当所有4条线长都已知时，如何推断梯形的最后两个坐标的可能值 [英] How to infer the possible values for the final two coordinates of a trapezoid when all 4 line lengths are known

问题描述

我有一个梯形，在那里我知道所有4条线的长度，而且我知道两个角的坐标.如何在JavaScript中找到其余两个角的坐标?

I have a trapezoid where I know the length of all 4 lines, and I know the coordinates of two of the corners. How do I find the coordinates of the remaining two corners in JavaScript?

注意:所有4条线的长度都不同，即这不是等腰梯形

NOTE: All 4 lines are of different length, ie this is not an isosceles trapezoid

这里有一个小图，以防万一:

Here's a little diagram in case it helps:

       L3  
    D ____ C
     /    \
 L4 /      \ L2 
    --------
  A    L1    B

我知道A和B的坐标，以及L1，L2，L3和L4的长度(都不同).我只需要获取D和C的所有可能的坐标集即可！

I know the coordinates of A and B, and the length of L1, L2, L3, and L4 (Which are all different). I just need to get all of the possible sets of coordinates for D and C!

推荐答案

将垂直线从点C和D拖放到AB线以找到它们在AB的E和F上的投影:

Drop vertical lines from points C and D to the line AB to find their projections on E and F on AB:

现在，AED和BFC是具有相同高度的直角三角形.我们将高度称为 h .从毕达哥拉斯:

Now AED and BFC are right triangles with the same height. Let's call the height h. From Pythagoras:

a² + h² = L4²     and     b² + h² = L2²

从另一个方程中减去一个方程:a²-b²=L4²-L2²

Subtracting one equation from the other you get: a² - b² = L4² - L2²

还可以将L1分为段AE，EF和FB，因此L1的长度必须为:

Also you can divide L1 into segments AE, EF, and FB, so the length of L1 must be:

L1 = a + L3 + b    =>    a + b = L1 - L3

因此，我们在a和b中有一个方程组:

Therefore we have a system of equations in a and b:

a² - b² = L4² - L2²
a + b  = L3 - L1

使用a²-b²=(a + b)(a-b)的事实和上面的等式，您将得到:

Using the fact that a² - b² = (a+b)(a-b) and the above equation, you get:

(L3 - L1)(a - b) = L4² - L2²   =>   a - b = (L4² - L2²)/(L3 - L1)

(请注意，L1和L3不能相等.如果L1 = L3，则解数是无限的.)

(Note that L1 and L3 can not be equal. If L1 = L3, there is an infinite number of solutions.)

因此等式简化为:

a + b  = L3 - L1
a - b = (L4² - L2²)/(L3 - L1)

解决方案是:

a = (L3 - L1 + (L4² - L2²)/(L3 - L1)) / 2
b = (L3 - L1 - (L4² - L2²)/(L3 - L1)) / 2

梯形的高度为:

h = sqrt(L4² - a²) = sqrt(L2² - b²)

您现在可以使用 a 求解A点的角度，并使用 b 求解B点的角度，然后使用它们来计算C和C的坐标.D.或者您可以直接使用 a ， b 和 h .

You could now use a to solve for the angle at point A and b to solve the angle at point B, and use them to calculate the coordinates for C and D. Or you can use a, b, and h directly.

例如:假设A在原点(0,0)，而B在(L1，0).那么这两种解决方案是:

For example: suppose A is at the origin (0,0) and B is at (L1, 0). Then the two solutions are:

• C位于(a，h)，而D位于(L1-b，h)
• C位于(a，-h)，D位于(L1-b，-h)

• C is at (a, h) and D is at (L1 - b, h)
• C is at (a, -h) and D is at (L1 - b, -h)

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