# A cone has a height of 8 cm and its base has a radius of 6 cm. If the cone is horizontally cut into two segments 7 cm from the base, what would the surface area of the bottom segment be?

May 24, 2017

Area is $\frac{765}{8} \pi$ $c {m}^{2} \approx 300.42$ $c {m}^{2}$

#### Explanation:

Figure 1

So we have a cone that has been sliced into 2. The bottom cone consists of two circles and a frustum. The flattened frustum can be seen on the right side of the image below.

Figure 2

We can see that the area of the frustum is just the difference between the area difference of the sector 2 concentric circles.
This area difference is essentially the formula for the area of a frustum:
$A = \pi \left(R + r\right) \sqrt{{\left(R - r\right)}^{2} + {h}^{2}}$

Ignoring Figure 2 and its labels, $R$ here is the big circle's radius and $r$ is the smaller circle's radius, and $h$ is the height of the frustum.

Back to Figure 1, the height $h$ of the bottom portion of the cone is $7 c m$ as stated in the problem, and the radius $R$ of the bottom is $6 c m$. Now we just need to find $r$.

Recall $\tan \theta = \frac{o p p o s i t e}{a \mathrm{dj} a c e n t}$

In the cone, image $2 \theta$ as the angle at the vertex of the cone (refer to Figure 1). In that case, the opposite side would be the radius of the cone, $R$ and the adjacent side would be the height of the cone, $h$.

$\tan \theta = \frac{R}{h} = \frac{6}{8} = \frac{3}{4}$

However, the opposite side of $\theta$ can also be $r$, the radius of the top of the frustum and the adjacent side can also be $h - 7 c m$ or $1$. In this case:

$\tan \theta = \frac{r}{h - 7} = r$

Substitute $\tan \theta = \frac{3}{4}$:
$r = \frac{3}{4}$

Phew!

Now, we can finally calculate the area of the frustum! $A = \pi \left(R + r\right) \sqrt{{\left(R - r\right)}^{2} + {h}^{2}}$
where $R = 6 c m$, $h = 7 c m$, $r = \frac{3}{4} c m$
$A = \frac{945}{16} \pi$ $c {m}^{2}$

Now we can calculate the areas of the top and bottom circles easily and add it all together:

${A}_{\text{total}} = \frac{945}{16} \pi + \frac{9}{16} \pi + 36 \pi$
$= \pi \left(\frac{945}{16} + \frac{9}{16} + 36\right)$
$= \frac{765}{8} \pi$ $c {m}^{2}$
$\approx 300.42$ $c {m}^{2}$

$\therefore$ The surface area of the bottom portion is around $300.42$ $c {m}^{2}$