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As shown in the figure, in the right angle trapezoid oabc, AB / / OC, O are the origin of the coordinate graph, point a is on the positive half axis of Y axis, and point C is on the positive half axis of X axis
The title is incomplete
As shown in the figure, in the right angle trapezoid oabc, ab ∥ OC, O are the origin of coordinates, point a is on the positive half axis of Y axis, point C is on the positive half axis of X axis, point B coordinates are (2,2 times root sign 3), ∠ BCO = 60 °, oh is perpendicular to BC and point h. moving point P starts from point h, moves along line HO to point O, moving point Q starts from point O, moves along line OA to point a, and two points start at the same time, The velocity is 1 unit length per second. The time of set point motion is T seconds
(1) The length of Oh is calculated;
(2) If the area of triangle OPQ is s (square unit), find the functional relationship between S and T, and find out the value of T, what is the maximum area of triangle OPQ?
(3) Let PQ and ob intersect at point M. ① when △ OPM is an isosceles triangle, find the value of s in (2)
② Explore the maximum length of line segment OM, and write the conclusion directly
（1）
In RT △ HCO, because ∠ och = 60 °, the opposite side of ∠ och is 2 √ 3,
So CB = (2 √ 3) / sin60 ° = 4, OC = 2 + 2 = 4
So: Oh = ∠ the opposite side of och = 2 √ 3
(2)
According to the meaning: OQ = t, HP = t
So OP = oh-ph = (2 √ 3) - t
So: the area of triangle OPQ is s = (1 / 2) * OQ * sin60 ° * Op
Namely: S = t (1 / 2) [(√ 3) / 2] [(2 √ 3) - t]
S=[(-√3)/4]t^2+(3/2)t
Due to - √ 3
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