When an acute triangle is rotated about its longest side the resulting volume is 2880π, when it is rotated along its medium side the resulting volume is 3240π, and when it is rotated about its shortest side the resulting volume is
25920π/7. Find the perimeter of the triangle.
Let a, b, and c be the lengths of the sides of the triangle. Let Va, Vb, Vc be the volumes of the revolutions. Then Va=2880*pi, Vb=3240*pi, and Vc=25920*pi/7.
Then Va/Vb = (2880*pi)/(3240*pi) = 8/9 = b/a, and Vb/Vc = (3240*pi)/(25920*pi/7) = 7/8 = c/b,
Then there is some value k such that a=9k, b=8k, c=7k. The area of this triangle is k^2*sqrt(720).
Then Va = 4*pi*(k^2*sqrt(720))^2/(9k) = 2880*pi, which simplifies to k^3*320 = 2880. Then k=cbrt(9), which makes the perimeter equal to 24*cbrt(9).
Solution
