SOLUTION: What is the volume of a cone that has an inscribed sphere removed? Use a variety of sized cones. Generalize a pattern (if any) for volume remaining to the total cone volume.
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-> SOLUTION: What is the volume of a cone that has an inscribed sphere removed? Use a variety of sized cones. Generalize a pattern (if any) for volume remaining to the total cone volume.
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Question 327887: What is the volume of a cone that has an inscribed sphere removed? Use a variety of sized cones. Generalize a pattern (if any) for volume remaining to the total cone volume.
A triangle with two sides of equal length.
a = c
A = C
B + 2A = Pi radians = 180o
P = 2a + b
s = a + b/2
K = b sqrt(4a2-b2)/4 = a2 sin(B)/2 = ab sin(A)/2
ha = b sqrt(4a2-b2)/(2a) = a sin(B) = b sin(A)
ma = sqrt(a2+2b2)/2
ta = b sqrt(a[2a+b])/(a+b) = b sin(A)/sin(3A/2)
hb = mb = tb = sqrt(4a2-b2)/2 = a cos(B/2)
R = a2b/4K = a/[2 sin(A)] = b/[2 sin(B)]
r = K/s = b sqrt[(2a-b)/(2a+b)]/2
Here's a start. Find the radius of the enclosed sphere by dealing with the intersection of the cone and sphere and a plane thru the cone's axis.
r = K/s = b sqrt[(2a-b)/(2a+b)]/2
= b*sqrt(4a^2 - b^2)/(4a+2b)
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r above is the radius of the inscribed sphere.
The dimensions of a cone are usually the radius of the base and the height.
a in the triangle is the slant height of the cone, the length of the side.
It's: s = sqrt(h^2 + (b/2)^2) where s = a.
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Solve for h in terms of a and b:
a = sqrt(h^2 + (b/2)^2)
a^2 = h^2 + (b/2)^2
h^2 = a^2 - (b/2)^2
h = sqrt(a^2 - (b/2)^2) and the radius of the cone = b/2
Vol of the cone = pi*r^2*h/3
Vol = pi*(b/2)^2*sqrt(a^2 - (b/2)^2)/3
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For the sphere: Vol = 4*pi*r^3/3
= 4*pi*b^3*sqrt[(2a-b)/(2a+b)]/8
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Subtract the sphere's volume from the cone's.
