(1) When t is what value, two moving points meet for the first time.
(2) When t is what value, the area of a triangle with points P, Q and C as its vertices is 8√3 during the process from departure to first encounter? Friendly reminder: the right side of a right triangle with a 30-degree angle is equal to half of the hypotenuse.
2. As shown in the figure, in the rectangular ABCD, AB=4cm, BC=8cm, the moving point M starts from point D and moves along the dashed line DCBAD at a speed of 2cm/s, while the moving point N starts from point D and moves along the dashed line DABCD at a speed of1cm/s..
(1) If the moving points m and n start at the same time, how many seconds will they intersect at two points?
(2) If point E is on a straight line BC, BE=2cm, moving points M and N start at the same time and stop moving when they meet, then when point M moves for a few seconds, it can just form a parallelogram with points A, E and N?
Answer:
(1) solution: according to the meaning of the question, 2t= 10+ 10+t,
Solution: t=20,
A: When t is 20, the two moving points meet for the first time.
(2) Solution: △ABC is an equilateral triangle with a side length of 10cm.
∴∠C=60，
There are four situations: ① As shown in figure 1, q is QH⊥BC in H.
CQ=2t, ∠ hqc = 30, CH=t, which is obtained from Pythagorean theorem: qh = $ \ sqrt {3} $ t.
From the triangle area formula: $ \ frac {1} {2} $ (10-t)? $\sqrt{3}$t=8$\sqrt{3}$，
Solution: t=2, t=8 (truncation);
② As shown in Figure 2,
BQ=20-2t，BH= 10-t，QH=( 10-t)$\sqrt{3}$，
According to the triangle area formula: $ \ frac {1} {2} $ (10-t) $ \ sqrt {3} $ t = 8 $ \ sqrt {3} $,
Solution: t=2 or t=8,
When t=2, q is on AC, and it is discarded.
∴t=8；
③ As shown in Figure 3, QH⊥AC is in H, CP=t- 10, AQ=2t- 10, AH=t-5, QH=(t-5)$\sqrt{3}$,
∴$\frac{ 1}{2}$(t- 10)(t-5)$\sqrt{3}$=8$\sqrt{3}$
This equation has no solution;
④ As shown in Figure 4: CQ = 30-2t, CP=t- 10, CH = $ \ frac {1} {2} $ (t-10), and pH = $ \ frac {\ sqrt {3.
∴$\frac{ 1}{2}$(30-2t)？ $ \ frac { \ sqrt { 3 } } { 2 } $(t- 10)= 8 $ \ sqrt { 3 } $，
Solution: t=$\frac{25+\sqrt{89}}{2}$ or t=$\frac{25-\sqrt{89}}{2}$,
Q is not on BC at this time, so it is not appropriate to give up;
The value of ∴t is 8, 2,
A: During the period from departure to the first encounter, when t is 8 and 2, the area of the triangle with points P, Q and C as its vertices is 8$\sqrt{3}$cm2.
2. Solution: (1) If two points meet in t seconds, there is t+2t=24.
The solution is t = 8.
A: In 8 seconds, we will meet at 2 o'clock. (4 points)
(2) According to (1), point N always moves on AD, so when point M moves to the side of BC, points A, E, M and N can form a parallelogram.
Suppose that after x seconds, four points can form a parallelogram. There are two situations: (1 min)
(1) x-(2x-4) = 2, x=2, (4 points).
②2x-6-4=8-x, x=6, (4 points)
A: In the second or sixth second, points A, E, M and N form a parallelogram.