Math problems about diamonds, math problems about diamonds

Just look at the definition of the diamond to know how to do it, and you still have to think more about the math problem.

**Cannot be plugged in. Look at the text.

1) From the meaning of the question, BCA=60 = F= ACD, so AB=2BC=CF

In ADC, ac=de, acd=60

So ad=cd=ac=cf=af

Therefore, the quadrilateral AFCD is diamond-shaped.

2) The quadrilateral ABCG is rectangular.

After e, EF is perpendicular to AB, so AG EF BC is because BC=CE AC=2BC

So e is the ac midpoint.

So f is the midpoint of ab and e is the midpoint of bg.

So EF is the median line BC=AG=2EF

Because of BC AG

So the quadrilateral ABCG is a parallelogram.

And because AC and BG divide each other equally.

So the parallelogram ABCG is rectangular.

Rotate RT abc 60 degrees to get RT dec because dec e knows that the angle ACB is 60 degrees on the AB edge.

Getting the angle bac cde is 30 degrees and you can know that the side bc is half the side ac.

There is RT ABC flipped 180 degrees along the AB edge.

Get half of bc=fb=ac and ac=dc known.

So fc=dc and the angle fac=bac is twice the angle fac=60 degrees.

That is, FA parallel cd has fc=dc

So it is prismatic.

First of all, it is known that the circular motion along the edge of the diamond is 8 cm at a time, in the order of ABCDEFGA starting from point A.

So 2008 centimeters can be equivalent to 8 centimeters that are rarely walked?

It can be concluded that you can walk 334 times 8 cm, and the remaining 4, so that is to say, after 334 cycles, you have to walk another 4 cm, that is, four steps forward, and you will reach point E from A 4 times!

For this kind of question, you can first see how many times you can loop and then how many times you can remain, and then move forward the corresponding number of steps, then the point you reach is the answer!

I hope it can help you, if you have any questions, please ask them in the follow-up!

The answer is point A.

You can first calculate the length of the ant's walk along the track for a cycle, that is, a total of 8 centimeters from a cycle to a.

Using 2008 8, the number of cycles required is 251 rounds, and the ant stays at point A after completing the 251st cycle.

2008 divided by 8 = 251 flavors of the whole number, so it still stops at point a. 8 cm tastes a cycle, so divide the total by a cycle and see how much is left at which point you count on yourself.

You can connect 3 lines into one, that is, pull 2 points outward, so that you can do it:

Because the diagonal is 12, and the diamond is 60 degrees, get.

sin60=6/x

x = 4 root 3

8 bar, not sure, I don't remember the answer.

As shown in Fig. It is known that the point A is the midpoint of the side IH of the diamond ihdj, and there are three line segments inside it, ab=3, bc=4, cd=5, and ab bc, bc cd, find the side length of the diamond ihdj.

AB BC, BC CD, AE CD

By ab = 3, bc = 4, cd = 5

Get ac=cd=5, get cf ad, get isosceles the height of the bottom edge of the acd is the midline of the bottom edge ad, get af=df, get ae cd

This gives eaf= cdf, so that there is a right-angle aef right-angle dcf, and ae=cd=5

The quadrilateral AEDC is a parallelogram, which can be obtained by the same method.

ae=de, the quadrilateral aedc is a diamond.

There is de ac

From the point a is the midpoint of the edge ih, the point c is the midpoint of the line segment ej, by be=ae-ab=5-3=2, in the right triangle ceb, it is easy to obtain ce=2 5, cj=ce=2 5, from the point c is the midpoint of the line segment ej, cf=ef=ce 2= 5, fj=cf+cj=3 5

Easy to find df=2 5, in the right triangle dfj, easy to get dj= 65, the side length of the diamond ihdj is: 65.

The area of the rhombus is determined, because the circumference is 40, the side length is 10, and because the acute angle is 30 degrees, the height is 5, (the side opposite by the 30 degree angle is half the hypotenuse.) So the area is 10*5=50