# 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

#### Answer:

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.

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

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

$\text{SA} = 28.274 + 227.985$

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