The bases of a trapezoid have lengths $a$ and $b$, and its legs have lengths $c$ and $d$. A formula for the height is \begin{equation*} h = \frac{ \sqrt{(-a + b + c + d)(a - b + c + d)(a - b - c + d)(a - b + c - d)} } {2\vert a - b \vert} . \end{equation*} The formula is reminiscent of Heron's Formula. I would like to see a derivation of it.
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$\begingroup$Your formula "is reminiscent of Heron's Formula" because it is based on Heron's Formula.
One formula for the area of a triangle is
$$A=\frac 12bh$$
and Heron's formula gives
$$A=\sqrt{s(s-a)(s-b)(s-c)}$$
where $s$ is the semiperimeter given by
$$s=\frac{a+b+c}2$$
Here is a diagram for the derivation for the height of your trapezoid, assuming that $a>b$ (i.e. $a$ is the larger base and $b$ is the smaller one).
Note that I constructed a line segment (in green) parallel to side $c$ through the end of side $b$ that is not on side $c$. This line segment also has length $c$, of course, and it makes a triangle with sides $a-b,\ c,\ d$ that has the same height $h$ (dotted) as the trapezoid.
Using those sides of the triangle rather than $a,\ b,\ c$ gives us the equations
$$A=\frac 12(a-b)h$$
and
$$A=\sqrt{s(s-[a-b])(s-c)(s-d)}$$
where
$$s=\frac{(a-b)+c+d}{2}$$
Solving for $h$ in $A=\frac 12(a-b)h$, substitutions, and simplifications give us
$$\begin{align} h &= \frac{2}{a-b}A \\[2ex] &= \frac{2}{a-b}\sqrt{s(s-[a-b])(s-c)(s-c)} \\[2ex] &= \frac{2}{a-b}\sqrt{\frac{(a-b)+c+d}{2}\left(\frac{(a-b)+c+d}{2}-[a-b]\right)\left(\frac{(a-b)+c+d}{2}-c\right)\left(\frac{(a-b)+c+d}{2}-d\right)} \\[2ex] &= \frac{2}{a-b}\sqrt{\frac{a-b+c+d}{2}\left(\frac{-a+b+c+d}{2}\right)\left(\frac{a-b-c+d}{2}\right)\left(\frac{a-b+c-d}{2}\right)} \\[2ex] &= \frac{1}{2(a-b)}\sqrt{(a-b+c+d)(-a+b+c+d)(a-b-c+d)(a-b+c-d)} \\[2ex] &= \frac{\sqrt{(-a+b+c+d)(a-b+c+d)(a-b-c+d)(a-b+c-d)}}{2|a-b|} \\[2ex] \end{align}$$
which is your formula.
It is easily seen that if we assume $a<b$ we end up with the same result, thanks to the absolute value in the denominator. If $a=b$ this formula fails, but we then get a parallelogram whose height is not uniquely determined, so no formula is possible for $a=b$.
I tested this formula in Geogebra, and it checks.
$\endgroup$ 4 $\begingroup$Here is a link to a (french language) website mentioning this formula (go almost to the last page) under the form:
$$h=\dfrac{\sqrt{((b+d)^2-(a-c)^2)((a-c)^2-(b-d)^2)}}{2(a-c)}$$
Edit Translation of a small part of the site of Gérard Villemin:
Isosceles trapezoid
Characterization or necessary and sufficient conditions for a trapezoid to be isosceles
Two angles adjacent to the same base are equal.
Non parallel sides have the same length.
The measures of the two diagonals are the same.
The two bases have a common perpendicular bissector, which is besides an axis of symmetry of the trapezoid.
If the trapezoid is isosceles, the sum of opposite angles is $\pi$.
Alignment: theorem of trapezoid
Trapezoid : ABCD
Sides intersect in point : m
Diagonals intersect in : s
Bases midpoints : n and t.
Theorem : The four points m, n, s, and t associated with the trapezoid are aligned.
Avec les diagonales -> with the diagonals
Due to the fact that sides AB and CD are parallel, triangles APB et CPD are similar and diagonals intersect themselves in the same proportions p/p'=q/q'.
Relationship between P and Q diagonals and sides
A convex quadrilateral is a trapezoid if and only if the product of the areas of triangles "created" by one diagonal is equal to the product "created" by the other one: $(V+U)...$. Developing...But V=T, validation the affirmation.
Yes, the third column gives examples.
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