A junior high school math problem about moving point question 5

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Problem description: In ABC, angle b=90, ab=5, bc=7Point P starts from point A and moves along the edge of AB to point B at a speed of 1 s, and point Q starts from point B and moves along the edge of BC to point C at a speed of 2 s (1) If P and Q start from A and B at the same time, then after a few seconds, the area of Pbq is equal to 4? (2) If p and q depart from a and b at the same time, then after a few seconds, the length of pq is equal to 5?

3) Can the area of pqb in (1) be equal to 7? Explain the reason for the lack of clarification (process).

Analysis: Solution, after x seconds, the area of pbq is equal to 4, then there is, ap=x, pb=5-x, bq=2x, qc=7-2x, angle b=90, so 1 2*(5-x)*2x=4 The solution key gives x=1 or 4

2) The length of pq after x seconds is equal to, pb=5-x, bq=2x, qc=7-2x, angle b=90, so (5-x) 2+4x 2=5 is solved x=2

3) To 1 2*(5-x)*2x=7 to get x 2-5x+7=0 because b 2-4ac=-3<0 so the equation has no real solution, so it can't.

Analysis: Def moves at a constant speed from the position in Figure 1 along Cb to ABC at a speed of 1 unit per second, while point P moves from A to point B at a constant speed of 1 unit per second along AB, and the right-angled edge of AC and def intersects at Q

P, E move at the same velocity, and at the same time, the Q point also moves from C to Ca at the same velocity.

During the move, |pa|=|ec|=|cq|Remain the same when |eq|=|ed|, Q is the intersection of AC and ED, after which Q will become the intersection of AC and DF, and Q will no longer move towards A, but will return to C in the direction of AC

The above is the trajectory of the movement of the point q.

Movement speed v=1, s=t

ac=8,bc=6, ∠acb=∠edf=90°，∠def=45°，ef=10

pa|=|ec|=|cq|=t

aq|=8-t

When pqe=90, apq=90

At this point, |ap|=|aq|cos∠paq=(8-t)*4/5=t

The solution yields t=32 9<5

There is that pqe is a right-angled triangle, t=32 9 def starts from the starting position and moves at a constant speed along cb to abc at a speed of 1 unit per second, and the point p starts from a and moves at a uniform speed along ab at a constant speed of 1 UNIT PER SECOND TOWARDS POINT B, AND THE RIGHT-ANGLED EDGES OF AC AND DEF INTERSECT AT POINT Q

P, E move at equal velocity, and at the same time, point Q also moves from C along Ca to A at the same speed.

During the move, |pa|=|ec|=|cq|Unchanged, when|eq|=|ed|, Q is the intersection of AC and ED, after which Q will become the intersection of AC and DF, and Q will no longer move towards A, but will return to C in the direction of AC

The above is the trajectory of the movement of the point q.

Movement speed v=1

s=tac=8,bc=6, ∠acb=∠edf=90°，∠def=45°，ef=10

pa|=|ec|=|cq|=t

aq|=8-t

When pqe=90, apq=90

At this point, |ap|=|aq|cos∠paq=(8-t)4/5=t

The solution yields t=32 9<5

The presence of pqe is a right triangle, t=32 9

The def starts from the starting position and moves at a constant speed along CB towards ABC at a speed of 1 unit per second, and the point P starts from A and moves at a constant speed along AB at a constant speed of 1 unit per second towards point B, and the right-angled edges of AC and def intersect at point Q

P, E move at equal velocity, and at the same time, point Q also moves from C along Ca to A at the same speed During the movement, |pa|=|ec|=|cq|Unchanged, when|eq|=|ed|, Q is the intersection of AC and ED, after which Q will become the intersection of AC and DF, and Q will no longer move towards A, but will return to C in the direction of AC

The above is the trajectory of the movement of the point q.

Movement speed v=1

s=tac=8,bc=6, ∠acb=∠edf=90°，∠def=45°，ef=10

pa|=|ec|=|cq|=t

aq|=8-t

When pqe=90, apq=90

At this point, |ap|=|aq|cos paq = (8-t)4 5 = t solution t=32 9<5

The presence of pqe is a right triangle, t=32 9

See if it's this? (*Hee-hee.......)Solution: (1) ACB = 45°, DEF = 90°, EQC = 45°

ec = eq = t, be = 9 t i.e.: (

2) When dq = dp, 6 t = 10 3t, the solution is: t = 2s

When pq = pd, pass p and pass de to point h, then dh = hq=, by hp ef, then, solution s is obtained

When qp = qd, after q, and dp is crossed at the point g, then gd = gp=, and we get: dqg dfe, then, the solution is s

3) Suppose there is a certain moment t, so that the points p, q, and f are in the same straight line.

then, through p made, pass bf to the point i, pi de, then: then, solve :s

Answer: When s, the points p, q, and f are in the same straight line

I'm also in junior high school, and I hate moving problems the most

What about the figure? Give me a picture and I'll help you answer.

This seems to be some kind of analog volume.

Ethylene bustles with the number of partners and users with Phoenix.

It is to find the maximum area of the triangle PBQ.

Let the time be t, bp=12-2t, bq=4t, triangle area s=4t(6-t)=4(9-(t-3) 2)(t<6), t=3 when s is maximum, so after three seconds.

Solution: 1. Let the time of Q point motion be x, then when Q moves on OC, its abscissa is 2x*4 5=8x 5, and the ordinate is 2x*3 5=6x 5. When Q moves on CB, its abscissa return is 4+2 (, the ordinate is 3, and X is greater than or less than or equal to this time period.

2. (1) The circumference of the trapezoidal OABC is 5+10+3+14=32, and the velocity of the movement of the Q point is V, then there is VX+X=32 2 according to the title, and the solution obtains V=(16-X) X, so the distance traveled by the Q point is S=16-X

2) Assuming that the area of the trapezoidal OABC can be divided into the same two parts at the same time by the straight line PQ, then according to the title: 1 2*[(16-x-5)+x]*3=1 2*[(15-16+x)+(14-x)]*3, and the solution is: 11=13, which is the result of the contradiction, and it is impossible to divide the area of the trapezoidal OABC into the same two parts at the same time with the straight line PQ.

Analysis: (1) According to the properties of similar triangles, the coordinates of point Q on OC can be obtained; According to the distance, the abscissa of point Q on CB is 2T-5, and the ordinate is consistent with the ordinate of point C, which is 3;

Obviously, at this time, q is on cb, which can be obtained from the knowledge of the parallelogram, and only needs to be solved according to the equation of the op=cq column;

2) Let the velocity of q be v, and the sum of the spring branches of the distance traveled by p and the point q is exactly half of the circumference of the trapezoidal oabc, then the functional relationship can be established;

Obviously, Q should be on CB, according to the conclusion of the area and , the equation about T should be solved, and the solution is Solution: (1) The point Q is Q (8 5T, 6 5T) when the only roll OC is on the mountain

Point Q is on CB when Q (2T-1,3).

Obviously, q is on cb, which can be obtained from the knowledge of the parallelogram, just op=cq

So 2t-5=t gives t=5

2) Let the velocity of q be v, and find the circumference of the trapezoid as 32, so t+vt=16, so v=16-tt

The distance traveled by the point Q is 16-t

No, apparently q should be on cb, first find the area of the trapezoid is 36, you can get [(t-1)+(14-t)] 32=18, at this time the eligible t does not exist Comments: Be proficient in solving the problem of this type of motion according to the properties of similar triangles, the properties of parallelograms, and distance = speed time

Let t seconds after s=8

6-t）2t=16

Just solve it.

4. Assuming that there exists, then it satisfies 1 2*(6-2t)* 3t 2=1 6*s trapezoidal or 1 2*(6-2t)* 3t 2=5 6* s trapezoidal solution equation can solve the t should meet 0 less than or equal to t less than or equal to 3

The third question; sabpqd=s trapezoid—s triangle pcq

spcq=1/

1/2(6-2 t)(3-t).sin60

Give me the mailbox, I'm also in the third year of junior high school, and I just did these questions two days ago, which is good. High school entrance examination questions.

14 (2010, Hengyang, Hunan) It is known that the side length of the equilateral triangle ABC is 4 cm, and the line segment Mn with a length of 1 cm moves to point B at a speed of 1 cm and a second along the AB direction on the edge AB of ABC (at the beginning of the movement, the point coincides with the point, and the motion ends when the point N reaches the point), and the perpendicular line of the edge is made by crossing the point M and N respectively, and the other sides of the ABC intersect at the two points P and Q, and the time of the movement of the line segment MN is seconds

1) In the process of line segment mn, why is the value of the quadrilateral mnqp exactly rectangle? and find the area of the rectangle;

2) In the process of line segment Mn, the area of the quadrilateral MNQP is S, and the time of motion is t Find the functional relationship of the area of the quadrilateral MNQP S with the motion time, and write the value range of the independent variable T

Answer] (1) If the quadrilateral mnqp is rectangular, then there is mp=qn, at this time, since pma= qnb=90°, a= b=60°, so rt pma rt qnb, so am=bnAfter moving t seconds, there is am=t, bn=3-t, from am=bn, t=3-t is obtained t= at this time, rt amp, am=, a=60°, so mp= , and mn=1, so the area of the rectangle is

2) Still according to the idea of the above question, if m,n is divided into the two ends of the middle line of AB at the bottom of the triangle, because AM=T, then MP= T, because BN=4-T-1=3-T, so Nq= (3-T), because MN=1, the area of the trapezoidal MNQP is Mn (MP+QN)= (T+ (3-T))= is a fixed value (i.e., does not change with time). In this case, 1 is required if t<=1 or t 2 then m and n are both on the same side of the bottom midline, as shown in the second and third figures. In the second graph, BM=T, BN=1+T, so the trapezoidal area is S= 1 [ T+ (3-T))]= (2T+1), where 0 t 1

Similarly, the case for 2 t = 3 can be obtained, where the area is s= (7-2t).

I'm sorry, I can't help myself, I have a lot of review materials for the high school entrance examination, but it's not good to give it to you, and the questions with moving points have pictures, and I don't know how to send them to you