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Nets, Sectors and Cones (Posted on 2006-11-12) Difficulty: 2 of 5
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.

No Solution Yet Submitted by Ansh    
Rating: 4.3333 (3 votes)

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Solution solution | Comment 1 of 2

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 2006-11-12 15:22:28
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