Mathematical reference answer
Me. BCDBACAADDCB
Second, fill in the blanks
13.60 degrees
14.x^2- 1 =(x+ 1)(x-l)
15.{ x=2，y= 1。 .
16 12m。
17.0.5 .
18. 12 minutes.
Third,
( 19)2√2+ 1
(20)
Solution: (1) 360× (1-20%-50%) =108 (degrees);
(2)20÷50%=40 (person);
(3) Figure (omitted).
2 1.
Solution (1) ⊿ bed ∽⊿ AEC;
⊿DBE∽⊿DAB.
(2) Proof: ∫∠ DBE = ∠ DAC; ∠DAC=∠DAB。
∴∠dbe=∠dab；
And ∠D=∠D, so: ⊿DBE∽⊿DAB.
22.( 1) congruence.
Proof: ∫∠OBA =∠CBD = 60 (degrees);
∴∠OBA+∠ABC=∠CBD+∠ABC, that is ∠ OBC = ∠ Abd;
And OB = BABC=BD. So ⊿OBC≌⊿ABD.(SAS).
(2)
Solution: ⊿OBC≌⊿ABD, then ∠ bad = ∠ BOC = 60;
∴ ∠DAC= 180 - ∠BAD-∠BOC=60。
Therefore, tan ∠ EAO = OE/OA, tan 60 = OE/ 1, OE =.
That is, the position of point E will not change, and the coordinate is (0,).
23.
Solution: (1) w = 4.6× (300 t ÷12) =115t.
(2)P=4.95×(300t÷ 15)=99t。
(3) Order 1 15t-99t = 8000, t = 500 (days)
A: It will take 500 days to recover the cost.
24.
Solution: the image of (1) parabola y = ax2+4ax+t (a > 0) passes through (-1, 0).
Then 0=a-4a+t, t = 3a.
Therefore, y = ax2+4ax+3a = a (x+ 1) (x+3), (a > 0).
When Y=0, X=- 1 or -3.
The symmetry axis is: x =-(4a/2a) =-2;
The coordinate of point A is (-3,0);
The quadrilateral ABCP is a parallelogram.
It is proved that if the symmetry axis is X=-2, then PC = 2；; AB=- 1-(-3)=2。
Then PC = AB and PC∨AB. Therefore, the quadrilateral ABCP is a parallelogram.
If AC⊥PB, the quadrilateral ABCP is a diamond.
∴BC=BA=2,OC=√3, so a=√3/3.
The analytical formula of parabola is y = (/3/3) x 2+(4 √ 3/3) x+√ 3.
25.( 1) Explore from special circumstances. Suppose that in trapezoidal ABCD, AD∨BC, AB=6, BC=8, CD=4,
AD=2, and MN is the center line (Figure ①). According to the definition of similar trapezoid, please explain whether the trapezoidal AMND is similar to the trapezoidal ABCD.
(1) The trapezoidal AMND is not similar to the trapezoidal ABCD.
Because ad/ad =1; AM/AB = 1/2；
That is, the corresponding sides of the two trapezoidal groups are not all proportional, so they are not similar.
(2) General conclusion: The straight line parallel to the bottom of the trapezoid is "not similar" to the original trapezoid.
Question 2: Are the two small trapeziums obtained by cutting two waists in a straight line parallel to the bottom of the trapezium similar?
(1) Starting from special parallel lines, it is "impossible to judge the similarity" to cut the center line of the trapezoid into two sides.
(2) Solution: When PA=2, trapezoidal APQD is similar to trapezoidal PBCQ.
Make AE∨DC and PQ in F, as shown in the figure.
Then FQ = EC = AD = 2；;
PF/BE=AP/AB，PF/6=2/6，PF=2，PQ = 4；
And DQ/DC=AF/AE=AP/AB,
That is, dq/4 = 2/6 and dq = 4/3; QC = 4-4/3 = 8/3；
∴ap/pb=pq/bc=dq/qc=ad/pq= 1/2；
PQ∨AD, then ∠ DAP = ∠ QPB; ∠APQ =∠B；
∠PQD =∠C； ∠D=∠PQC。
So the trapezoidal APQD is similar to the trapezoidal PBCQ.
(3) General conclusion: For any trapezoid (Figure ③), it must exist.
The straight line PQ parallel to the bottom of the trapezoid makes the two small trapezoids similar.
If it exists, the conditions for determining the position of the parallel line are:
AP/Pb = √ (ab)/B. The number of prescriptions is AB.
Prompt: If they are similar, then A/PQ = PQ/B and PQ = √ (AB); Referring to fig. 2, AP/AB=PF/BE,
That is, PA/c=(√ab-a)/(b-a), so:
PA/PB =(√a b-a)/[(b-a)-(√a b-a)]=√ab/b。