Объяснение:
1) угол ACD = 90-CAD=60 => EDC = 90-CAD=30°. AB = 8/tg(BAC) = 8/√3, AO=AC/2 = 8/√3 => S(ABO) = (1/2)sin(60°)*AO*AB = (8*8)/(2√3) = 32/√3. S(BCO) = (1/2)sin(30°)*OC*BC = (8*8)/(2√3) =32/√3
2) CE=ED=BF=4 => AF=4/√3 => AD = AF+ED+BC = 7+4/√3. S(ABCD) = (AD+BC)/2 * CE = 2*(7+3+4/√3) = 20+8/√3
