Surface area of a cone

[Guest]

please help with this question.

[Guest]

The length of the arc is [tex]\frac{216}{360}[/tex]

Lateral surface area [tex]= \frac{216}{360} \pi \cdot 8^2[/tex]



Lateral surface area [tex]= \pi r l[/tex]
l = 8
the radius [tex]= \frac{\frac{216}{360} \pi \cdot 8^2 }{8\pi }=\frac{216}{360} 8[/tex]

[HallsofIvy]

I am puzzled by your asking this question. Do you not know that the circumference of a circle of radius r is $2\pi r$? Do you not know that the area of a a circle of radius r is $\pi r^2$? Do you not know that there are 360 degrees in an entire circle so that a sector with central angle 216 degrees is $\frac{216}{360}= \frac{3}{5}$ of an entire circle? So the base circumference of this cone is $\frac{3}{5}(2\pi (8))= \frac{48}{5}\pi$. From that the radius of the base is $\frac{\frac{48}{5}\pi}{2\pi}= \frac{24}{5}$ cm.

The problem asks you to find the "slant height" of the cone but what I would call the slant height is given as 8 cm. They must mean what I would call just the "height". Imagine looking at the cone from the side. you have two right triangle with base of the triangle the radius of the cone, $\frac{24}{5}$ cm and hypotenuse 8 cm. Use the Pythagorean theorem to find the height.

[GeradHum]

Very good question, I'll figure it out and come back with the answer.