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Question 1159597: This is my dilemma . . . A cone inscribed in a hemisphere, vertex of cone at center of hemisphere, cone has height 12 cm and radius of cone makes a 60 degree angle with the slant height of the cone. What is the volume of the region inside the hemisphere and outside the cone. I don't think there is enough information to answer the problem.
Found 2 solutions by ikleyn, jim_thompson5910:
Answer by ikleyn(52790)   (Show Source):
Answer by jim_thompson5910(35256)  (Show Source):
You can put this solution on YOUR website!
If you were to look at the side profile 2D cross section, then you'd get this
Triangle AFE is a right triangle
A = center of hemisphere = vertex of cone
F = center of conical base (base is a circle)
AF = height of cone = 12 cm
EF = radius of cone = y
AE = slant height of cone = x
AB = radius of hemisphere = x
AB and AE are radii of the same semi-circle, which is why they are equal.
From here, you would use the tangent ratio to find the value of y
hint: tan(60) = 12/y
Also, you would use the sine ratio to find the value of x
hint: sin(60) = 12/x
Assuming you found y first, you could also use the cosine ratio
cos(60) = y/x
or you could use the pythagorean theorem
12^2+y^2 = x^2
There are a number of ways to determine x assuming you found y first
After computing x and y, you will then have enough information to find the volumes of the cone and hemisphere. After this point, you subtract the volumes to get the leftover space that isn't taken up by the cone. You can think of this as the volume of the "empty air" inside the hemisphere so to speak.
Useful formulas
Volume of cone = (1/3)*(area of circular base)*(height)
Volume of cone = (1/3)*(pi*r^2)*(h)
Volume of hemisphere = (1/2)*(volume of sphere)
Volume of hemisphere = (1/2)*( (4/3)*pi*r^3 )
Volume of hemisphere = (2/3)*pi*r^3
Hopefully this clears things up and you are able to finish the problem. If not, then please let me know. Thank you.
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