(1) When point P reaches the end point C, find the value of t and point out the length of BQ at this time;
(2) When point P moves to AD, why does the value of t make PQ∨DC? ?
(3) Let the area of light QK sweeping the trapezoidal ABCD be S, and find out the functional relationship between S and T when point E moves to CD and DA respectively; (Don't write the range of T)
(4) Can △ PQE become a right triangle? If yes, write the range of t; If not, please explain why.
Solution: (1)t? = (50+75+50) ÷ 5 = 35 (seconds), and point P reaches the end point C. ............................................ (1 min).
At this point, QC=35×3= 105, and the length of ∴BQ is135-105 = 30 ............................................. (2 points).
(2) As shown in figure 1, if PQCD, AD∨BC, then the quadrilateral PQCD.
Is a parallelogram, so that PD=QC, from QC=3t, BA+AP=5t.
50+75-5t = 3t, and t = is the solution.
After inspection, when t=, PQ ∑ DC ....................................... (4 points)
(3)① When point E moves on the CD, as shown in Figure 2, it passes through point A and point D respectively.
Let AF⊥BC be at point F, DH⊥BC be at point H, and then a quadrilateral.
ADHF is rectangular, while △ ABF △ DCH, therefore,
FH=AD=75, so BF = CH = 30. ∴ DH = AF = 40。
And QC=3t, so QE = QC tanc = 3t = 4t.
(Note: Similar triangles can also be used to solve)
∴S=S⊿QCE？ = QE QC = 6 T2； .................................... (6 points)
(2) When point E moves on DA, as shown in figure 1. Passing through DH⊥BC at point D is point H, from ①, DH=40, CH=30, QC=3t, so ED = QH = QC-CH = 3t-30.
∴S=S trapezoidal QCDE? = (ed+QC) DH =120t-600 .......................... (8 points).
(4)△PQE can be changed into a right triangle, and ........................................................................................................................................................ (9) scored 9 points.
When △PQE is a right triangle, the value range of t is 0 < t ≤ 25 and t≠ or t = 35...( 12 minute).
(Note: (4) If you don't answer t≠ or t=35 in the question, deduct 1 point, and the rest of the writing methods are given as appropriate. )
The following is the answer to question (4) for teachers' reference only:
(1) when point p is on BA (including point a), that is, 0 < t ≤ 10, as shown in figure 2. If point P is PG⊥BC of point G, then PG = Pb SINB = 4t, QE=4t? =PG, it is easy to get that the quadrilateral PGQE is a rectangle, and at this time △PQE can always become a right triangle.
② When both point P and point E are on AD (excluding point A but including point D), that is, 10 < t ≤ 25, as shown in figure 1.
According to QK⊥BC and AD ∨BC, at this time △PQE is a right triangle, but points P and E cannot coincide, that is,
5t-50+3t-30 ≠ 75，t≦。
③ When point P is on DC (excluding point D but including point C),
That is, when 25 < t ≤ 35, as shown in figure 10, ed > 25× 3-30 = 45,
It can be seen that point P is outside the circle with QE = 40°, so
∠EPQ will not be a right angle.
From ∠PEQ < ∠ dirk, we can know that ∠PEQ must be an acute angle.
For ∠PQE, ∠PQE≤∠CQE, only when points p and c
Coincidence, that is, when t = 35, as shown in Figure 4, ∠ PQE = 90, △PQE.
This is a right triangle.
To sum up, when △PQE is a right triangle, the value range of t is 0 < t ≤ 25 and t≠ or t = 35.