Trapezoidal math problems for junior 2, trapezoidal math for junior 2.

The area of a triangle CDE is 192/5 square centimeter.

The area of trapezoidal ABCD is 96 square centimeters.

Ed is the median line of triangular ABC, so ED is parallel to BC, and the length is 1 2 of BC (i.e., ED length is equal to BC length). Then the quadrilateral EBFD is trapezoidal.

The ED is parallel to the base of the BC extension line CF, and the DF is parallel to the EC. So the quadrilateral EDFC is a parallelogram. So DF parallel is equal to EC.

And because the triangle ABC is a right-angled triangle, it is correct to know that the EC length is half of AB, which is equal to EB and EA. So the DF length is equal to the EB length. Therefore, the quadrilateral EBFD is an isosceles trapezoid.

ce=be (the middle line on the hypotenuse of a right triangle is equal to half of the side of the oblique combustion chamber);

e. The midpoint of the segment beats d, so de is parallel to bf;

Because DF is parallel to EC, DF=EC;

So df=be; Slim shirts.

So EBFD is isosceles trapezoidal.

Answer: 1. If you cross point A as a parallel line of DC, and cross BC at point E, then the quadrilateral AECD is a parallelogram, AD=EC=3, BE=7 3=4, AE=DC=4, and the trapezoidal ABCD is an isosceles trapezoid, AB=DC=4, and ABEs are equilateral, B=60°.

2. The perpendicular line of AB is made through the points D and C respectively, the perpendicular feet are respectively E and F points, the quadrilateral AEFD is a rectangle, AD=EF=2, it is easy to prove that Abe DCF, BE=CF=3, in the right-angle split width ABE, it is obtained by the Pythagorean theorem:

AE = 4, trapezoidal ABCD area = 2 8 4

1.Take a point e in BC so that EC=3

then AECD is a parallelogram AE=CD

Because ABCD is isosceles trapezoidal, AB=CD

Hence ab=be=ae=4

then ABE is an equilateral triangle.

So the angle b is 60 degrees.

2.Take point d to make a vertical line on ab and point e as e

then the ade is a right-angled triangle punch key.

ab-dc)/2=3

Buried with a bucket ad=5 now ae=3

According to the formula of the side length of a right triangle a square + b square = c squared.

ae square + de flat empty ant square = ad squared.

Result de=4

Trapezoidal area = (dc + ab) multiplied by de divided by 2 = 20

1.In the triangle ABE, the angle AEB is equal to 90 degrees of Li Jian, AB is 4, and BE is 2According to the angle function of the three noisy Chang, the angle bae is 30 degrees, so the angle b is 60 degrees.

2.Make a perpendicular line from points c and d downwards to divide the trapezoid into two triangles and a rectangle. Both triangles have an area of 6, a rectangle has an area of 8, and a trapezoid has an area of 20

You can use the theorem of geometry to do this problem, Li Pengshu is equal to half of the hypotenuse in the right angle and the 30-degree angle in the stove triangle. Make two perpendicular lines from point A and point D downward, the perpendicular foot is E, F, and form a rectangular AEFD, at this time, you can find be 2, because ab 4, so the angle bae 30, then the angle b 60 degrees (this problem is easy to do with trigonometric functions, which defeat is not yet learned in the second year of junior high school).

You have drawn the auxiliary lines, from m, n are ad, bc midpoint, get bm-an=bm-be=cm-nd=cm-cf, that is, em=mf, and b=40°, c=50°, and ab en, cd nf, nem=40°, nfm=50°, enf=90°, m=ef midpoint, there is rt triangular property: nm=em=fm, so it is proven.

If D is passed as de bc and bc is crossed to e, then p is the smallest when p is the midpoint of be.

It is easy to get de=4=ab, so ap=root number 17

Then the height on the AP = 2 * 4 root number 17 = 8 root number 17