The bottom edge of isosceles ABC is 8cm, the waist length is AB 5cm, and the moving point is P 20

With point A as the perpendicular line of BC, it is clear that the isosceles ABC can be seen as consisting of two right triangles.

According to the Pythagorean theorem, it can be seen that the height of isosceles ABC on the edge of BC is 3 cm, so there is: sin b=sin c =3 5=; cos b = cos c = 4/5 = ;

There are two positions where point P moves to the position where Pa is perpendicular to the waist, and the one near point B is B'(In this case, Pa is perpendicular to AC), and the one closest to C is C'(At this point, PA is perpendicular to AB), obviously, BB'=cc'=bc-b'c = 8-5/cos c = 8 - 5/ = = cm

So, when p moves perpendicular to ac with pa, the time is: 7 seconds.

And since p moves from b to point c the total time is 8 = 32 seconds, 32 - 7 = 25 (seconds).

Therefore, the time of p movement should be 7 seconds and 25 seconds.

There are two answers. One is perpendicular to AC. The time is seconds. The other is perpendicular to ab and the time is seconds.

You can use the area equality to find it!

The blind area of the triangle ABC is equal to the sum of the areas of the triangle ABP and ACP! Using the properties of the stupid Shenbi isosceles triangle and the Pythagorean theorem to obtain the height on the edge of bc is 3, let bp=x, then pc=8-x using the Pythagorean theorem to find that AP is equal to the square of x under the root sign minus 25, and the length of AB and AP is used to express the area of the three-belt lifting angle ABP, and then the equation can be listed! I did the math and the answer seemed to be 25 seconds!

Give it a try!

[Analysis].

1) According to the perimeter formula, the functional relation of y to x y=10-2x can be obtained;

2) the range of independent variables is calculated by using the triangle side length as a positive number and the trilateral relationship;

3) Directly substitute the formula of x represented by y into the range of values of x to directly solve the inequality group.

Answer] Solution: (1).

Circumference = length of the bottom edge of BC + length of waist AB + length of waist AC = y + 2x = 10, i.e.:

y=10-2x；

2) x 0, y 0, 2x y (according to the sum of the two sides of the triangle is greater than the third side) There is a system of inequality equations:

10-2x＞0

2x＞10-2x

The value range of x is as follows

5/2＜x＜5

3）∵y=10-2x

2x=10-y

i.e. x=5-y2

5 2 x 5

5/2＜5-y/2＜5

The value range of the solution y is:

0＜y＜5

The Handsome Wolf Hunting team will answer for you:

y=10-2x，⑶0

(1) y=10-2x

2) Column two inequalities: 1) 2x>y (the sum of the two sides is greater than the third side) Substituting the function of (1) into the solution to get x>

2) x>0 ;y>0 (the side of the triangle is greater than zero) solves 0-2x>-10

5>10-2x>0

The value range of y is 0

Let P1A vertical AC; then the triangle ACP1 is similar to ACD;

That is, ac:cp1=dc:ac 5:cp1=4:5;i.e. cp1 = 25 4;

then bp1=8-25 4=7 4;then the time of p to p1 is 7 4 seconds;

That is, after 25 seconds of P movement, PA is perpendicular to AC;

In the same way, when P2A is perpendicular AB, BP2=25 4;

That is, after 25 seconds of P movement, PA is perpendicular to AB;

Solution: Two scenarios.

as AD BC

ab=ac=5，bc=8

then AD=3When PA AC, according to APD CPA, it can be obtained: PC=25 4BP=8-25 4=7 4

That is, when the p-point movement is noisy for 7 seconds, the pa ac

Similarly, when pa ab, bp = 25 4

That is, when the p-point is moving for 25 seconds, the finch touches the rubber pa ab

In summary, when P moves for 7 seconds or 25 seconds, Pa is perpendicular to the waist.

If you learn the cosine theorem, you can also find the cos value of the angle b first, and then the length of ab (ab=ac, bc=10), which is simpler.

Solution: In BCD, 6 +8 = 10

BCD is a right triangle, cd ab, then: (ab-6) +8 = ab

Sorted out ab=25 3

Let P1A vertical AC; The triangular ACP1 is similar to ACD;

That is, ac:cp1=dc:ac 5:cp1=4:5;i.e. cp1 = 25 4;

then bp1=8-25 4=7 4;then the time of p to p1 is 7 4 seconds; Nahong.

That is, after 25 seconds of P movement, PA is perpendicular to AC;

In the same way, when P2A is vertical AB, it can be accompanied by BP2=25 4;

That is, after 25 seconds of P movement, PA is perpendicular to AB;