First of all, I deeply regret for supplying the wrong answer choices, which were meant for some other question. Please consider the following answer choices for this question:
A. 150
B. 180
C. 210
D. 250
E. 270
Please draw on your own according to the following explanation:
∆ABC has AB = 6, BC = 10, and CA = 8, and consequently it is right angled at A. Mathsbuddy did justice to the situation that I described in the question. Now drop perpendicular from A on BC to meet BC in D. This AD would serve as the common radius in the resulting solid so generated. Area of ∆ABC is ½ AB.AC = ½ BC.AD = 24, hence AD = 4.8; for approximation purposes we'll later on take AD = 5. Revising properties of right triangles anybody could come to believe that the three triangles so formed are all similar. Set up the following similarity relation
AB/AC = BD/AD = AD/DC= 6/8 (*This step is actually not very necessary, this we'd realise towards the end of our calculations.)
Since AD = 4.8, hence 4.8/DC = 6/8 or DC = 6.4 and hence BD = 3.6, this solves Uva@90's 'H' issue. Yes, it does generate two cones surmounted over one another as correctly and beautifully pictured by Mathsbuddy.
The collective volume = 1/3 π (5)^2 (3.6) + 1/3 π (5)^2 (6.4) [Let's take π = 3 only]
This would result in
The collective volume = (5)^2 (3.6) + (5)^2 (6.4) = (5) ^2 (*3.6 + *6.4) = [spoiler](25) (*10) = 250. It's D.[/spoiler]
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
