A circle of radius 12 is cut out of a piece of paper. A 90 degree sector is then cut from this circle. This 90 degree sector is rolled into a cone. What is the volume of the cone?

Guest Jul 4, 2020

#1**-1 **

The circumference of the original circle is: 2·pi·12 = 24pi

Since a 90^{o} sector is cut out, one-fourth of the circle is cut out, leaving three-fourths.

Three-fourths of 24pi = (3/4)·24pi = 18 pi.

Therefore 18pi is the circumference of the cone.

Finding the radius of the cone: 18pi = 2·pi·r ---> r = 9

The radius of the original circle is now the slant height of the cone.

To find the height of the cone, use the right triangle whose hypotenuse is 12 and whose base is 9.

This gives a height of: h^{2} + 9^{2} = 12^{2} ---> h = sqrt(63)

Volume of a cone: (1/3)·pi·r^{2}·h ---> (1/3)·pi·9^{2}·sqrt(63) ---> ....

geno3141 Jul 4, 2020

#2**+1 **

Hi, Geno! Please take the time and read the question again.

A circle of radius 12 is cut out of a piece of paper. A 90-degree sector is then cut from this circle. **This 90-degree sector is rolled into a cone. What is the volume of the cone?**

**r = 3 h = 11.619**

**V = pi * r ^{2} * h / 3 ≈ 109.51 u^{3} **

Dragan
Jul 5, 2020