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If h, C, V are respectively the height, the curved surface area and volume of a cone, then 3πV3 – C2h2 + 9V2 is equal to(a) 0 (b) 1 (c) 2 (d) 3 |
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Answer» Answer: (a) = 0 C = πrl = \(\pi r\sqrt{h^2+r^2}\) and V =\(\frac{1}{3}\pi r^2 h\) where, r and l are respectively the radius of the base and slant height of the cone. ∴ 3πVh3 – C2h2 + 9V2 = 3π × \(\frac{1}{3}\pi r^2 h\) ×h3 – π2r2(h2+r2)h2 + 9 × \(\frac{1}{9}\pi^2 r^4 h^2\) = π2r2h4 – π2r2(h2+r2)h2 – π2r4h2 = 0. |
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