| A Volume Problem |
In triangle ABC, AB = 25, BC = 16, and AC = 39. If ABC is rotated about its shortest side, what is the volume of the resultant solid? |
Problem Moderated by: Sasha |
| Problem Solution |
Draw a perpendicular from A to line BC. I labeled the intersection D. Let x = AD and y = BD. By the pythagorean theorem , we know that
x2 + y2 = 252 and
x2 + (y+16)2 = 392
Subtract these equations to obtain an equation in y yields:
(y+16)2 - y2 = 392 - 252.
y2 + 32y + 256 - y2 = 896
32y = 640
y = 20
Plug this into the second equation to obtain:
x2 + (20)2 = 252
x2 = 225
x = 15
Now with these two numbers find the volume of the cone when A is revolved about CD (cone1).
V(cone1) = Bh/3
V(cone1) = π(15)2(16+20)/3
V(cone1) = 225π(36)/3
V(cone1) = 2700π
Now find the volume of the cone when A is revolved about BD (cone2):
V(cone2) = Bh/3
V(cone2) = π152(20)/3
V(cone2) = 225π(20)/3
V(cone2) = 1500π
Subtracting the volume of cone2 from cone1 yielded the volume of A revolved about CB.
2700π - 1500π
1200π
Solution submitted by Roth |
|
|
