Geometry A primary 2 isosceles trapezoidal problem. Urgent

I don't understand what you mean.

Make an isosceles trapezoidal two high.

So it is divided into 2 congruent triangles and a rectangle.

The sum of the bases of two congruent triangles is 16-6=10

So the bottom edge of the triangle is 5

There are trigonometric functions because the three lines are one.

The trapezoidal waist is 10, and the overall idea is probably this.

You'll know if you draw it.

From the question you gave, you can see that ab and dc are the two waists of an isosceles trapezoid, then let ab=dc=x, now the vertex a and vertex d are high for the bottom edge bc, and the intersection points with the bottom edge are n and n respectively'Because it is an isosceles trapezoid, the answer is 42

Do you know the length of which side? Is it an isosceles triangle or an isosceles trapezoid?

It's good to calculate the waist length by passing the perpendicular line of A to do BC.

(1) ADE is an equilateral triangle with a side length of 2 cm. Solution: As can be seen from the question, ABC and DBC are right triangles, and AE and DE are the middle lines on their hypotenuses, respectively, so AE=DE=BC 2=AD=2CM

2) From the meaning of the title, ad=be, ead= bea, ae=ea, so, bea is equal to dae, so, ab=de=2cm

3) AC and DE are bisected perpendicular to each other. From (2), it can be seen that ab=cd=ec=ad=ae, therefore, the quadrilateral aecd is a rhombus, ac and de are its two diagonals, and the diagonals of the rhombus are bisected perpendicular to each other, so ac and de are bisected perpendicular to each other.

(1) ADE is an equilateral triangle with a side length of 2 cm. Solution: From the meaning of the title, ABC and DBC are right triangles, and AE and DE are the middle lines on their hypotenuses, respectively, so AE=DE=BC 2=AD=2cm

2) From the meaning of the title, ad=be, ead= bea, ae=ea, so, bea is equal to dae, so, ab=de=2cm

3) AC and DE are bisected perpendicular to each other. From (2), it can be seen that ab=cd=ec=ad=ae, therefore, the quadrilateral aecd is a rhombus, ac and de are its two diagonals, and the diagonals of the rhombus are bisected perpendicular to each other, so ac and de are bisected perpendicular to each other.

The isosceles trapezoidal is symmetrical left and right, so the angle between the diagonal and the bottom is 60°, and then the two bases and the two diagonals form two equilateral triangles respectively, so the diagonal length is 2, so the height is the root number 3

The area is the root number 3

Another way to compare that is to change the limit of the isosceles trapezoidal into an equilateral triangle with 0 at the top and 2 at the bottom, and the area will come out directly.

== Missing a situation...

It can also be a fatter trapezoid, and a similar method to find the area is 3 thirds of the root

Extend the CD to point E so that Be is parallel to AD, and reconnect.

Because the acute angle formed by two diagonals is 60

So the angle BCE = angle DAB = angle BEC

Make another high.

Because the sum of the upper and lower bases of the isosceles trapezoidal ABCD is 2

So the high is the root number 3 3

So s=1 2*(root number 3 3)*2=root number 3 3 talks about the properties and trigonometric functions that should be trapezoidal.

Passing through the intersection of the diagonal lines to make the perpendicular line of the two bottoms The perpendicular foot is the midpoint of the two bases The two triangles formed by the diagonal and the bottom are regular triangles Then use the Pythagorean theorem to find the height If you have learned trigonometric functions, you can directly use the tangent to find the height Then use the area formula The answer is the root number 3 Please use it I'm doing a task.

If the isosceles trapezoidal ABCD, ab=4, AD=5, DC=10, respectively through the point A and point B as the perpendicular line of the DC edge, and the intersection of the DC edge at the point E point F, then the edge of the blade ae=bf, the angle AED angle bfc 90 degrees. According to the Rt triangle congruence determination method HL, it can be verified that the slip match triangle ADE is all equal to the triangle BCF, so DE=FC, and because AB=4, so EF 4, so DE+FC 6, so DE 3, and because in the RT triangle ADE, AD=5, DE=3, so AE 4, so the height of the isosceles trapezoidal is 4

Please, landlord, give some hard points!

Fh is perpendicular to ab

Intersection ab is h, because the triangle abc is an isosceles right triangle and af bisects cab, fh ab

AF is equal to FA so the triangle ACF AHF is based on. hl so. cf=fh

Because AF divides cab

So caf= cab

cfa=90°—½cab

cef=∠aed

aed=90°—½cab

So cfa= cef

CE=CFSO.

CE=FH proves the congruence of the triangle CEG and FHB.

cg=bf to prove cf=gb

Cross the point F to do the perpendicular line FH of AB

AF is the angular bisector, CF is perpendicular to AC, FH is perpendicular to AB, so CF=FH, D is the midpoint of AB, according to the similarity of triangles, AED is similar to AFH, so ED's, do GI perpendicular to AB

gi=edgi=bg root number 2

fh = 2ed = 2bg root number 2 = root number 2 multiplied by bg, so fh > bg

Because cf=fh

So cf > bg

Draw according to me. Draw an isosceles trapezoidal abcd, ab=cd,ad,bc=37cm. Crossing point A as AE vertical BC crosses BC at point E, and the same method is used as DF vertical BC.

The easily verified triangle ABE is fully equal to the triangle DCF. Then be=cf=12cm. According to the right-angled edge of 30 degrees, which is equal to half of the hypotenuse, the length of ab and cd can be obtained.

The answer is 98cm.