side length of trapezoid after cut Given a symetrical trapezoid with a bottom length (a) and a top lenght (b) and a height (h)  that I would like to cut off at a certain height (h'), how do I calculate the new top lenght (b')?

 A: Let $w$ be the required width. The area of the bottom part of the trapezoid is $$\frac{1}{2}(a+w)h'.$$ 
The area of the top part of the trapezoid is 
$$\frac{1}{2}(w+b)(h-h').$$ 
Together, they make up the whole trapezoid, which has area 
$$\frac{1}{2}(a+b)h.$$ 
After multiplying by $2$, we obtain the equation
$$(a+w)h'+(w+b)(h-h')=(a+b)h.\tag{$1$} $$
This equation is linear in $w$, and not difficult to solve.
For fun, we solve the equation. Things will look nicer if we write $p$ instead of $h'$, and $q$ instead of $h-h'$. Then $h=p+q$. Equation $(1)$ becomes
$$(a+w)p+(w+b)q=(a+b)(p+q).$$
A little algebra now gives
$$w=\frac{qa+pb}{p+q}=\frac{q}{p+q}a+\frac{p}{p+q}b.\tag{$2$}$$
Note the beautiful symmetry revealed by the change of notation. Note also that $w$ is a weighted average of $a$ and $b$, with weights $\frac{q}{p+q}$ and $\frac{p}{p+q}$. If one thinks about it for a while, the formula $(2)$ becomes geometrically self-evident. 
Remark: Note that nowhere did we use the assumption that the trapezoid is symmetrical. We get exactly the same value of $w$ for non-symmetrical trapezoids. 
A: 
So ,now  $$\frac{a-b}{2h}=\frac{x}{h-h^'}$$
And your answer will be $b+2x$
