Solution: Make a straight line L pass through point B, pass through the edge of the separator paper at points M and N respectively, and pass through point L5 at point Q,
Then BM/AB=sin∠a, ab = BM/sin ∠ a.
Because the width of the separator is 10mm, it can be seen from the figure that BM = 20 mm
According to the formula (Sina) 2+(COSA) 2 = 1, and sin32 =0.6.
∴ AB = 20 ÷ 0.6 =100/3mm. ∴ AM = AB× COS32 =100/3× 0.8 = 80/3mm.
It can be proved that △AMB is similar to △BQC, and BQ = 40 mm.
∴AB/BC=AM/BQ, that is, (100/3)/BC=(80/3)/40.
∴BC=50mm
So the rectangular perimeter C=2AB×2BC=500/3.
Question 2 1
(1) proof: link AD, BD, OD,
∫D is the midpoint of the lower arc AB. ∴AD=BD,∠AOD=∠BOD。
And ≈AOB = 120 and ∠AOD=∠BOD.
∴∠AOD=∠BOD=60 .
OD and OB are both radii of circle o.
∴OD=OB
∫∠ BOD =60
∴△BOD is an equilateral triangle.
∴BD=OB
It can also be proved that AD=AO.
AD = BD
∴AD=BD=BO=OA, that is, the quadrilateral AOBD is a diamond.
(2) proof: link AC. ∫∠AOB = 120, BP is the extension of BO.
∴∠AOC=60 .
AO and CO are both radii of circle o, ∴ ao = co
△ AOC is an equilateral triangle ∴AC=OC。
BP = 3OB，OB=OC。
∴OB=OC=AC=CP
∫△AOC is an equilateral triangle
∴∠ACO=60, ∠ Cao =60 .∴∠ACP= 120.
AC = CP .∴∠PAC=30 .∴∠ OAP = 90, that is, OA⊥AP.
And ∵A is the point on the circle, that is, point A is on the circle.
∴AP is the tangent of circle O.
22 questions
(1) solution: ∵ the image of linear function Y=Kx+b and the image of inverse proportional function Y=m/x intersect with a (-2, 1) b (1, n).
∴A is an inverse proportional function, that is, 1=m/(-2). ∴m=-2.
The analytical formula of inverse proportional function is y =-2/x.
∵B is also on the inverse proportional function, ∴n=-2/ 1, ∴n=-2, ∴B( 1, -2).
Both a and b are on linear functions,
∴ (simultaneous expression) 1=-2k+b
-2=k+b
The solution is k=b=- 1.
The analytical formula of the linear function is Y=-x- 1.
(2) Solution: Let the intersection of the linear function and the X axis be M and the intersection with the Y axis be N. ..
∴S△AOB=S△AOM+S△MON+S△BON
∫△AOM and △BON are the absolute values of the ordinate of point A and the abscissa of point B respectively, that is, 1.
The cardinal number is the absolute value of the abscissa of point A and the absolute value of the ordinate of point B, which is 2.
∴s△aom=s△bon=( 1/2)×2× 1= 1
M and n are the intersections of linear functions with the x and y axes.
∴ The coordinates of points m and n can be found as M (- 1, 0) and N (0, 1) respectively.
∴s△mon=( 1/2)× 1× 1= 1/2
∴s△aob= 1+ 1+ 1/2=2.5
(3) Solution: From the figure, when -2
23 questions
(1) Solution: If X apartments A and Y apartments B are set, then
(Simultaneously) 50x+40y≤ 13000
20x+25y≤70 10—— The unequal sign can be solved as an equal sign, and the solution value is the largest integer.
The maximum integer value of x is 92, and the maximum integer value of y is 2 10.
∴ Can decorate 92 apartments A and 2 10 apartments B.
(2)=-= This question requires you not to say anything.
24 questions
(1) solution: s △ AMN = 3x 2/8
(2) Solution: Draw a vertical line from point A to BC, where BC intersects with points H, ∫Mn//BC, ∴AN⊥MN, and let the vertical foot be Q.
∫MN//BC and ∠ A = 90.
∴ When p happens to fall on BC, point P coincides with point H, △AMN and △HMN are congruent, and AQ=HQ= 1/2AH.
Q is the midpoint of AH.
∫Mn//BC，AN⊥MN，AQ=HQ= 1/2AH。
∴△AMQ is similar to △ABH, ∫AQ = HQ = 1/2ah.
∴M is the midpoint of AB.
∵AB=4,∴AM=2
When x=2, point P falls right on BC.
(3) Solution:
When 0 < x ≤ 2, y = s △ AMN = 0.