1. The area of the quadrilateral AEBD is SAEBD=SAEBO+SAOD, and the area of the triangular AOD is easy to find. Just use AO*DO/2, ao = ad * sin60 = 5 √ 3 (5th root number 3), and DO = AD * SIN 30 = 5. Don't tell me that I don't know the sum of 30. To use similar triangles's theorem, because triangle CFE and triangle COB are similar triangles, so OB:FE=CO:CF, knowing the length of AO, we can know that the length of AC is 10√3. As I said just now, the angle BAD = 60, the natural angle BAC = 30, and the angle EAB =15, so the full angle EAC.

, that is, 10√3, OB=OD=5, CO=5√3, from the similar triangle formula just now, OB:FE=CO:CF, substitute two data, 5√3? EF=5？ CF, that was out of order. Anyway, CF is FC, as long as you know it. There are two simultaneous expressions, EF+FC= 10√3 5√3? EF=5？ The two unknowns of the two equations of CF can be solved by EF, but I haven't worked out the exact number. There is no pen and paper, and then use AC perpendicular to EF to calculate the area of large triangle AEC. Because the area of triangle AOD is equal to the area of triangle COB, is the area of quadrilateral required in the topic equal to the area of big triangle? Hmm. How interesting

2. the first question, first of all, look at the water inflow, from 4 to 8, 4 hours, 20 cubic meters, that is, 5 cubic meters per hour.

From 8: 00 to 12: 00, the time is also 4 hours, and the water is in and out, which is 10 cubic meter in total, which means that the water is 2.5 cubic meters per hour. The second question, y=kx+b, with x=8, x= 12 as the formula, find k and b, and find it yourself. Without pen and paper, the domain seems meaningless, just (8, 12). The third question is the same as the second one. It's up to you.

Answer so tired, ~ ~ ~ at least give some!