.113)# #S=322.196+223.402# #S=545.598"cm"^2# #S="surface area of bottom segment"# #=545.60"cm"^2"to the nearest 2 decimal places"#

Avery Allen · Answer

#A = (640/9 + 256/9 sqrt(13)) pi ~~ 173.7 pi " "cm^2 ~~549.6 " "cm^2#

Given: cone of #h= 12 cm, b = 8 cm#. Cut #8cm# 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_"lateral" + A_"bases"#

#A_"lateral" = pi (r_1 + r_2) sqrt((r_1 - r_2)^2 + h_f^2)#

#A_"bases" = pi((r_1)^2 + (r_2)^2)#, where

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

#h_f = "the height of the frustum" = 8 cm#

#h_"top cone"/h_"bottom come" = (12-8)/12 = 4/12#

#color(blue) "Find the radius of the top"# using proportions of the two cones:

#h_"top cone"/h_"bottom come" = r_2/r_1; " "4/12 = r_2/8#

Use the cross-product: #12 r_2 = 4*8 = 32#

#r_2 = 32/12 = 8/3 cm#

#A_"lateral" = pi(8 + 8/3) sqrt((8-8/3)^2 + 8^2)= 32/3 pi sqrt((16/3)^2 + 64)#

#A_"lateral" = 32/3 pi sqrt(256/9 + 576/9) = 32/3 pi sqrt(832)/3#

#A_"lateral" = 32/9 pi sqrt(16) sqrt(4)sqrt(13) = 32/9 pi *8 sqrt(13)#

#A_"lateral" = 256/9 sqrt(13) pi " "cm^2#

#A_"bases" = pi(8^2 + (8/3)^2) = 640/9 pi " "cm^2#

#A = (640/9 + 256/9 sqrt(13)) pi ~~ 173.7 pi " "cm^2 ~~549.6 " "cm^2#

Mason Adams · Answer

undefined To find the surface area of the bottom segment of the cone, you can use the formula for the lateral surface area of a cone and subtract the area of the smaller circular base.

Given:
Height of the cone, $ h = 12 $ cm
Radius of the cone's base, $ r = 8 $ cm
Distance from the base to the cut, $ d = 8 $ cm

First, we need to find the radius $ R $ of the circular base of the bottom segment after the cone is cut.

Using similar triangles, the ratio of the radius $ R $ of the bottom segment to the radius $ r $ of the cone is equal to the ratio of the distance from the base to the cut $ d $ to the height $ h $.

$$ \frac{R}{r} = \frac{d}{h} $$
$$ \frac{R}{8} = \frac{8}{12} $$
$$ \frac{R}{8} = \frac{2}{3} $$
$$ R = 8 \times \frac{2}{3} $$
$$ R = \frac{16}{3} $$
$$ R = 5.\overline{3} $$ cm

Now, let's calculate the lateral surface area $ S_l $ of the cone:

$$ S_l = \pi r l $$

Where $ l $ is the slant height of the cone, which can be found using the Pythagorean theorem:

$$ l = \sqrt{h^2 + r^2} $$
$$ l = \sqrt{12^2 + 8^2} $$
$$ l = \sqrt{144 + 64} $$
$$ l = \sqrt{208} $$
$$ l = \sqrt{4 \times 52} $$
$$ l = 4\sqrt{13} $$

Now, compute $ S_l $:

$$ S_l = \pi \times 8 \times 4\sqrt{13} $$
$$ S_l = 32\pi\sqrt{13} $$

Next, calculate the area $ A $ of the smaller circular base:

$$ A = \pi R^2 $$
$$ A = \pi \left(\frac{16}{3}\right)^2 $$
$$ A = \pi \times \frac{256}{9} $$
$$ A = \frac{256\pi}{9} $$

Finally, the surface area $ S $ of the bottom segment is:

$$ S = S_l - A $$
$$ S = 32\pi\sqrt{13} - \frac{256\pi}{9} $$

$$ S = \pi(32\sqrt{13} - \frac{256}{9}) $$
$$ S = \pi(32\sqrt{13} - 28.\overline{4}) $$
$$ S = \pi(32\sqrt{13} - 28.\overline{4}) $$
$$ S = \pi(32\sqrt{13} - 28.\overline{4}) $$
$$ S \approx \pi(32\sqrt{13} - 28.44) $$
$$ S \approx \pi(32\sqrt{13} - 28.44) $$
$$ S \approx 108.54\pi - 89.6\pi $$
$$ S \approx 18.94\pi $$

Thus, the surface area of the bottom segment of the cone is approximately $ 18.94\pi $ cm^2.