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In trapezoidal ABCO, OC ∥ AB establishes a plane rectangular coordinate system with o as the origin. The coordinates of a, B and C are a (8,0), B (8,10) and C respectively In trapezoidal ABCO, OC ‖ The coordinates of a, B and C are a (8,0), B (8,10) and C (0,4) respectively. Point d (4,7) is the midpoint of line BC, and the moving point P starts from point O and moves along the route of broken line OAB at the speed of 1 unit per second. The moving time is T seconds (1) Find the analytical formula of the straight line BC; (2) Let the area of △ OPD be s, the functional relationship between S and t is obtained, and the value range of independent variable t is pointed out; (3) When t is, the area of △ OPD is the same as that of trapezoidal oabc
(1) Let the analytic expression of the line BC be y = KX + B,
Substituting C (0,4) and B (8,10) into the equation, the following results are obtained
4=0×k+b10=8×k+b,
The solution is: K=
34b=4,
That is, y = 34x + 4,
So the analytical formula of the line BC is y = 34x + 4
(2) There are two cases
① When P moves on OA;
The height of the edge op of △ OPD is 7,
The area of ∧ OPD is as follows:
S=12×t×7
That is s = 72t (0 < T ≤ 8),
② When P moves on AB:
∵A（8,0）,B（8,10）,C（0,4）,D（4,7）,
The area of △ ODC is:
S1=12×4×4=8,
The area of △ OPA is:
S2=12×8×（t-8）=4t-32,
The area of △ DBP is:
S3=12×{10-（t-8）}×（8-4）=36-2t,
The area of quadrilateral oabc is:
S4=12×（4+10）×8=56,
The area of ∧ ODP is as follows:
S=S4-S1-S2-S3=56-8-（4t-32）-（36-2t）=-2t+44,
That is s = - 2T + 44 (8 < T < 18),
∴S=72t(0＜t≤8)-2t+44(8＜t＜18)；
(3) It can be seen from (2)
a：72t=38×56,
The solution is t = 6 seconds,
b：-2t+44=38×56,
The solution is t = 11.5 seconds,
T = 6 seconds or T = 11.5 seconds
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