1)abcda1b1c1d1 прямоугольный параллелепипед ab=4 aa1=6 a1c=корень из 56. Найти: АД,...

$|a_1c|^2 = |aa_1|^2 + |ab|^2 + |bc|^2\\ |bc^2| = |a_1c|^2 - |aa_1|^2 - |ab|^2 = 56 - 36 - 16 = 4\\ |bc| = 2 = |ad|\\ V = |aa_1|*|ab|*|ad| = 6 * 4 * 2 = 48\\ S = 2*(|aa_1|*|ab| + |aa_1|*|ad| + |ab|+|ad|) = \\ =2(6*4 + 6*2 + 4*2) = 2(24+12+8)=2*44 = 88$

$|ab|=4; |ad|=3;\\ |df|=|ab|\sin 60^0 = 2\sqrt{3}\\ |de| = |cd|\cos 60^0 = 2\\ |ac| = \sqrt{|ae|^2 + |ec|^2} = \sqrt{25+12}=\sqrt{37}\\ |bd| = \sqrt{|df|^2+|bf|^2} = \sqrt{12 + 1} = \sqrt{13}\\ |b_1d|^2 = |bd|^2+h^2 = 13 + h^2 = |ac|^2 = 37\\ h = \sqrt{37-13} = \sqrt{24} = 2\sqrt{6}\\ V = |ad|*|df|*h = 3*2\sqrt{3}*2\sqrt{6} = 36\sqrt{2}\\ S = 2(3h + 4h + 3|df|) = 2(24\sqrt{6} + 6\sqrt{3}) = 48\sqrt{6}+12\sqrt{3}$