Cau 8:
a/ Hạ BH _|_ AD => tan(^BAC) = BH/AH = 0,75 = 3/4 <=> BH^2/AH^2 = 9/16 <=> BH^2/(AH^2 + BH^2) = 9/25 <=> BH^2/AB^2 = 9/25 <=> BH/AB = 3/5 <=> BH = 3AB/5
=> S(ABCD) => 2S(ABD) => AD.BH = 3AB^2/5 = 117/5 (cm^2)
b/ Đặt AM = x (0 =< x =< 7)
Ta có: MN/AC = BM/AB = (AB - AM)/AB = 1 - AM/AB = 1 - x/7 (1)
MQ/BD = AM/AB = x/7 (2)
(1) nhân (2) : MN.MQ/AC.BD = x(7 - x)/49 <=> S(MNPQ)/2S(ABCD) = x(7 - x)/49
<=> S(MNPQ) = 2x(7 - x).S(ABCD)/49 = 6x(7 - x)/5 = (6/5)(7x - x^2) = (6/5)[49/4 - (49/4 - 2.(7/2)x + x^2)] = (6/5)[49/4 - (x - 7/2)^2] =< (6/5).(49/4) = 117/10 (cm^2)
=> S(MNPQ)max = 117/10 (cm^2) = S(ABCD)/2 khi x = 7/2 = AB/2 hay M là trung điểm AB