Question

A conical paper funnel of radius 4 and height 6 units is needed to make a good cup of coffee. If this funnel is made out of a sector of a circle, what must be the radius of this circle and what must be central angle of the sector?

Answer 1

Answer:

The radius of this circle is $2\sqrt{13}$ and the central angle of the sector is $\theta=\frac{2}{\sqrt{13}}\times360^\circ$.

Step-by-step explanation:

Consider the provided information.

If a conical paper funnel is made out of a sector of circle then the radius of the sector becomes the slant height of the cone, and the length of the curved part of the sector becomes the circumference of the base of the cone.

First find the slant height: $l=\sqrt{r^2+h^2}$

$l=\sqrt{4^2+6^2}$

$l=\sqrt{16+36}$

$l=2\sqrt{13}$

Thus, the radius of the sector is $2\sqrt{13}$ which is same as the slant height of the cone.

Now, the circumference of the base of the cone is same as the length of the curved part.

$C=2\pi r$

$C=2\pi \times4=8\pi$

The length of the curved part = $\frac{\theta}{360\circ}\times2\pi r$

$\frac{\theta}{360\circ}\times2\pi \times2\sqrt{13}=8\pi$

$\frac{\theta}{360\circ}\times\sqrt{13}=2$

$\theta=\frac{2}{\sqrt{13}}\times360^\circ$

Hence, the radius of this circle is $2\sqrt{13}$ and the central angle of the sector is $\theta=\frac{2}{\sqrt{13}}\times360^\circ$.