A cone has a height of 36 cm and its base has a radius of 9 cm. If the cone is horizontally cut into two segments 24 cm from the base, what would the surface area of the bottom segment be?

May 23, 2017

The surface area of the bottom segment $= 256.259 {\text{ cm}}^{2}$

Explanation:

To explain this, I will show it with a diagram of a triangle as if the cone was vertically split.

$\text{Surface area of a cone} = \pi {r}^{2} + \pi r s$

$\text{SA" = pi "DG"^2 + pi r"DG" xx"DE}$

We can find out the length of line $D G$ because since lines $C E$ and $B E$ are of constant gradient, $\text{CH"/"CE" = "DG"/"DE}$

$\text{CH"/"CE" = "DG"/"DE}$

$\frac{4.5}{\text{CE" = "DG"/"DE}}$

${\text{CE}}^{2} = {4.5}^{2} + {36}^{2}$ (Pythagoras theorem)

${\text{CE}}^{2} = 20.25 + 1296$

$\text{CE} = \sqrt{1316.25}$

$\text{CE} = 36.28$

$\frac{4.5}{\text{36.28" = "DG"/"DE}}$

$\text{HE"/"CE" = "GE"/"DE}$

$\text{36"/"36.28" = "24"/"DE}$

$\text{36"/"36.28" = "24"/"24.19}$

now we know that:

$\text{DE} = 24.19$

So we can now use the pythagorean theorem again to find out what $\text{DG}$ is.

${\text{DG"^2 = "DE"^2 - "GE}}^{2}$

${\text{DG}}^{2} = {24.19}^{2} - {24}^{2}$

"DG" = sqrt(585 - 576

"DG" = sqrt(9

$\text{DG} = 3$

We can also do the same equation as above to find out what $\text{DG}$ is, but this was what first came to my mind when answering this question

Now that we have those measurements, we can substitute them in the Surface area equation.

$\text{SA} = \pi {3}^{2} + \pi 3 \times 24.19$

$\text{SA} = 28.274 + 9.425 \times 24.19$

$\text{SA} = 28.274 + 227.985$

${\text{SA" = 256.259" cm}}^{2}$