已知四边求梯形面积

hbghlyj

$$\text{梯形面积}=\frac{a+c}{4(a-c)}\sqrt{(a+b-c+d)(a-b-c+d)(a+b-c-d)(-a+b+c+d)}$$ Wikipedia

证明:作$▱AA'CD$,在$\triangle A'CB$中使用海伦公式,$$\frac12(a-c)h=\frac14\sqrt{(a-c+b+d)(a-c+b-d)(a-c-b+d)(-a+c+b+d)}$$所以$$h=\frac1{2(a-c)}\sqrt{(a-c+b+d)(a-c+b-d)(a-c-b+d)(-a+c+b+d)}$$代入$\text{梯形面积}=\frac12(a+c)h$即得

hbghlyj

在少儿漫画书"可怕的科学经典数学系列·超级公式 by [英]卡佳坦·波斯基特"中称为"胡一杰公式"
这本书有一个网站, 抄录如下:
This is probably all you ever need to know about the area of trapeziums, so if you want to leave now, that's fine.
Goodbye!
You now continue at your own risk.
So far it's all been simple BUT...
...we were asked if we could work out the area if we knew the lengths of the four sides, but we DIDN'T know the height! Obviously the usual formula wasn't going to help much, so with the help of some of our regular visitors to this site, we came up with another formula.
The first sensible suggestion came from our Singapore friend Hu Yi Jie. He suggested splitting up the trapezium up into three bits and then attacking the problem like this:

You chop off the two triangles at the end to leave a rectangle in the middle. You then put the two triangles together to make one big triangle. We know the lengths of all three sides of this triangle.
The bottom of the triangle measures $a-c$ because the bottom of the trapezium was $a$, and we have just removed a section that measures $c$. The other two sides measure $b$ and $d$.

If you've read The
Perfect Sausage
you'll know about
semi-perimeter cows!

You can work out the area of the triangle using the "semiperimeter" formula.
This brilliant formula lets you work out the area of a triangle if you know the lengths of the three sides! The "semiperimeter" is half the distance around the triangle so if the lengths of the triangle sides are $X,Y$ and $Z$ then $s=\frac{X+Y+Z}2$. Once we have $s$, we get the triangle area with the formula:
            TRIANGLE AREA= $\sqrt{s(s-X)(s-Y)(s-Z)}$

We know that for OUR triangle the side lengths are $b,d$ and $a-c$ so first you work out the semiperimeter $s=\frac{b+d+a-c}2$ and then plonk everything into the area formula to get:
    TRIANGLE AREA = $\sqrt{s(s-b)(s-d)(s-(a-c))}$.
Nice and easy isn't it?

The most common formula for the area of a triangle is AREA = base x height/2. As we now know the area and we already know the base=$a-c$ we can work out the height $h$ of the triangle.
So $h = \frac2{a-c}\sqrt{s(s-b)(s-d)(s-a+c)}$

And here's the good bit... the height of the triangle is also the height of the rectangle, so you can then work out the area of the rectangle.
    The rectangle area is $c\times h$ therefore the RECTANGLE AREA = ${2c\over a-c}\sqrt{s(s-b)(s-d)(s-a+c)}$

Add together the areas of the triangle and rectangle... and that's it!
    AREA OF TRAPEZIUM = $\left(1+\frac{2c}{a-c}\right)\sqrt{s(s-b)(s-d)(s-a+c)}$
where $s=\frac{b+d+a-c}2$

And when you swap every $s$ in the main formula with $\frac{b+d+a-c}2$ and boil it down you get our formula for the area of a trapezium...$$\frac{a+c}{4(a-c)}\sqrt{(a+b-c+d)(a-b-c+d)(a+b-c-d)(-a+b+c+d)}$$