Xuexueba helped solve this math problem in Grade Three, thank you! Set the height of the flagpole to h.

Then BC = (AB-CD) * CTG35 = (H-0.8) *1/0.7.

BF =(a b-EF)* ctg 30 =(h- 1.6)* 1/0.6

And BF-BC= 1.5, h =12.7m.

Help solve this math problem in grade three? ∵ Quadrilateral ABCD is a parallelogram.

∴AO=CO

∫AE‖CF

∴∠EAC=∠ACF

∴△AEO≌△CFO(ASA)

AE‖CF and AE=CF

∴ Quadrilateral AECF is a parallelogram

∵EF⊥AC

Perpendicular bisector whose ef is AC.

AE = CE

∴ The parallelogram AECF is a diamond.

Xuexueba helped solve this math problem in Grade Three. Thank you. The idea is the same as that of a square, that is, to make the other diagonal of the diamond. The two diagonals intersect at the O point, and the straight line passing through the O point and the P point divides the diamond ABCD into two parts with equal areas. The proof is the same as that in the example (2), that is, the square ABCD is regarded as a diamond ABCD. First, it is proved that two triangles are congruent, and then the area of quadrilateral is half that of rhombus.

Learn to help with math problems in grade three. 23. There are AD = BE+EC = 8, K = AE/AD = BE/AF = 5/8, 3/5.

24. Two straight lines are introduced, and a quadratic equation about a is obtained at the same time if and only if M >；; =3/2 has a solution, and the solution is a= 1/2m-3 or 1/2m- 1.

When a= 1/2m-3, A-M

When 1/2m- 1, 3/2; =0，y 1 & lt； y2； M>2 o'clock, y1>; y2

Learn to be a bully in the math problem of grade three! Urgent! Thank you. A positive N-polygon is to divide a circle into positive N points, so:

(1) 120 degrees

(2)90 degrees

(3)360/N degrees

Math problems in grade three, learn to solve them, thank you! Just a hint, no matter where M is, the coordinates of MN must satisfy the inverse proportional function. Suppose two coordinates arbitrarily, and then use the calculated length, a ratio should be constant.

Learn from God and help solve math problems in Grade Three. Thank you, thank you, thank you! Do it with Pythagorean theorem.

First, prove the following equation:

Af * af+cf * cf = BF * BF+DF * DF. (Point F is located at any point on the plane, which is a beautiful conclusion)

Analytic geometry is simple, so is Pythagorean theorem. )

Learn to help the third grade math problem. The third question, 1, is sunshine and lighting. Generally, because the angle of sunlight is relatively high, the difference in shadow length is relatively small.

Math problems in grade three, learn to master solution 24, (1) solution: starting from the meaning of the problem.

y=(8-3)*x+( 12-5)*(500-x)

y=-2x+3500

(2) solution: from the meaning of the question.

-2x+3500 & gt； =3200

x & lt= 150

So buy 150 toys at most.

25 (1) proves that the quadrilateral ABCD is a square.

So AD=AB=BC

Angle ABD= angle CBD=45 degrees.

Angle ABC= angle BCD=90 degrees.

Because BE=BE

So triangle ABE and triangle CBE are congruent (SAS)

So AE=CE

(2) Proof: Because OC=OH

So angle OCH= angle OHC

Because angle ABH+ angle OHC+ angle BCE= 180 degrees.

Angle ABH= angle ABC=90 degrees

So angle BAE+ angle OHC=90 degrees.

Because triangle ABE and triangle CBE are congruent (proved)

So angle BAE= angle BCE

So angle BCE+ angle OCH=90 degrees.

Because angle BCE+ angle ECO+ angle OCH= 180 degrees.

So the angle ECO=90 degrees.

Because OC is the radius of circle o.

So EC is the tangent of circle O.

(3) Solution: Because the angle BCD=90 degrees (proof)

So the triangle BCF is a right triangle.

So BF 2 = CF 2+BC 2.

Because angle FGC= angle FHC= 1/2 arc cf.

Tangent angle FGC=3/4

So tan angle FHC=FC/CH=3/4.

Because the angle ABC=90 degrees

So the triangle ABH is a right triangle.

So tan angle FHC=AB/BH.

Because AB=AD=BC (authentication)

AD= 12

So BH= 16.

BC= 12

Because BH=BC+CH= 16.

So CH=4

So angle ABC+ angle BCD= 180 degrees.

So AB parallel CD

So CF/AB=CH/BH.

Because CF=3

In a right triangle BCF, the angle BCD=90 degrees.

So BF 2 = BC 2+FC 2.

So BF=3 times the root number 17

Because angle BCD= angle BGH

Angel CBF= angel CBF

So triangle CBF is similar to triangle GBH (AA).

So CF/HG=BF/BH.

So HG= 16 times the root number 17/ 17.

Help solve this problem on the third grade math test paper? (1) For DF‖AB to BC at point F, connect the DE line to BC at point G..

∫AB = CD AD‖BC

∴DF=DC means △DFC isosceles.

∫D E is symmetric about BC

∴DE⊥BC

∴BG=CG DG=EG

∴DF is parallel and equal to the Chief Executive.

∴AB is parallel and equal to the Chief Executive.

∴ABCE is a parallelogram

(2) extend the intersection of BA CD and h respectively.

∫AD‖BC，AD= 1/2BC。

△ HBC equilateral

And AB= 1/2BC=HB.

∴CA⊥HB means ∠ cab = 90.

∵ quadrilateral ABEC is a parallelogram.

∴ parallelogram ABEC is a rectangle.