A school takes an equilateral △ ABC corner with a side length of 2A The corner of an equilateral △ ABC with a side length of 2a is divided into two parts with equal area by de. D is on AB and E is on AC (1) Let ad = x (x ≥ a), ed = y, find the functional relation of Y expressed by X;? (2) If De is the location of irrigation water pipe, in order to save money, I hope it is the shortest, where should de be? If De is the tour line, I hope it is the longest, where should de be? The answer is (1) y = √ (x ^ + 4 * (a ^ 4) / x ^ - 2A ^) (2) When ad = (√ 2) * A and de ‖ BC, De is the shortest; when D is the midpoint of AB, e coincides with C or D coincides with B, e is the midpoint of AC, De is the longest Mainly the first question

A school takes an equilateral △ ABC corner with a side length of 2A The corner of an equilateral △ ABC with a side length of 2a is divided into two parts with equal area by de. D is on AB and E is on AC (1) Let ad = x (x ≥ a), ed = y, find the functional relation of Y expressed by X;? (2) If De is the location of irrigation water pipe, in order to save money, I hope it is the shortest, where should de be? If De is the tour line, I hope it is the longest, where should de be? The answer is (1) y = √ (x ^ + 4 * (a ^ 4) / x ^ - 2A ^) (2) When ad = (√ 2) * A and de ‖ BC, De is the shortest; when D is the midpoint of AB, e coincides with C or D coincides with B, e is the midpoint of AC, De is the longest Mainly the first question

Because the side length of ABC is 2a, as shown in Figure D, on AB, the area of ADE = &

As shown in the figure, the park has an equilateral △ ABC corner with a side length of 2a, which is now built into lawn. In the figure, de divides the lawn into two parts with equal area, D on AB and E on AC
(1) Let ad = x (x ≥ 0), ed = y, find the functional relation of Y expressed by X;
(2) If De is an irrigation water pipe, in order to save cost, where is the location of de? If De is a tour line, where is the location of de? Please prove it

(1) The area of △ ade is half of the area of △ ABC, that is √ 3 / 2 * a ^ 2. Let the height of △ ade in AD be h, then its area = XH / 2, and the solution of H. angle a = 60 °, AE = h * sin 60 ° and ae. △ ade is obtained. Given ad and AE, De is obtained by cosine theorem, and y
(2) Y = f (x), find the derivative f '(x), Let f' (x) = 0 to get the extreme point x, and then add the boundary point y = f (x = 2A) into y = f (x), where there must be a maximum value of Y and a minimum value of Y

As shown in the picture, there is a tree on each corner of a lawn in the park. Now the gardeners want to double the area of the lawn, but they still need four trees, and make the expanded lawn a parallelogram. Can this idea be realized? If you can design a sketch? Otherwise, please give reasons

Can be realized, as shown in the figure, through a, C, B, D respectively as parallel lines of BD, AC, and these parallel lines intersect in pairs at e, F, G, h, then the quadrilateral efgh is a qualified parallelogram. From the properties of the parallelogram, it can be concluded that the area of the lawn is doubled, four trees do not move, and the expanded lawn is a parallelogram