The cross section of a section of Zhanghe reservoir dam is isosceles trapezoid, the width of dam crest is 6m, the width of dam bottom is 126m, the slope gradient is 1: root 3, then the slope angle and height of the dam here are?

The cross section of a section of Zhanghe reservoir dam is isosceles trapezoid, the width of dam crest is 6m, the width of dam bottom is 126m, the slope gradient is 1: root 3, then the slope angle and height of the dam here are?

Because the isosceles trapezoid can be divided into a quadrilateral and two triangles
In a triangle, the slope angle is a and the height is h
Then TGA = 1 / radical 3 = H / 60
H = 20 root sign 3, slope angle is 30 degrees

It is known that the circumference of the isosceles trapezoid ABCD is 120, AD / / BC, ad: ab: BC = 1:4:3, then the median length of the trapezoid is?

AD:AB：BC=1：4：3
So AB = CD = 4AD, BC = 3aD
So perimeter = AB + BC + CD + ad = (1 + 4 + 4 + 3) ad = 12ad = 120
AD=10
BC=3AD=30
So median = (AD + BC) / 2 = 20

As shown in the figure, in the isosceles trapezoid ABCD, AD / / BC, ad = AB, BD ⊥ CD, then the degree of ∠ A is
The answer is 120 degrees,

：∵AD//BC,AD=AB
∴∠ABD=∠ADB,∠ADB=∠DBC,∠ABC=∠BCD
Let ∠ abd = X
Then ∠ DBC = x, ∠ ABC = ∠ BCD = 2x
Again BD ⊥ CD
Then ∠ BDC = 90 °
∴∠DBC+∠BCD=180°-∠BDC=180°-90°=90°
That is, x + 2x = 90 degree
∴X=30°
∵AD//BC
∴∠A+∠ABC=180°
The degree of a = 180 ° - ABC = 180 ° - 2x = 180 ° - 2 * 30 ° = 180 ° - 60 degrees = 120 °

As shown in the figure, in trapezoidal ABCD, ad ∥ BC, diagonal AC ⊥ BD, if ad = 3, BC = 7, then the maximum area of trapezoidal ABCD______ ．

Solution 1: make de ‖ AC intersection BC extension line at e through D, ∵ ad ‖ BC, de ‖ AC, ∵ quadrilateral aced is parallelogram, ∵ ad = CE, ∵ according to the equal area of triangles with equal base and height, it is concluded that the area of △ abd is equal to the area of △ DCE, that is, the area of trapezoid ABCD is equal to the area of △ BDE, ∵ AC ⊥ BD, de ‖ AC