The question of quadrilaterals and polygons, the question about quadrilaterals

The distance and the smallest point from the vertices of the convex quadrilateral in the plane are the intersection of the diagonal lines, which is proved by "the sum of the two sides of the triangle is greater than the third side", and in the concave quadrilateral, the distance from the four vertices and the smallest point is its concave point; in other convex five or six ......The distance from each vertex and the smallest point in the polygon is its center of gravity.

The concave quadrilateral is better said to be at the vertex of the concave one. This can be easily drawn with two sides of a triangle and one that is larger than the third side.

2 2)a (i.e., a of 2 times the root number of 2).

Solution: Extend the extension line of AN and BC to intersect with point E

Because AE bisects DAB, Δabe, ΔAMD, and Δcne are all right-angled isosceles triangles.

Then there are: dm=adsin45°, cn=cesin45°, dm+cn=(ad+ce)sin45°, and because ad=bc, so.

dm+cn=(bc+ce)sin45°=besin45°=absin45°=（2/2）a2.

I have done this problem: the problem is: verify: ap=ab+pc(2) if ab=8, find the length of pc.

Here's how to solve it. 1) Take the midpoint E of BC and connect AEThe extension line of the extension AE and DC intersects at the point F, which proves: adq abe fce, so daq = bae cf = ab

2=2 1 pae= bae = 1= f, so ap=pf=fc+pc=ab+pc

2) From question 1, we can see that Pa=Pf, AE=EF, so PE is perpendicular to AF and set PC=X, then PF=8+X

In the right triangle PEF, EF=4, PC=X So PE 2=4 2+X 2 In the right triangle PEF, PF=8+X, EF=AE=4 times the root number 5 So PE 2=PF 2-EF 2=(8+X) 2-(4 times the root number 5) 2

So 4 2+x 2=(8+x) 2-(4 multiplied by the root number 5) 2 gives x=2

Hope it helps.

Can you tell me what you're asking for?

What to ask!!

Don't you write it all?

5-sided. The sum of the inner angles of this polygon is.

The sum of the inner angles of the quadrilateral is 360°

The sum of the inner angles of the pentagon is 540°

The sum of the inner angles of the polygon is (n-2)*180

n is the number of sides of the polygon.

If you don't understand. Hi me. Have fun.

Answer: 1. It is not possible to form a quadrilateral.

2. Because: you can imagine this: a 24 cm long line segment, there are three interconnected three line segments on one side, and the beginning and end of these three lines are connected to the end of the 24 cm line segment, because this line segment should be curved, so he should be 24 cm in his socks, and the result is that the three line segments are only 24 cm, so they can't.

3. The sum of any three sides of the quadrilateral cannot be less than or equal to the other side).

<> a closed plane figure or three-dimensional figure enclosed by four line segments that are not on the same straight line in turn end to end is called a quadrilateral.

All parallel and equal ah ab parallel to cd

BC is also parallel to AD

Parallelograms are equal diagonally, opposite sides are parallel and equal, and adjacent angles are complementary (additive angles of 180 degrees).

It is known that lb+lc=180, which proves dc ab, and the same way as ad bc, can prove that abcd is a parallelogram, so ab=cd

This quadrilateral should be a parallelogram!!

Because of ab cd, the angle odc = angle obe, and because the angle doc = angle boe, the triangle (doc) is similar to the triangle (boe).

And because E is the end point of AB, so BE=1 2AB=1 2DC, so the similarity ratio is 1 2, so the area triangle (BOE) Area triangle(doc)=1 4

So m = 4 times the root number under 5