1. In trapezoid ABCD, AD / / BC, diagonal AC is perpendicular to BD, AC = 6, BD = 8, find the height of trapezoid 2. In the trapezoid ABCD, AD / / BC, ad and BC intersect at one point, AC = 12, BC = 8, BD = ad = 5, find the height of the trapezoid 3. In the isosceles trapezoid ABCD, AD / / BC, ab = CD = 2, ∠ bad = 120 ° AC bisects ∠ bad to find the perimeter of the trapezoid ABCD 4. The cross section of a reservoir dam is trapezoidal ABCD, AD / / BC, ∠ ABC = 45 ° and the back water surface is ab. now the bus is renovated. The slope ∠ ABC is changed from 45 ° to ∠ AEC = 30 ° and the measured AB = 20m. The length of the ground width be after the renovation is calculated 5. In the quadrilateral ABCD, one set of opposite sides AB = DC = 4, the other set of opposite sides ad is not equal to BC, m, N and H are the key points of AD, BC and BD respectively, ∠ abd = 30 ° (1) find the length of MH; (2) find the length of Mn

1. In trapezoid ABCD, AD / / BC, diagonal AC is perpendicular to BD, AC = 6, BD = 8, find the height of trapezoid 2. In the trapezoid ABCD, AD / / BC, ad and BC intersect at one point, AC = 12, BC = 8, BD = ad = 5, find the height of the trapezoid 3. In the isosceles trapezoid ABCD, AD / / BC, ab = CD = 2, ∠ bad = 120 ° AC bisects ∠ bad to find the perimeter of the trapezoid ABCD 4. The cross section of a reservoir dam is trapezoidal ABCD, AD / / BC, ∠ ABC = 45 ° and the back water surface is ab. now the bus is renovated. The slope ∠ ABC is changed from 45 ° to ∠ AEC = 30 ° and the measured AB = 20m. The length of the ground width be after the renovation is calculated 5. In the quadrilateral ABCD, one set of opposite sides AB = DC = 4, the other set of opposite sides ad is not equal to BC, m, N and H are the key points of AD, BC and BD respectively, ∠ abd = 30 ° (1) find the length of MH; (2) find the length of Mn

1) The extension line of cross BC is at e ∵ AC ⊥ BD, the extension line of cross BC is at e ∵ AD / / BC, the extension line of cross BC is at e ∵ AD / / BC, the extension line of cross BC is at e ∵ AD / / BC, the extension line of cross BC is at e ∵ AD / / BC, the extension line of cross BC is at e ∵ AD / / BC, the extension line of cross BC is at e ∵ AD / / BC, the extension line of cross BC is at e ∵ AD / / BC, and the extension line of cross BC is at e ∵ AD / / BC

As shown in the figure, it is known that the perimeter of the quadrilateral ABCD is 36cm. From the obtuse vertex d to AB and BC, two high lines, de and DF, are introduced, and de = 4cm and DF = 5cm, so the area of the parallelogram can be calculated (sorry, I didn't draw it well,

The area of a quadrilateral will not change, so BC * DF = AB * de = area
Because de = 4 √ 3cm, DF = 5 √ 3cm, AB: BC = 5:4
The perimeter of the flat quadrilateral ABCD is 36cm, so AB = 10, BC = 8
Area = AB * de = 10 * 4 √ 3 = 40 √ 3 (cm ^ 2)

1. Calculate (5 + 1) (5 & sup2; + 1) (4th power of 5 + 1) (8th power of 5 + 1) (16th power of 5 + 1) + & frac14;
2. (1 / 2 of 1-2) (1 / 2 of 1-3) (1 / 2 of 1-4) (1 / 2 of 1-9) (1 / 2 of 1-10)

Question 1: multiply the formula by (5-1), and then divide by 4, the result remains unchanged, but you can use the square difference formula, so as to simplify!
The second question: 1 - (1 / N & sup2;) = (n & sup2; - 1) / N & sup2; = (n + 1) (n-1) / N & sup2;
So in the formula (1-2 of the square) (1-3 of the square) (1-4 of the square) (1 / 2 of 1-9) (1 / 2 of 1-10) reduction, which can reduce the amount of computation!
I just said a general idea, if there is a problem, please ask me, I am online~~~