### Instruction: A circular sheet of paper of radius 6 inches is cut into 3 equal sectors, and each sector is formed into a cone with no overlap. What is the height in inches of the cone?### Response: I know that the circumference of the circle is 2π times the radius, so the original circle had a circumference of 12π inches.
When I cut it into 3 equal sectors, each sector has an arc length of 4π inches, which becomes the circumference of the base of the cone.
To find the radius of the base of the cone, I can use the formula 2πr = 4π, where r is the radius of the base.
Solving for r, I get r = 2 inches.
Now I have the radius of the base of the cone, but I need to find the height.
I can use the Pythagorean theorem to relate the height, the radius, and the slant height of the cone.
The slant height of the cone is the same as the radius of the original circle, which was 6 inches.
So I have a right triangle with hypotenuse 6, one leg 2, and the other leg h, where h is the height of the cone.
Using the Pythagorean theorem, I get 6^2 = 2^2 + h^2, or 36 = 4 + h^2.
Subtracting 4 from both sides, I get 32 = h^2.
Taking the square root of both sides, I get h = √32.
I can simplify this by factoring out the largest perfect square from 32, which is 16.
So h = √16 √2, or h = 4√2.
Therefore, the height of the cone is 4√2 inches.