# A cone has a height of #12 cm# and its base has a radius of #2 cm#. If the cone is horizontally cut into two segments #6 cm# from the base, what would the surface area of the bottom segment be?

:.Pythagoras:

:.*r^2+pi*r*L#

:.S.A.

:.S.A.

:.Total S.A.

:.Pythagoras:

:.S.A. top part

S.A. top part

S.A. top part

S.A. top part

:.S.A. Botom part

The total surface area of the bottom part got to include

the surface area of the circle of the top part.

:.S.A. Botom part:.=3.142+66.755=69.897 cm^2#

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To find the surface area of the bottom segment of the cone after it is cut horizontally 6 cm from the base, you can use the formula for the surface area of a cone's lateral surface:

[ A = \pi r \ell ]

where:

- ( r ) is the radius of the base of the cone,
- ( \ell ) is the slant height of the cone.

First, we need to find the slant height (( \ell )) of the cone using the height (( h )) and the radius (( r )) of the cone. We can apply the Pythagorean theorem to find ( \ell ):

[ \ell^2 = h^2 + r^2 ]

Given that the height (( h )) is 12 cm and the radius (( r )) is 2 cm:

[ \ell^2 = 12^2 + 2^2 = 144 + 4 = 148 ] [ \ell = \sqrt{148} ]

Now, we can calculate the surface area (( A )) of the bottom segment of the cone:

[ A = \pi r \ell = \pi \times 2 \times \sqrt{148} ]

[ A \approx 12.566 \times 12.165 ]

[ A \approx 152.870 ]

Therefore, the surface area of the bottom segment of the cone, when cut horizontally 6 cm from the base, is approximately 152.870 square centimeters.

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