In the quadrilateral ABCD, what is the range of values of AB 6, BC 8, CD m, and DA 5?

1. Proof of:

In the quadrilateral ABCD, if AC is connected, then in ABC, 0 AC AB+BC, AB=6, BC=8, and 0 AC14,

ADC, 0 cd AC+DA,

Yes, 0 AC+CD 14+AC+DA

Then 0 cd 14+da, da=5, 0 cd 19, and cd=m, i.e., 0 m 19.

2. The above proves that the convex quadrilateral, as shown in the figure, is well understood. For concave quadrilaterals, 0 m+5 6+8 0 m 9;, 0 m+8 6+5 0 m 3: just the graph is different, the principle is the same.

For concave quadrilaterals, be complex, m+8 5+6 m3;⑸，m+5＞8+6→m＞9。

3. Above the comprehensive, 0 m is obtained

I'm from the Mathematics Department. Let's start with the answer (0,19). I don't know what grade you are in, or let's be clear. The meaning of this symbol is that any real number between 0 and 19 can be taken (excluding 0 and 19).

Because they can be enclosed in a triangle, they can form a closed figure on three sides even without a CD. So cd can be an arbitrarily small positive number. Let's look at the maximum.

If you put these three sides in a straight line, which is 19, then as long as they are slightly angled, you can connect the two endpoints to form a quadrilateral. And the connected line is the CD. Therefore, he cannot exceed 19 at most

I don't know if you know.

To be clear, you look for three chopsticks to compare. One left untouched. Holding two outside one after another, connecting one end of each one with the one that is placed, isn't it like a trichale?

Next, connect the other end of the chopsticks on both sides with a string. You will clearly see that the shortest rope can be 0, and the longest can not exceed the sum of the lengths of the three chopsticks.

If you haven't learned the concave quadrilateral, it's (3,19).

Between two points, the line segment is the shortest.

m<5+6+8=19

The value range of 5+6-8m is 3

ab=bc=5, b=60°, abc is an equilateral triangle, ac=ab=5 In acd, according to the trilateral relationship of the triangle, it is obtained.

The value range of AD is 2 and 12

p is on a straight line ac.

Because P is on EF and EF is on face ABC, P is on face ABC in the same way that P is also on face ACD.

So, P is at the intersection of the surface ABC and the surface ACD.

The intersection line is also known as AC

Proofed with mathematical notation is:

P ef, EF surface ABC

P-plane ABC

The same goes for p-plane ACD

P-plane abc face-acd = ac

i.e. Pac

The diagonal intersection of the quadrilateral ABCD.

If you connect the quiet god AC, then AC=10, and it can be seen from the Pythagorean theorem that ACD is a right triangle CAD=90

So s=s abc+s acd=1 2*8*6+1 2*10*24=144

It can connect two points, A and C, and use the diagonal AC as an auxiliary line to obtain two triangles, ABC and ADC

In this question, the range of AD values is calculated.

According to the fact that the sum of the two sides of the triangle is greater than the third side and the difference between the two sides is less than the third side, the range of AC in ABC is: 2 AC 6

In addition, according to the same principle in the ADC, the value range of AD can be obtained: 5 AD 13

It can connect two points, A and C, and use the diagonal AC as an auxiliary line to obtain two triangles, ABC and ADC

In this question, the range of AD values is calculated.

According to the fact that the sum of the two sides of the triangle is greater than the third side and the difference between the two sides is less than the third side, the range of AC in ABC is: 2 AC 6

In ADC, according to the same principle, the range of values of AD can be obtained: 5 AD 13 Mathematics has been lost for a long time, but it should be like this

Hope it helps