Analytic Geometry Math Problems for Middle School, Analytic Geometry Problems for Junior High School

1.According to the known, the coordinates of point A can be calculated as (0,6) and point B as (8,0).

Extending be, the y-axis is at the point k, then in the triangle AKB, the angular division line is perpendicular to the bottom edge, AKB is an isosceles triangle, and the coordinates of the k point can be obtained as (0,-4).

Therefore, the angular equinox ae is the bottom midline, ke=eb, the abscissa of e is 4, and m is the ordinate of the ob midpoint, which is also easy to find, is -2, then the parabolic analytic formula of the parabola with e as the vertex and the point a is:

y=1/2(x+4)^2-2

Extend the Y axis of the be intersection at point F.

It can be seen that δaeb δaef and rtδbme rtδbof is: be ef=bm mo

Because: be=ef

So: bm=mo

The advantages of the above solution: no need to find the analytic formula, no need to calculate. The conclusion can be proved only on the basis of triangular knowledge of elementary geometry.

It's more intuitive and the steps are simpler. Moreover, it is not limited by analytical, as long as there are conditions that are similar to those in the question, such a conclusion can be obtained.

1.Winding length, then s=4 3=48 (cm) winding width, then s=3 4=36

Large volume of rotation around the length.

2.When the vertices are the intersection of faces of different colors at most, then each opposite face is of the same color.

There are 8 such vertices;

When the vertices are the least intersections of faces of different colors, every two adjacent faces are of the same color.

There are 2 such vertices.

3.Two straight lines in a plane intersect to have (1 or an infinite number of (at this time the two lines coincide)) intersection points, and the two planes intersect to form (1 or countless (at this time the two planes coincide)) straight lines.

1. The volume of the cylinder is h r 2

If the length is h=4 r=3 v=36 (cubic centimeters), then the width is h=3 r=4 v=48 (cubic centimeters), and the volume of the width is large.

Each color of the polygons has two polygons in space relative to all the points, so that there are up to 8 points that meet the requirements.

Faces of 3 colors Each two are adjacent so that the points that meet the requirements are at least 2 points in total 3, two straight lines in the plane intersect with (only 1) intersection points, and two planes intersect to form (only 1) straight lines.

Hope it helps.

1. There is a rectangle with a length of 4 cm and a width of 3 cm, and rotate around the straight line where its length and width are located to obtain different cylinders, and what are their volumes? Which one is larger?

a. Rotate around the line where its length is located v= cubic centimeters, and rotate around the line where its width is located v= cubic centimeters.

b. Obviously, it is large to rotate around the straight line where its width is.

2. There is a cube, red, yellow, and blue faces each have two sides, in this square, to make some vertices are the intersection points of faces of different colors, how many vertices are there at most? How many are there at least?

There are up to 8 such vertices? At least 2?

3. There are () intersection points between two straight lines in the plane, and the intersection of the two planes forms () straight lines.

Two straight lines in a plane intersect with one (1) intersection point, and two planes intersect to form one (1) straight line.

1.Cylinder volume with 4 diameter: 16 3=48 Cylinder volume with 3 diameter: 9 4=36 The former is larger.

2.A maximum of 8 and a minimum of 2.

3.1 intersection point in a straight line.

1) Extend GFAD to the point H, easy to prove the triangle DFH triangle BCF (AAS), so BC DH will give you the answer tomorrow are some simple geometry problems Ha for a while, add. I'm a master 398103955

The title must be wrong, if the EBCF is an isosceles trapezoid, then AE=DF=CE, then the diagonal of the rectangle is equal to OA=OC, and the ACE on the E** segment OA is an isosceles triangle, so the ACE and AOC of the same method coincide, so ACE is AOC, then the isosceles trapezoidal EBCF is actually BCD or COD. The contradictory topic must be wrong!

Is EBCF isosceles trapezoid?? Wrong, right? 、

It's easy to swap C and D in rectangular ABCD, you make a mistake, and you have to mark the letters in one direction in order.

I don't think your graph AE can be equal to DF.

Solution: abc is a regular triangle.

Positive abc b= a=60°, ab=ac

and ad=ec

ac-ec=ab-ad

i.e. db=ae

In DBF vs. EAD.

ae=dbb=∠a

bf=ad△dbf≌△ead（sas）

The same is true for DF=ED

df=fe=de

ABC is a regular triangle.

If I use the method of junior high school, it is as follows:

First of all, this question is not fully formulated without conditions: Is D on AB? Didn't say it, just default?

The third year of junior high school already involves the method of counter-evidence, if it will be very simple, but the method of counter-evidence many people feel even more clueless.

Set the angle b = angle c = x

Because in the triangle ABC the angle A + angle C + the angle B = 180°, the angle BAC = 180-2X

So the angle dac = 180-2x-40

Similarly, the angle AED = (180-(180-2x-40)) divided by 2=x+20

Because the angle AED is the outer angle of the triangle.

So the angle AED = angle CDE + angle c

Because the angle c = x

So the angle cde=20°

Let the three corners be x, x, y

2x+y=180

The first thing that can be determined is that y cannot become the top angle after the division, otherwise it will be the original triangle.

Then, there's only one line, and it's past the vertices. So the original horns were either divided, or became apical or isosceles. And since there are two X's, there must be a triangle after the division, either the apex angle is X, or the isosceles angle is X.

At the same time, if y is an isosceles angle, then there must be an x angle that is divided into a y and an angle of another triangle.

Possibility 1: Divide and complete two, isosceles angle is x. Then y is divided into two x angles, 4x=180, x=45, y=90

Possibility 2: Divide into two, one of which has an isosceles angle of x and the other apex angle of x, then x+x+x+(180-x) 2=180, x=36, y=108

Possibility 3: Divide into two, one isosceles angle is x, the other isosceles angle is y, then 2y+x+(180-2x)=180, x=2y, y=36, x=72

Possibility 4: One vertex angle is x and the other isosceles angle is y, then 2y+(180-x) 2=180, resulting in y=60 and x=60, but this is an equilateral triangle, and it is impossible to divide into two equilateral triangles.

Possibility 5: Both vertex angles are x, then 2x+2(180-x) 2=180, 3x=0, x=0, it is impossible to do. In fact, if both top angles are x, it will be a diamond shape.

So far, all the possibilities have been enumerated.

Solution: Connect the PC and hand over EF to point H. Pass the midpoint g of AC and connect GE, GF, GH.

AE BCACE is at right angles

and g is the midpoint of ac.

eg=1/2ac=

In the same way: fg=

i.e. gef is isosceles

pecf is a parallelogram.

Point h is the midpoint of EF.

i.e., GH bisects EF vertically, HE=HF=2

Get: gh= point g is the midpoint of AC, and point h is the midpoint of PC.

GH is the median line of APC, and the process of AP=2GH=3 should be pushed like this, and the format should be changed by yourself.

The parallel lines of the point P as EF intersect Cd and G

then PEFG is a parallelogram.

fg=pe=fc，pg=ef=4

Because of the af cd

So AF bisects CG vertically

So ac=ag=5

Extended EP, FP

Because AE BC AF CD, PE parallel CF, PF parallel EC, EP AF, FP AE

So ap ef (triangle three high lines intersect at one point).

i.e. AP PG

In RT APG, the Pythagorean theorem gives ap=3

Is this a three-dimensional figure or a flat figure?

Let the triangle ABC bottom bc=a, and the height is H

Then in the triangle DMN, DM=A4, the height is H6, the area is AH 48, the triangle ADE is DE=A2, the height is H2, the area is AH 8, the area of the quadrilateral ANME is AH 8-AH 48=5AH 48, then the area ratio is 1 5