Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

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.

share|cite|improve this answer
+1 for the extra bonus of supporting asymmarical trapezoids and for the detailed deduction. – bitbonk Aug 7 '12 at 11:29

So ,now $$\frac{a-b}{2h}=\frac{x}{h-h^'}$$ And your answer will be $b+2x$

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.