A junior high school geometry math problem for teaching!

Even oo', then boo' is a regular triangle, and aoo' is a right-angled triangle with three sides, and the area of the quadrilateral ao'bo is 4 3+6. Similarly, turn OC 60 degrees clockwise, and then connect AO'' to get a side length of 5 regular triangles and right triangles, with an area of (25 roots, numbers, 3) 4+6, and the same AO turns 60 degrees to obtain quadrilaterals (9 roots, numbers, 3) 4+6

Let aob=s1 boc=s2 aoc=s3s1+s2=4 3 6

s2+s3=25√3/4+6

s1+s3=9√3/4+6

It is OK to solve a system of ternary equations

Connect 0'0 is the auxiliary line, the angle obo'=60 degrees bo'=bo=4 >> angle oo'b is a regular triangle.

Horn o'bo=60=angular abc>> o'BA = Angle OBC >> Triangle O'BA is full equal to the triangle OBC >>AO'=oc=5

ao=3,oo'=4, ao=5, the angle aoo is obtained by the Pythagorean theorem'= 90 degrees The rest is simple, dear, you can do the math yourself.

1) Certifiable BOC and BO'aCongruent.

Even oo'boo'is an equilateral triangle, and AOO is determined by the Pythagorean theorem'It is a right-angled triangle.

It is easy to get ...... later

2) From 1), the angle AOB is 150 degrees, S AOB = 1 2AO*ob*SIN AOB

Find the other: it can be rotated similar to the original problem, and it can be obtained by using the area of the quadrilateral aob.

Connect o o', easy to know bo bo' oo'Equality is 4 easy to know ao = 3 ao'=5 then angular aoo'= 90 degrees, and then you understand.

Solution: From the question, we can see that be=ae

Let ce=x, then ae=be=8-x

In RT BCE, the Pythagorean theorem is solved by 6 +x =(8-x), x=7 4

In BEF and AEF { BE=AE BEF= AEF EF=EF

So, BEF AEF (SAS).

So, bfe = afe = 90°

, AFE is a right-angled triangle.

In RT BCA vs. RT AFE, { BAC= EAF BCA= EFA=90°.

So, BCA AFE

then, ef:bc=ae:ab

As can be seen from the question, ab=10

So there is, ef:6=(8-7 4):10

Solution, ef=15 4

To sum up, the length of the EF is 15 4

90°。This is similar to the evidence and was learned in the third year of junior high school.

Proof: Because ace= ace

And because ac cg=ce ac=(self-counting).

So aec= cag

Because cag+ agc=45

So AEC+ AGC=45

then ACB+ AEC+ AGC=90

Because ac=ce, acb=45°

So ace=135°

CEA = Wait, I want to go down.

Hehe: There is.

ae⊥bc∠aeb=90°

bae==90°-∠b

Similarly, EAC = 90°- C

AD is the angular bisector of A, and A+ B+ C=180°180°-(B+ C) 2=(90°- B)+ DAE, resulting in DAE=(B- C) 2

Let the side length of the square be a, then ad=a, cf=a, ac= 2a, cg=2a, and acg= fca, acg fca

Repentance and (Slip before answer2) Xinhui ACG FCA, 2= CAG

ah∥bg， ∴1=∠gah.

gah+∠cag=45°， 1+∠2=45°.

Solution: (1) Let the side length of these three squares be a, then cf=a, nabub ac= 2a, cg=2a, acf is a common corner hole spike, cf:ac=ac:cg=1: 2, so the triangle acf is similar to the triangle acg.

2) From the conclusion in (1), the triangle ACF is similar to the triangle ACG, Jing Wu.

Then 2= cag, and cag+ 1= the outer angles of the ACB triangle are equal to the sum of the two inner angles that are not adjacent) so 1+ 2=45°

Similar, the balance of repentance cf:ac=ac:ag, angle acg = angular gca, one angle is equal, the corresponding edge cost ratio is known for example, so it is similar. Angle 1 + Angle 2 = Angle Hga + Angle Cag = Angle CAD The final silver is equal to 45°

1.FCA = ACG, AC slag cf = CG AC, ACF ACG

2.The only way to get a sale from acf acg is caf= 1, and the loss is 1+ 2= caf+ 2=45

In the triangle ABC, AB=AC=8, A=150°, find the area of the triangle ABC

Pass C to do CD BA on BA extension line D

Because bac=150, cad=30

So cd=ac 2=4

then s=ab*cd2=16

Let the vertex be aThe bottom two points are b and c. First, extend the BA and make a perpendicular line from the extension line from C to BA. The following knowledge of right triangles will do. The area of the large triangle minus the small one.