Geometry math problems, math geometry problems

I don't think it's the right thing to do upstairs, it's supposed to rotate the diagonal at the same angle, turn it to any position, and get the same area of the four polygons.

Or do it as follows:

Let the center of the parallelogram ABCD be O, find a point E on AD, let AE:ed=m:n, and then find a point F on AB so that BF:

fa=m:n, connect eo, fo and extend, let g, h, the two lines eg, fh are obtained, and the condition is satisfied.

Time is running out, and tomorrow will give proof.

As a reminder, the quadrilateral efgh is a parallelogram, and it is only necessary to prove that the triangles on the four corners of the original parallelogram are equal in area, using the area formula s=(absinc) 2

Check out the proof below!

Passing through the intersection of the diagonal lines of this quadrilateral to make two straight lines perpendicular to each other, the four figures divided into must be congruent, so the area is equal.

Arbitrarily cross two diagonals of the focal point of two vertical straight lines! ~

Take advantage of: Divided into four figures must be congruent! ~

Hehe! That's all !

It's very simple, the dots connect the diagonal, make 2 lines that are perpendicular to each other and intersect the intersection of the diagonal lines, so that the 4 4 side rows that are separated must be the same area, you draw it first, and you will understand it after you finish drawing, no matter how you do it, the shape made has 4 congruent 3 corner rows, and the remaining parts are equal in area, thank you for sharing it with me.

: On the surface of the earth, from place A (45°N, 120°E) to B(45°N, 150°W).

The radius of the latitude circle of the corresponding points of A and B is r=rcos45°=root number 2 2 r The longitude difference is 360°-(120°+150°)=90°, so ab=root2r=root2r=root2(root2 2 r)=r AOB is an equilateral triangle, and the spherical angle is AOB = pie 3 The spherical distance between A and B is pie r 3

These two places are on a circle at 45 degrees north latitude, east longitude 120 west longitude 150 is on the circle inferior arc 90 reading, the diameter of the circle at 45 north latitude is the root number 2 times r, then the distance between the two earth planes is the root number 2 times r divided by 4.

The angle between two places on a circle of radius (root number 2) 2r is 2.

So the distance between two places is 1 4 circumference, and the circumference l=(root number 2) r, so the distance between two places is s = ((root number 2) r) 4

<> make the radius and h below the vertical of a, and let the center of the circle be o

It can be seen that oh=acosx

ah is obviously which of the circles below.

If aoh=y oh=acosx ob=a is known in oah, then the cosine theorem can be used to find bh 2=oh 2+ob 2-2oh*obcosy (connected).

Because ah underside.

ah bhabh is at right angles

ah=asinx

ab=(ah²＋bh²)^1/2)..#

In OAB, we know ob=aoa,=a ab=( cosz=(ob 2+oa 2-ab 2) (2ob*oa) in oab

1-cosxcosy)

z=arccos(1-cosxcosy)

There is no solution to the question of how many copies to give, because the maximum angle relationship between the two circles cannot be determined in the given parameters.

You may still lack some known conditions here, and there is also a problem with the annotation in the figure, how can the radius and perimeter be a, omit this problem, if the two sides of the angle x are perpendicular to the intersection line of the two circles or the words "angle x between two circles" are mentioned in the known, then you can calculate cosz=cosx multiplied by cosy (solid geometry is a bit forgotten, it seems to be this formula).

Rotate 45°, and one side of the moving square coincides with the diagonal of the moving square.

Side length = 1 diagonals = 2

Overlap area = 1 2-( 2-1) 2 = ( 2)-1

The meaning of the title is not clear.