Connecting EG, EH, FG and FH, it is obvious that these four lines are equal in length, and because of symmetry, the quadrilateral EGFH is a square (which is a bit troublesome to prove), so ef = √ 2 * eg
Then find out the length of EG
Connect EA, EB, ED, HA. Then: EA=EB=AB= 1(EA, EB is the radius).
So △ABE is an equilateral triangle, so ∠ EAB = 60, so ∠ DAE = 30. Similarly ∠ GAB = 30.
So ∠ gab = ∠ gae = ∠ EAD = 30.
So GE=ED
△EAD is an isosceles triangle with a vertex angle of 30 and a waist length of 1. You can calculate EG=ED=(√6-√2)/2.
So EF=√2*EG=√3- 1.
Who has 20 1 1 solution in the problem-solving process of mathematics test 8 in Benxi senior high school entrance examination: make PF perpendicular to AE and M through P, cross AC to F, and then QF.
If it is proved that △AFM is all equal to △APM, then PM=FM.
Then it is proved from the edge that △PMQ is equal to △QMF, then PQ=FQ.
As long as you find the minimum value of DQ+FQ, you can get what you want.
Obviously, when D Q F is on the same straight line, DQ+FQ finds the minimum value.
Just find DF.
This is a concrete idea. Please let me know if you need a more specific problem-solving process!
20 1 1 How to solve the DE⊥x-axis of 25 math questions in Tianjin senior high school entrance examination?
From ∠AOD=β, Tan∠aod = Tan∠β ∠βis obtained.
So de: OE = 3: 4.
So let DE=x, we can find OE, AE=OA-OE. AE available
Given that AD=3, DE=X, AE can make equations according to Pythagorean theorem. Find x, and then the corresponding coordinates of point d will be known.
At the same time, point A and point D can be used to find the analytical formula of AD. According to AD⊥CD, (here, the vertical slopes of two straight lines are multiplied by-1, that is, k of y=kx+b is multiplied by-1. ), you can find the slope of the straight line CD (that is, k), and then substitute it into the coordinates of point D, that is, you can find the analytical expression of CD.
Look again, when rotating to a certain size, there is also a triangle in the third quadrant that is symmetrical with the triangle ACD found here. According to the nature of symmetry, you only need to multiply k and b of the analytical formula of CD found earlier by-1.
If you rotate clockwise, as shown in the figure, point D is DE⊥OA in E, and point C is CF⊥OA in F.
∠∠AOD =∠ABO =β，
∴tan∠AOD= Germany /OE= 3/4,
Let DE=3x, OE=4x,
AE=3-4x，
In Rt△ADE, ad 2 = AE 2+de 2,
∴9=9x^2+(3-4x)^2，
∴x= 24/25，
∴D( 96/25，72/25)，
∴ The analytical formula of linear AD is: y= 24/7x- 72/7,
The straight line CD is perpendicular to the straight line AD and passes through the point d,
Let y=- 7/24x+b,
Then b=4,
The analytical formula of ∴ linear CD is y=- 7/24x+4.
If rotated clockwise, the analytical formula of straight CD can be obtained as y = 7/24x-4.
20 10 Harbin senior high school entrance examination mathematics examination questions 20 problem-solving process The rotation mentioned in this question does not mean clockwise and counterclockwise, so there should be two situations:
(1) When ⊿DCE rotates 60 degrees clockwise, as shown in the left figure:
If the extension line of E'H⊥BC is at H, then E 'ch = 60 and CE 'h = 30.
∴ch=( 1/2)ce'=3,e'h=√(e'c^2-ch^2)=3√3；
BE' = √ (BH 2+E' H 2) = 14。 (so calculating the length of Be can avoid cosine theorem')
Let AQ⊥CM be in Q and D 'P ⊥ cm be in P; And CN⊥BE', then ∠CBN=∠ACQ.
And CB = CA∠ CNB =∠ Q = 90, so ⊿ CBN ≌ δ ACQ (AAS), AQ = CN, CQ = BN;
It can also be proved that: ⊿ CPD' ≌ δ E 'NC (AAS), PD' = CN = AQ, CP = E 'n 。
AQ‖PD', then QM/MP=AQ/PD'= 1, so QM=MP.
∴cm=(cp+cq)/2=(e'n+bn)/2=be'/2=7.
According to the area relationship, CB * e 'h = be' * CN, 10 * 3 √ 3 = 14 * CN, CN = 15 √ 3/7.
Therefore: Mn = cm-cn = 7-15 √ 3/7;
(2) When ⊿DCE rotates 60 degrees counterclockwise, as shown in the right figure, the same result can be obtained: cm = 7；;
CN= 15√3/7。 At this time MN=CM+CN=7+ 15√3/7. (Because the method is similar, I won't go into details. )
So the length of MN is 7- 15√3/7 or 7+ 15√3/7.
20 10 Shandong Laiwu senior high school entrance examination mathematics exam 17 problem-solving steps This is a combined problem.
C (up 6, down 10)
=( 10×9×8×7×6×5)÷(6×5×4×3×2× 1)
=2 10
Liaocheng senior high school entrance examination 20 10 math test question 17, which eldest brother can give a problem-solving process or idea? Thanks that point B' makes B'F perpendicular to CA, and the extension line of intersecting CA is at point F, then triangle B'FA≌ triangle B'C=, so AF = 3, B' f = 3 times the root number, FC = 6, then from Pythagorean theorem, we can get that B' c = under the root number (27+36) = 3 times the root number 7.
20 1 1 How can I give you the problem-solving process of the third sub-question of the last question of the mathematics in Mudanjiang senior high school entrance examination?
Suzhou 20 1 18 extends the problem-solving process of Suzhou senior high school entrance examination mathematics test 18. From the question, it is concluded that all triangles ADE are equal to triangle CFE, then CF is equal to 5, AE is equal to EF, in right triangle ABF, if available, AF is equal to 13, then AE is equal to half AF, which is equal to 6.5.
20 1 1 Hangzhou senior high school entrance examination mathematics 14 problem solving process 48
According to radian, the angle COD is 84 degrees, so the angle OCD is 48 degrees.
And because the angle ABD is equal to the angle ACD.
Then from OA=OC, angle CAO equals angle ACO.
So the sum of the two angles is the angle OCD=48 degrees.
It is proved that: (1) As shown in the figure, extend AC, so that the intersection of FD⊥BC is D, FE⊥AC, and the intersection is E, and the ∴ quadrilateral CDFE is square, that is, CD=DF=FE=EC, ∫ at an isosceles right angle △ABC, AC = BC = 60.