Coffee Anyone?

Consider the first figure in my diagram. Note that it is a circular sector with radius of $5$ and arc length of $c$ . But the circular sector is two layers because it is an inflated right circular cone. So the circumference of the base of the cone (figure 2) is twice the value of $c$ or $y=2c$ . The value of $c$ is $\dfrac{1}{4} \times \text{circumference of a circle of radius 5}$ , we have

$c=\dfrac{1}{4}(2)(\pi)(5)=\dfrac{5}{2}\pi$

Now we consider figure $2$ . Let $x$ be the radius of the base of the cone and $h$ be the height. Note that the slant height of the right circular cone is the radius of the circular sector in the first figure. Since we know that $y=2c$ , we have

$y=2\left(\dfrac{5}{2}\right)(\pi)$

$2\pi (x)= 2\left(\dfrac{5}{2}\right)(\pi)$

$x=\dfrac{5}{2}$

By the theorem of Pythagoras of Samos, we have

$h=\sqrt{5^2-\left(\dfrac{5}{2}\right)^2}=\sqrt{\dfrac{75}{4}}=\boxed{\dfrac{5}{2}\sqrt{3}}$

This solution is a bit tricky because it uses the length of the "mouth" of the coffee filter in two different ways. First, the length of the mouth is calculated as the circumference of half a circle with radius 5 (half a circle because the quarter-circle coffee filter is 2 layers of paper).

$L_{\text{mouth}} = \frac{1}{2} \pi \times \text{diameter} =\frac{1}{2} \pi \times 10 = 5 \pi$

Next, consider the mouth of the coffee filter when it is the full circumference of the 3D cone, filled with grounds. We can use knowing the circumference of full filter to find the radius of the cone.

$5 \pi\ = 2\pi \times R_{\text{cone}}$ $R_{\text{cone}}\ = \frac{5}{2}$

Finally, the radius of the cone and the radius of the paper (which is now the length along the side of the cone) to find the height of the cone. This last step is done with the Pythagorean theorem. The height of the cone, $H_{\text{cone}},$ and the radius across the top are the two legs of a right triangle with the side of the cone as the hypotenuse.

$A^2 + B^2 = C^2$ $H_{\text{cone}}^2 + R_{\text{cone}}^2 = 5^2$ $H_{\text{cone}}^2 = 5^2 - \left(\frac{5}{2}\right)^2$ $H_{\text{cone}}^2 = \frac{100}{4} - \frac{25}{4} = \frac{75}{4}$ $H_{\text{cone}} = \sqrt{ \frac{75}{4}} = \frac{5\sqrt{3}}{2}$

Considering the empty conical coffee filter:

The length of arc (consider the figure) is $\dfrac{1}{4}(2)(\pi)(5)=\dfrac{5}{2}\pi$ . Since it is two layers thick, the circumference of the base is twice the length of arc: $2\left(\dfrac{5}{2}\pi \right)=5\pi$ .

When the coffee filter is filled with coffee grounds:

Let $r$ be the base radius, then we have

$5\pi=2\pi r$

$r=\dfrac{5}{2}$

By pythagorean theorem, the height must be

$h=\sqrt{5^2-\left(\dfrac{5}{2}\right)^2}=\boxed{\dfrac{5}{2}\sqrt{3}}$