The second year of junior high school mathematics is a similar problem, and it is another second yea

1. It is known that the rectangular piece of paper ABCD, AB 2, AD 1, fold the paper piece so that the vertex A coincides with the point E on the edge BC.

Connect de*de=fe*fe-fd*fd=

ae*ae=ad*ad+de*de=1*

ae=2*de

Angle Fag = Angle Feg = 90, so AFEG four points are round, Angle DAE = Angle FGE, Angle ADE = Angle FEG = 90

So the triangle ade is similar to the triangle gef.

fg：fe=ae：de

fg=2*fe=2*af=

2) If the crease FG intersects with CD, AE and AB respectively and the point Fog, the distance from the point O to BC is equal to 1 2AE, and the length of the crease FG is obtained.

Points A and E are symmetrical with respect to Fg, so Fg is perpendicular to AE and AO=OE.

In the right-angled triangle APO, the angle AGF=45 is due to op=1 2AE=AO, and FG=

Is the value of af known? And cos(2 ) = 1 - 2(sin) (sin ) You probably wouldn't have to?

I guess I can't solve it.

The conditions are not exhaustive, please make it clear.

Pythagorean? I don't know, the conditions are less.

（1）10/50=

So it is 20cm long and 12cm wide

4) These two ratios are equal.

2 Because ab a'b'=ab/b'c'=ca/c'a'=3 5 so the circumference of abc: a'b'c'The circumference of = 3:5 so the circumference of ABC is 30cm

Surely, I hope it helps.

1. 1) 10m=1000cm 6m=600cm1000*(1/50)=20 cm600*(1/50)=12 cm2) 20:12=5:

33) 10:6=5:34) The ratio of the distance on the graph is the same as the ratio of the actual distance2

ab b in the question'c'That's right.

Press the computer yourself, it's all calculation questions.

Solution: Use m,n to denote the coordinates of the out points a,b,p;

y=x+n, let y=0,x=-n, a(-n,0)y=-2x+m, let y=0, x=m 2, b(m 2,0) substitute y=x+n into y=-2x+m to get: x=(m-n) 3, y=(m+2n) 3, i.e.: p((m-n) 3, (m+2n) 3).

If the point q is the intersection of the pa and y-axis, and the area of the quadrilateral pqob is five-sixths, ab=2, try to find the coordinates of the point p, and find the analytic formula of the straight line p.

y=x+n, let x=0, y=n, that is, oq=n let the line y=-2x+m, and intersect the y-axis at point c, so that x=0, y=m, i.e., oc=m ab=|ob|+|oa|=2

m/2+n=2， m+2n=4

s quadrilateral pqob=s obc-s pqc=5 6 1 2*m*m 2-1 2*(m-n)*(m-n) 3=5 6 simplified: (m+2n) -6n =10

i.e. 16-6n =10

n=1m=2

p（1/3,4/3)

y=x+1y=-2x+2

Solution: For the straight line Pa, when y=0, x=-n, so a(-n,0), when x=0, y=n, so q(0,n); For the straight line pb, when y=0, x=m 2, so b(m 2,,0)y=x+n and y=-2x+m simultaneous equations solve x=(m-n) 3 y=(m+2n) 3

Therefore, p((m-n) 3,(m+2n) 3)ab=n+m 2=2 (1) pass the point p as pc ab, cross ab at the point c, the area of the quadrilateral pqob is the sum of the area of the trapezoidal pqoc and the triangular pcb = 1 2[n+(m+2n) 3)*(m-n) 3]+1 2(m+2n) 3*[m 2-(m-n) 3]=5 6(2), substituting (1) into (2) gives n 2=1 because n >0, so n=1, so m=2, so the analytic formula of the straight line pb is y=x+1, and the analytic formula of the straight line pb is y=-2x+2

In the title, "right-angled side ab length" should be" right-angled side ab length.

The method is to make an angular bisector (45°) through the right-angled vertice, and the hypotenuse is the parallel line of two right-angled sides respectively after the point d and d.

The side length is 6, 7 meters.

Analytic geometry.

y=(-4/3)x+2

The intersection point of y=x is (6 7, 6 7).

∵ce⊥ab，bf⊥ac

BEFC four-point contour (e,f on a semicircle with BC diameter) AEF= ACB, AFE= ABC AEF ACB