LPQ is the sector of a circle of radius x cm centered at L, where PQ is the arc portion of its boundary. Angle PLQ is 120 degrees.
The sector is turned into a cone, with PQ forming the circular base. The curved surface area of the cone is A cm^2, and its volume is V cm^3. The height of the cone is h cm. Given that V = 3 A, find h.
The area of the curved surface of the cone is just the area of the sector, so A = pi x^2 / 3. The circumference of the cone's base is the arc length of the sector, or 2 pi x / 3.
The radius, r, of the cone's base is its circumference divided by 2 pi, and so is x/3. As the volume of the cone is V = pi r^2 h / 3, it is also then V = pi x^2 h / 27.
Setting V = 3 A,
pi x^2 h / 27 = pi x^2
so h = 27.
By the way, as a vertical cross section of the cone through the apex forms an isosceles triangle, divisible into two right triangles, r^2 + h^2 = x^2, or
(x/3)^2 + 27^2 = x^2
8x^2/9 = 27^2
x^2 = 9 (27)^2 / 8
x = 81/sqrt(8) = 81 sqrt(2) / 4 = 28.637824638....

Posted by Charlie
on 20061112 15:22:28 