Middle school math problems urgent to solve

Under the root number (square +

Make ep perpendicular to ab, cq perpendicular to ab, so qb=5, because e is the midpoint, so bp=, so ap=, so ep=root number under (square + so ae = ap square + ep square = root number under (square +

Make an ab perpendicular line from point c and cross ab to f

bf=5;bc=x;The angle BFC is a right angle.

af squared = x 2-25

Make a perpendicular line from E to AB and cross AB to G

ag=eg=

If two sides of a right triangle are known to solve the third side, the expression can be simply written.

Let the height of the trapezoidal be h, and it is easy to know h= (x 2-25) as a right-angled triangle, and if the point e is the high-intersection ab perpendicular to ab in f, then ef=h 2, af=19 2, so the right-angled triangle aef can be solved:

ae=√[（19/2）^2+h^2/4]=√(361/4+h^2/4)

That is, (361 4+H 2 4) can be expressed as ae

First make a few auxiliary lines, 1 do EF perpendicular to AB to AB to F, 2, do CG perpendicular to AB to AB to G, at this time the triangle BEF is similar to the triangle BCG, we can get BF =, EF can be expressed as.

x 2) 2-(5 2) 2], in the right triangle AEF, AE can be expressed as.

x/2）^2-（5/2）^2+（19/2）^2]

It is all 120°, and this point is called the Fermat point.

Let's prove that the Fermat point satisfies the minimum of PA+PB+PC:

Rotate APC 60° counterclockwise around point A, point P around point P' and point C around point C'

Therefore, when bp+pp'+p'c=pa+pb+pc=bc, the degrees of the three angles in the original question are all 120°

(1) Connect CF, CF is perpendicular AB because AC=BC.

Since E is the midpoint of AC and the angle E=45°, the angle A=.

Because the angle acb = 45°, ac=bc, the angle fac=, so the angle e = the angle fac

Because CA=CA, Angular ADC=Angular AFC, Angular E=Angular FAC, so ADC is all equal to AFC

So ad=af

Because BG AC, F is the AB midpoint.

So af=fg

Because fac= dac, ag=ag, ad=af, adg is equal to afg

So dg=fg

So ad=af=fg=dg

So the quadrilateral AFGC is a diamond.

(1) Connect CF, because AC=BC, therefore, CF is perpendicular to AB, and then the easy to prove triangle ADC, AFC, BFC congruence, then the first problem is solved.

The area of the diamond is equal to the triangle ABG, while the triangle ABG is similar to the triangle AFC i.e. AG times 1 2 of GB

ag et al. ac minus gc bc fraction root 2], gb is equal to gc can be obtained.

Extending FO, AB intersects with G, and the quadrilateral AEDF is rectangular. df=ae, ad=ef, odfOBG congruence. df=bg.

eg=ab=ef,o.is the eg midpoint, so eo is perpendicular fg, i.e. fo perpendicular eo

∠daf＝90º-∠aeb＝∠abe.∴⊿adf≌⊿bae（asa）,∴fd＝ea＝ad/2＝cd/2

f is the midpoint of the cd. , extend AD to GMake mg mbConnect fg, fb, note bm=dm+cd. There is DG DC BC

FDG FCB (SAS), DFG CFB, B, F, G collinear.

mbc＝∠amb＝2∠agb＝2∠gbc＝2∠abe[∵⊿abe≌⊿cbf（sas）]

1) Proof : In square ABCD, ab=ad=cd=bc

Because AF is perpendicular to BE, the angle FAD + angle AEB = 90 degrees, and because the angle FAD + angle AFD = 90 degrees, the angle AEB = angle AFD

In the triangle bae and adf, because the angle aeb = angle afd, the angle bae = angle adf, ab = ad, so these two triangles are congruent, so ae = df. And because ae=1 2ad, ad=cd, df=1 2cd, i.e., f is the midpoint of cd.

Proof ABE ADF

af⊥be ∴∠fad＋∠adb＝90°

bad＝90° ∴abe＋∠aeb＝90°∠fad＝∠abe

ab＝ad ∠bad＝∠d＝90°

abe≌△adf ∴df＝ae

DF Fc point F is the midpoint of DC.

It's a little annoying, it's too much, I don't write o(o

Solution: (1) As shown in the figure, the passing point P is PM BC, and the perpendicular foot is M, then the quadrilateral PDCM is rectangular, PM=DC=12, QB=16-t, S=1 2 12 (16-t)=96-6T(0 t 16);

2) From the figure, it can be seen that cm=pd=2t, cq=t, if the triangle with b, p, q as the vertices is an isosceles triangle, it can be divided into three cases:

If pq=bq, in rt pmq, pq2=t2+122, from pq2=bq2, t2+122=(16-t)2, and t=7 2;

If BP=BQ, in RT PMB, BP2=(16-2T)2+122, from BP2=BQ2, (16-2T)2+122=(16-T)2, that is, 3T2-32T+144=0, δ=-704<0, 3T2-32T+144=0 has no solution, BP≠BQ;

If pb=pq, from pb2=pq2, (16-2t)2+122=t2+122, sorted, 3t2-64t+256=0, solution t1=16 3, t2=16 (not in line, rounded), based on the above discussion, we can know:

When t=7 2 or 16 3, the triangle with b, p, q as vertices is an isosceles triangle;

The third question is shown in the figure below

Figures 1 and 2 are as follows:

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Solution: (1) From the meaning of the title, there is dp=2t, cq=t

So bq=16-t(t

2) When ap=bq, the quadrilateral abqp is a parallelogram, at this time, 21-2t=16-t, and the solution is t=5

3) When BP=QP, the perpendicular lines of AD through B and Q are E and F respectively

then PE=PF

i.e. (21-2t)-5=2t-t(t

The solution yields t=16 3

When bp=bq,