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2019年3月2日 · HKDSE Core Math 2012 Paper 1 (Solutions) 4 12. HKDSE 2012 Q.12 1 (a) Volume = 3 𝜋(48)2(96) 73728𝜋cm3 (b) (i) Volume= 2 3 𝜋(60)3=144 000𝜋 cm3 2 (ii) The height of the frustum under the surface of the milk =√602−48=36 cm Let 𝑟 cm be the base radius of
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3 = cm3. Let h cm be the height of the frustum under the surface of the milk and cm be the base radius of the circular cone above the surface of the milk. h. = √602 − 482 = 36. 96 − 36 =. 48 96 = 30 The volume of the milk remaining in the vessel. = 144000. − (73728. 0)2(96 − 36)) = 88272cm3. 277.
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