Nine on math problem 5

Proof: 1. From the known: afg= afe= b=90°ab=af=6,ag=ag

abg≌△afg （1）

Fe=DE=1 3CD=2, EC=CD-DE=6-2=4 Reg = Fg=X, CG=6-X, Ge=Fg+Fe=X+2 in RT GCE, GE = CG +EC

That is: (x+2) =(6-x) +4, and the solution is x=3BG=FG=3, CG=3, BG=GC=3 (2)ge=fg+fe=x+2=3+2=5

Pass f for FM cd for m, pass f for fn bc for n fng ecg

fg/ge=fn/ce,3/5=fn/4∴fn=12/5s△fgc=1/2×fn×cg=1/2×12/5×3=18/5≠3 （4）

The same goes for EMFG ECG

EF eg=fm cg, 2 5=fm 3 fm=6 5 In rectangular MCNF, cn=fm=6 5

In RT FNC, tan FCN=FN CN=(12 5) (6 5)=2

In RT ABG, Tan AGB=AB BG=6 3=2Tan FCN=Tan AGB, i.e., FCN= AGBag, CF (3).

1) (2) (3) True, (4) False.

123 is correct, 4 should be [1 is hl, that is, the angle AFG = angle ABG = 90 °, and AB = AF = AD = 6] [2 let BG = FG = X, in the triangle ECG there are: (2 + X) square = (6-x) square + 4 square, x = 3] [3 CG = FG, so angle GFC = angle GCF, angle BGA = angle AGF, angle BGF = GFC + GCF].

According to the definition of root number, there are: root number 3a-6 = root clear wheel number 3 * root number a-2> = 0, root number 2-a> bend mark = 0

Only a-2=0, a=2

b-4=0,b=4

a=2, b=4, and 1 side can't = 2, then the 3rd side = 4 perimeter = 4 + 4 + 2 = 10

The bottom edge is buried and high = root number [4 2-(2 2) 2] = root number 15 area = 1 2 * 2 * root number 15 = root number 15

There are two unknowns in this equation, so it is impossible to solve them all at once. But you can take a=2, b=4 into the validation by looking at it and find that this equation is satisfied. Therefore, the length of the three sides of the triangle is changed to .

The other group does not form a triangle and must be rounded), so the circumference of the triangle is 4 + 4 + 2 = 8, and the high skin of this isosceles triangle on the bottom edge 2 can be calculated by the Pythagorean theorem, and the height is 15 under the lead number, so the area is 15 under the base * height * half = 15 under the root number.

Are you sure it's a cubic root number? The cubic root number can satisfy a solid number, so that the problem cannot be solved. It's going to be pushed.

3a-6 under the root number can be converted into root number 3 multiplied (root number a-2), because the number under the root number must be greater than the isokinetic guess ruler at 0, and the root number 3 is greater than 0, so the root number a-2 is greater than the equal dispersion higher than 0, thus obtaining the inequality a-2 is greater than or equal to 0, and then a is greater than or equal to 2

1.If it is a triple root number, it can only be calculated according to the second floor.

2.If it is 2-a under the 2nd root number, then according to the above idea, a is less than or equal to 2, so we get a=2, followed by b=4.

Assuming that a is a waist, then the three sides are 2,2, which does not satisfy the trilateral relationship of the triangle.

Assuming b is the waist, then the three sides are 4, 4, 2Satisfy the trilateral relationship of the triangle. Then the perimeter is 10, and the area is made by the Pythagorean theorem, and the megabase height on the bottom edge is 15So the area is the root number 15 times 2 divided by 2 = root number 15

Using the method of doubling the midline, extend AD to E and connect BE

It is easy to prove that Bed CAD(SAS) is be=ac ad=de

So be-ab 2 is therefore 1, so the value range of the midline AD is 1

Imagine that if you flatten ab and ac, the length of the midline is 0, and if you pull the angle of ab and ac to the smallest and smallest, close to 0, that is, (6+8) 2=7

So 0 ad 7

The idea of greater than 0 and less than 7 limits.

1) BAC=30°, ACB=90°, ABC=60°=BAE, and ACB= EAF=90°,AB=AE, tease ABC AEF, AC=EF

2) DAE= DAC+ CAB+ BAE=120°, AEB=60°, Shan Douming.

DAE+ AEB=180°, AD EF, EF=AC=AD, parallelogram ADFE

Proof: (1) In RT abc, BAC=30°, AB=2BC, and ABEe is an equilateral triangle, EF AB, AEF=30°

AE=2af, and AB=2AF, af=CB, while ACB= AFE=90°, in RT AFE and RT BCA, af=BCAE=BA, AFE BCA(hl), AC=EF;

2) From (1) we know that ac=ef, and acd is an equilateral triangle, dac=60°ef=ac=ad, and ad ab, and ef ab, ef ad, cherry blossom rising.

The quadrilateral ADFE is a parallel ridge orange old quadrilateral jujube shape.

1) BAC=30°, ACB=90°, ABC=60°=BAE.

and acb= eaf=90°, ab=ae, abc aef, socks.

ac=ef

Pass o as og ab at g, and hand cd at point h

So mg=gn, eh=hf

And because the quadrilateral adhg is rectangular.

So ad=hg

Because eh=ef 2=4

So: dh=de+eh=1+4=5 so ag=5 and because ag=am+mg=2+mg=5

So mg=3

So mn = 2mg = 2 times 3 = 6

After reading high school, I still feel a little bit about this kind of math problem, doing this kind of problem is like reasoning, and if you think about it more, you will improve, hopefully.

I don't know if there is a mistake in the question, I get 1+ from the question! Solution: Connect CG, CG AB

bc=ab= bridge stool source 2

cg=bg=1

The radius of the bottom surface of the cone is r=135 360=3 8

h= [1-(3 8) 2]= 55 Sensitive Crude Book 8 Answer: ......