A triangle ABC has sides of length a, b and c.
A solid composed from two cones is produced by rotating the triangle by 360° about the side of length a. This process is then repeated for sides b and c to produce two more solids, both formed from pairs of cones.
Find the ratio of the volumes of the resulting solids.
I'm not sure what the solver is looking for but it probably:
The extended ratio of the volumes is the same as the extended ratio of the triangle's three altitudes.
let h,i,j be the altitudes to sides a,b,c respectively.
A=ah/2=bi/2=cj/2
ah=bi=cj
For the revolution about side a (assuming at first an acute triangle*) the result is a pair of cones attached at the bases with radius h and heights that sum to a. Total volume V(a) = (1/3)pi(h^2)a
Likewise V(b)=(1/3)pi(i^2)b, V(c)=(1/3)pi(j^2)c.
Form the extended ratio and reduce the equal parts and you are left with h:i:j.
(*It doesn't matter if the triangle is right or obtuse. You just get a single cone of height a or the difference of the heights is a)
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Posted by Jer
on 2024-07-20 18:16:12 |
Solution?
