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If a right circular cone is separated into solids of volumes V1, V2 , V3 by two planes parallel to the base, which also trisect the altitude, then V1 : V2 : V3 is
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- 1: 2 : 3
- 1: 4 : 6
- 1: 6 : 9
- 1: 7 : 19
- 1: 2 : 3
Correct Option: D
Let FQ = r1, DP = r2 and BO = r3
Also, AQ = QP = PO = h/3
From ∆AFQ and ∆ADP,
| = | |||
| BO | AO |
| ⇒ | = | ⇒ r3 = 3r1 | ||
| r3 | 3 |
| ∴ V1 : V2 : V3 = | πr1² × | : | π | (r1² + r2² + r1r2) | : | π | (r2² + r3² + r2r3) | ||||||
| 3 | 3 | 3 | 3 | 3 | 3 |
= r1² : (r1² + 4r1² + 2r1²) : (4r1² + 9r1² + 6r1²)
= r1² : 7r1² : 19r1²
= 1 : 7 : 19
