Let AF=X, then EP =16-x AE = 32-x.
The area of the whole graph can be divided into four parts, namely triangle AEF, EPC, CDF and FEC. The sum of the areas of these four triangles is equal to the area of the whole quadrilateral APCD, while the S triangle FCE=200.
So: 0.5 * (1 6-x) *16+0.5 *16 * x+0.5 * x * (32-x)+200 = 384 (replace 0.5 in the equation with/kloc-0
If you solve the equation, you get X=4 or X=28. Because an edge has only 16, X=28 is not the meaning of the question, so X=4 BE= 16-4= 12.
So BE= 12.
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2. Draw a diamond ABCD and diagonal. Let the vertices of the two angles of 120 degree be A and C, with point A above, point C below, point B on the left and point D on the right, and two diagonal lines AC and BD intersect at point E ... According to various theorems of diamond, it is axiomatic or characteristic. It is known that the angle ABC is 60 degrees, the angle BAC and the angle BCA are both 60 degrees, so the triangle ABC is an equilateral triangle. The diagonal of the topic is 4, which is divided into two situations, one is the long diagonal, that is, BD is 4, and the other is the short diagonal AC is 4.
In the first case, BD is 4: according to the diamond theorem, we can know the diagonal bisection, so BE=2. In the right triangle ABE, the angle ABE is 30 degrees, so AB=2AE. Because the square of AB plus the square of AE =2 is the square of AB plus the square of half AB =4, and AB =4 times the root number 3, the side length of the diamond is obtained.
So the side length of the diamond is 16 times the root number 3.
In the second case, AC is 4, which is relatively simple. Because the triangle ABC is an equilateral triangle, the three sides are equal, so AB=AC=4, and the side length of the diamond =4 * 4 = 16.
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3. Draw a diamond ABCD, where point A is at the upper point B, point C on the left and point D on the right, connecting two diagonals and intersecting at point E. According to the diamond theorem, the two diagonals are perpendicular to each other and equally divided. In other words, half of two diagonal lines and a diamond form a right triangle. For example, the triangle ABE is a right triangle. Because a diagonal is 8cm long, half of the diagonal is 4cm. In a right-angled triangle ABE, the hypotenuse AB is 5cm, one right-angled side is 4cm and the other right-angled side is 3cm, that is, half of the other diagonal is 3cm, so the total length of this diagonal is 6cm.
So the other diagonal of this diamond is 6 cm long.