Current location - Training Enrollment Network - Mathematics courses - 20 13-20 14 Mid-term examination paper for the first semester of the school year (math answer sheet for Grade 2)
20 13-20 14 Mid-term examination paper for the first semester of the school year (math answer sheet for Grade 2)
20 13-20 14 mid-term examination paper for the first semester of the school year (junior two mathematics answer sheet) 26. ( 1)( 1)y=-4/3x+8,

Let x=0 and y=8.

Let y=0 and x=6.

∴A(6,0),B(0,8),

∴OA=6,OB=8 AB= 10,

∫A B ' = AB = 10,

∴O B'= 10-6=4,

The coordinates of ∴B' are: (-4,0).

(2) Solution: y=-4/3x+8

Get a (6,0), b (0,8)

BA= 10 from Pythagorean Theorem

According to the meaning of the question, m is a point on OB. If △ABM is folded along AM, the point B falls on the point B' on the X-axis.

Then AM is ∠BAO bisector

Let MD⊥AB become D.

Let mo = MD = H.

From the area of 6h/2+ 10h/2=6*8/2.

The solution is h=3.

That is, the coordinate of point m is m (0,3).

Let the analytical formula of straight line AM be y = kx+b.

Then 6k+b=0 and 0k+b=3.

The solution is k=-0.5 and b=3.

Therefore, the analytical formula of straight line AM is y=-0.5x+3.

27. Proof: (1) As shown in the figure, ∫∠BAC = 90, AF⊥AE.

∴∠EAB+∠BAF=∠BAF+∠FAC=90,

∴∠EAB=∠FAC,

∵BE⊥CD,

∴∠BEC=90,

∴∠EBD+∠EDB=∠ADC+∠ACD=90,

∫∠EDB =∠ADC,

∴∠EBA=∠ACF,

∴ in △AEB and △AFC,

∠EAB=∠FAC

AB=AC

∠EBA=∠ACF

∴△AEB≌△AFC(SAS)

∴ae=af;

(2) As shown in the figure, the intersection point A is AG⊥EC, and the vertical foot is G. 。

∵AG⊥EC,BE⊥CD,

∴∠BED=∠AGD=90,

∵ Point is the midpoint of AB,

∴BD=AD.

△ ∴and△ △AGD on the bed,

∠ Bed =∠AGD

Bromdiphenyl ether = ADG

BD=AD

∴△BED≌△AGD(AAS),

∴ED=GD,BE=AG,

AE = AF

∴∠AEF=∠AFE=45

∴∠FAG=45

∴∠GAF=∠GFA,

∴GA=GF,

∴CF=BE=AG=GF,

∫CD = DG+GF+FC,

∴CD=DE+BE+BE,

∴CD=2BE+DE.

28.( 1) ∠ CMQ = 60 remains unchanged.

In an equilateral triangle, AB=AC, B = Cap = 60.

And the condition AP=BQ,

∴△ABQ≌△CAP(SAS),

∴∠BAQ=∠ACP,

∴∠cmq=∠acp+∠cam=∠baq+∠cam=∠bac=60。

(2) Let time be t, then AP=BQ=t and Pb = 4-t..

① When ∠ PQB = 90,

∫∠B = 60,

∴PB=2BQ,4-t=2t,t=4/3。

② When ∠ BPQ = 90,

∫∠B = 60,

∴BQ=2BP,t=2(4-t),t=8/3。

When t=4/3 or t=8/3, △PBQ is a right triangle.

(3) ∠ CMQ = 120 remains unchanged.

In an equilateral triangle, AB=AC and B = Cap = 60.

∴∠PBC=∠ACQ= 120,

According to the conditions, BP=CQ,

∴△PBC≌△QCA(SAS)

∴∠BPC=∠MQC

∠∠PCB =∠MCQ,

∴∠cmq=∠pbc= 180-60 = 120

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2. 1:3

3.2

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(2).5/4

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