As shown in the following figure, in Δ ABC, ∠ B = 90 ° and point P moves from point a to point B at the speed of 1cm / s along the edge of AB, and point Q starts from point B and moves along the edge of BC to point C Cm / sec.) (1) If P and Q start from a and B at the same time, and P to B continue to move on the BC side, and Q to C continue to move on the CA side. After a few seconds, the area of Δ PCQ is equal to 12.6 cm? 2 Point Q starts from point B and moves along the edge of BC towards point C at a speed of 2 cm / s

As shown in the following figure, in Δ ABC, ∠ B = 90 ° and point P moves from point a to point B at the speed of 1cm / s along the edge of AB, and point Q starts from point B and moves along the edge of BC to point C Cm / sec.) (1) If P and Q start from a and B at the same time, and P to B continue to move on the BC side, and Q to C continue to move on the CA side. After a few seconds, the area of Δ PCQ is equal to 12.6 cm? 2 Point Q starts from point B and moves along the edge of BC towards point C at a speed of 2 cm / s

As shown in the figure, in △ ABC, ab = 6cm, BC = 8cm, ∠ B = 90 °, point P starts from point a and moves along side AB at a speed of 1cm / s, and point Q starts from point B and moves along BC edge point C at a speed of 2cm / s
(1) If points P and Q start from a and B at the same time, and P reaches point B and then continues to advance on edge BC, and Q reaches point C and continues to advance on edge Ca, after a few seconds, the area of △ PCQ is equal to 12.6cm2?
According to Pythagorean theorem, AC = 10 cm
It takes 6 / 1 = 6S for P to move to B
It takes 8 / 2 = 4 s for Q to move to C
A. When t

In the right triangle ABC, ∠ B = 90 degrees, ab = 6cm, BC = 12cm, and point P moves from point a along the edge of AB to point B at the speed of 1cm per second Point Q starts from point B and moves along BC at the speed of 2cm per second. If P and Q start at the same time, after a few seconds, triangle PBQ is similar to triangle ABC

There are two situations,
1、BP/BC=BQ/AB,
Let two triangles be similar after T seconds,
(6-t)/12=2t/6,
t=1.2(s),
2、BP/AB=BQ/BC,
（6-t)/6=2t/12,
t=3(s)..

In △ ABC, ab = 6cm, BC = 8cm, ∠ B = 90 °, point P starts from point a and moves along the edge of AB at a speed of 1cm / s, and point Q starts from point B and moves along point B In △ ABC, ab = 6 cm, BC = 8 cm, ∠ B = 90 °, point P moves from point a along the edge of AB to B at a speed of 1 cm / s, and point Q moves from point B along the edge of BC to point C at a speed of 2 cm / S. if P and Q start from a and B at the same time, after several seconds, the area of △ PBD is the largest? What is the largest area

According to the question, AP = t, BQ = 2T
So Pb = 6-t
So the s triangle PBQ = 2T (6-T) / 2 = - t squared + 6T
So when t = 3, s = 9cm2

As shown in the figure, in RT △ ABC, ∠ B = 90 °, ab = 6cm, BC = 3cm, point P starts from point a and moves along the edge of AB to point B at a speed of 1cm / s, and point Q starts from point B along the edge of BC to point C at a speed of 2cm / s. if P and Q start at the same time, after a few seconds, the distance between P and Q is equal to 4 2cm？

After x seconds, PQ = 4
2cm
According to the meaning of the title: (2x) 2 + (6 − x) 2 = (4)
2) 2
The results are as follows: (5x-2) (X-2) = 0,
The solution is: X1 = 2
5，x2＝2
ν 3x = BC
A: 2
After 5 seconds, PQ = 4
2（cm）

As shown in the figure, ∠ B = 90 ° in △ ABC, point P starts from point a along the edge of AB and moves at a speed of 1cm / s to B, and point Q moves from point B along the edge of BC to point C at a speed of 2cm / s (1) If P and Q start from a and B at the same time, after a few seconds, the area of △ PBQ is equal to 8cm2? (2) If P and Q start from a and B at the same time, and P to B continue to move on the BC side, and Q to C and then continue to move on the CA side, after a few seconds, the area of △ PCQ is equal to 12.6cm2?

(1) Let p be on AB and Q on BC for x seconds, and the area of △ PBQ is 8cm2,
According to the meaning of the title, 1
2 (6-x) · 2x = 8, then X1 = 2, X2 = 4,
After 2 seconds, point P is 4 cm away from point B and point q is 4 cm away from point B;
Or after 4 seconds, the area of △ PBQ is 8cm2 from point P to 2cm from point B and from point Q to 8cm from point B,
In conclusion, the area of △ PBQ is 8cm2 after 2 or 4 seconds;
(2) When p is on AB, the area of △ PCQ is 1
2×PB×CQ=1
2×（6-x）（8-2x）=12.6，
The solution is: X1 = 25 + 2
Eighty-five
5, X2 = 25 − 2
Eighty-five
5，
After x seconds, point P moves to BC with CP = (14-x) cm, point Q moves to Ca, and CQ = (2x-8) cm,
QD ⊥ CB is made by passing Q, and the vertical foot is D, and QD is obtained from ⊥ CQD ≁ cab
2x−8＝AB
AC，
That is, QD = 6 (2x − 8)
10，
According to the meaning of the title, 1
2（14-x）•6(2x−8)
10 = 12.6, the solution is X1 = 7, X2 = 11
After 7 seconds, point P is 7cm away from point C on BC and 6cm away from point C on Ca, so that the area of △ PCQ is equal to 12.6cm2
After 11 seconds, point P is 3cm from point C on BC, point q is 14cm from point C on Ca, point q is beyond the range of Ca, and this solution does not exist
To sum up, after 7 seconds and 25 − 2
Eighty-five
At 5 seconds, the area of △ PCQ is equal to 12.6cm2

As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 6, BC = 8, the moving point P starts from point a, and moves along the edge AC to point C at a rate of 1 As shown in the figure, in RT △ ABC, ∠ C = 90 °, AC = 6, BC = 8, the moving point P starts from point a, and moves along the edge AC to point C in 1 unit per second   Moving point d starts from point a and moves at a speed of 5 / 3 unit length per second along side AB to point B, and it can keep the straight line PD ⊥ AC connecting the two moving points all the time. Moving point Q starts from point C and moves along CB to point B at the speed of 2 unit length per second. Connecting PQ, points P, D and Q starts from point a and C respectively. When one of the points reaches the end point, the other two points also stop moving, Let the movement time be T seconds (t ≥ 0) (1) When t is the value, the area of the quadrilateral bqpd is half of the area of △ ABC (2) Is there a value of t that makes the quadrilateral pdbq parallelogram? If so, find the value of T; if not, explain the reason (3) Is there a value of t to make the quadrilateral pdbq rhombic? If so, find out the value of T; if not, explain the reasons, and explore how to change the speed of point Q (uniform motion), so that the quadrilateral pdbq is diamond at a certain time, and calculate the speed of point Q  

Please accept!

As shown in the figure, in the RT triangle ABC, the angle B = 90 degrees, point P starts from point B and moves along side Ba to point a at a speed of 1cm / s, and at the same time, point Q starts from point B along the edge BC moves to point C at a speed of 2cm / s. After a few seconds, the area of triangle bpq is 36cm square

Let the area of triangle bpq be equal to 36 square centimeters after T seconds
1/2╳t╳2t=36
T = 6
So six seconds later, the area is 36 square centimeters

As shown in the figure, ∠ C = 90 ° AC = 4, BC = 3 in △ ABC. The center P of a circle with radius 1 moves along AC direction from point a at a speed of 1 unit / s, and the moving time is t (unit: s) (1) When t is the value, ⊙ P is tangent to ab; (2) Let PD ⊥ AC intersect AB at point D, if ⊙ P and line segment BC intersect point E, it is proved that if t = 16 Pdbe is parallelogram

(1) When ⊙ P is tangent to AB in motion,
If the tangent point is m and connected with PM, then ∠ amp = 90 °,
∴△APM∽△ABC，
∴AP
AB=PM
BC，
∵AP=t，AB=
AC2+BC2=5，
∴t
5=1
3，
∴t=5
3. (4 points)
(2) Proof: BC ⊥ AC, PD ⊥ AC,
∴BC∥DP，
When t = 16
At 5 s, AP = 16
5，
∴PC=4-16
5=4
5，
∴EC=
PE2-PC2=
12-(4
5)2=3
Five
∴BE=BC-EC=3-3
5=12
5，
∵△ADP∽△ABC，
∴PD
BC=AP
AC，
∴PD
3=16
Five
4，
∴PD=12
5，
∴PD=BE，
When t = 16
Pdbe is parallelogram

As shown in the figure, in △ ABC, ∠ C = 90 °, AC = 3cm, BC = 4cm. Starting from point B, the moving point P moves to point C at the speed of 2cm / s, and the moving point Q moves from C to point a at the speed of 1cm / s. if the moving points P and Q start at the same time, how many seconds does it take to make △ CPQ similar to △ CBA?

If the two triangles are similar after T seconds, the solution can be divided into the following two cases,
① If RT △ ABC ∽ RT △ QPC, AC
BC＝ QC
PC, i.e. 3
4＝ t
4 − 2T, t = 1.2;
② If RT △ ABC ∽ RT △ PQC, PC
QC＝ AC
BC，4−2t
t＝ 3
4, t = 16
11；
From the velocity of point P on the side of BC is 2cm / s and that of point Q on the side of AC is 1cm / s, the value range of t should be 0 < T < 2,
Therefore, it can be known that the time required for △ CPQ to be similar to △ CBA is 1.2 or 16
11 seconds

As shown in the figure, the two right angle sides of RT △ ABC are ab = 4cm and AC = 3cm. Point d moves from a to B along AB at a speed of 1cm / s. at the same time, point e moves along BC from B to C at a speed of 2cm / s. when moving point e reaches point C, the movement stops. Connect De, CD and AE (1) When the moving point moves for several seconds, △ BDE is similar to △ ABC? (2) Let the area of △ ade be s when the moving point moves for T seconds, and then find the function analytic formula of S and t; (3) Is there a certain time t in the process of movement, which makes CD ⊥ de? If it exists, find the time t; if not, please explain the reason

If the movement time of point D is t, then ad = t, BD = 4-T, be = 2T, CE = 5-2t (0 ≤ t ≤ 5)
2），
(1) When ∠ BDE = BAC, i.e., ed ⊥ AB, RT △ BDE ∽ RT △ BAC,
ν BD: Ba = be: BC, i.e. (4-T): 4 = 2T: 5,
∴t=20
13；
When BDE = BAC, i.e. de ⊥ AB, RT ⊥ BDE ∽ RT ⊥ BCA,
ν BD: BC = be: Ba, i.e. (4-T): 5 = 2T: 4,
∴t=8
7；
So when the moving point moves 20
13 seconds or 8
At 7 seconds, △ BDE is similar to △ ABC;
(2) Use e to make ef ⊥ AB in F, as shown in Fig,
It is easy to prove that RT ∽ bef ﹤ RT △ BAC,
/ / EF: AC = BF: ab = be: BC, that is, EF: 3 = BF: 4 = 2T: 5,
∴EF=6t
5，BF=8t
5，
∴S=1
2AD•EF=1
2•t•6t
5=3
5t2（0≤t≤5
2）；
(3) Exist
DF=AB-AD-BF=4-t-8t
5=4-13
5t，
If CD ⊥ De,
It is easy to prove that RT △ ACD ∽ RT △ FDE,
/ / AC: DF = ad: EF, i.e. 3: (4-13)
5t）=t：6t
5，
∴t=2
13．