Question

Очень срочно решение, даю максимум баллов!!!

Answer 1

$A_1C = 2AA_1 = 2 * 6 = 12\\ \sin 30 = \frac{AC}{A_1C} = \frac{ \sqrt{3} }{2} \\ AC = \frac{CA_1 \sqrt{3} }{2} = \frac{12 \sqrt{3} }{2} = 6\sqrt{3} \\ AO= \frac{AC}{2} = \frac{6 \sqrt{3} }{2} = 3\sqrt{3} \\ \sin 60 =\frac{AA_1}{DB_1}=\frac{\sqrt{3}}{2} \\ DB_1=\frac{6\sqrt{3}}{2}=3 \sqrt{3} \\ \cos 60 = \frac{BD}{B_1D} = \frac{1}{2} \\ BD = \frac{1}{2} B_1D = \frac{3 \sqrt{3}}{2} = 1.5 \sqrt{3} \\ OD=\frac{BD}{2} = \frac{1.5 \sqrt{3}}{2} = 0.75 \sqrt{3} \\ AD = \sqrt{AO^2+OD^2} = \sqrt{ {(3\sqrt{3} )}^{2} + {(0.75 \sqrt{3})}^{2} } = \frac{3\sqrt{51}}{4}\\ AD=AB,так \: как \: ромб \\ V=AD^2*AA_1 \\ V= {(\frac{3\sqrt{51}}{4})}^{2} *6 = 172.125$