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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')?

symetrical trapezoid, given, wanted

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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.

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+1 for the extra bonus of supporting asymmarical trapezoids and for the detailed deduction. –  bitbonk Aug 7 '12 at 11:29
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So ,now $$\frac{a-b}{2h}=\frac{x}{h-h^'}$$ And your answer will be $b+2x$

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So the actual equation would be $$b'=b + \frac{2(a-b)(h-h')}{2h}$$ which can be simplified to $$b'=b + \frac{(a-b)(h-h')}{h}$$ Is this correct? –  bitbonk Aug 7 '12 at 11:32
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