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I'm trying to prove that length of the line $AB$, parallel to both bases of a trapezoid, that cuts a trapezoid into two trapezoids of equal area is the Root Mean Square of the bases. In other words, if the length of the top base is $a$ and the length of the other base is $b$, then :

$AB=\sqrt{\frac{a^2 + b^2}{2}}$

I've been stuck on this for a couple of days. Can someone give me a hint? Please don't just state the proof. I've already tried using the areas of the two trapezoids to get the RMS.

  • What does it mean for a "line" (line segment?) to be "the Root Mean Square"? Are you describing the length of such a line segment? In adjuring us not to "just state the proof", are you emphasizing that you want only a hint, or is there another implied request? – hardmath Jan 15 '14 at 03:21
  • Sorry, I'd like a hint. I'll amend the question to make it more clear. –  Jan 15 '14 at 03:22
  • Let the length of $AB$ be $c$, let the height of the trapezoid be $h$, and let the height of the sub-trapezoid between bases $a$ and $c$ be $k$ (so that the height of the sub-trapezoid between bases $b$ and $c$ is $h-k$). Proportionality arguments show that $c$ and $h$ are related by $$c = a + \frac{k}{h} ( b - a )$$ (Sanity check: Try $k=0$ and $k=h$.) Use this relation to write $k$ and $h-k$, and then also the areas of the two sub-trapezoids, in terms of $a$, $b$, $c$, and $h$. Setting the areas equal should give the relation you seek. – Blue Jan 15 '14 at 05:11
  • Use the fact that surface areas of two subtrapezoids are equal.On the other hand the sum of their surface areas is equal to the total surface area.From these two you can obtain the length of the required segment. – Nasibabuba Jan 15 '14 at 05:27
  • P.s. I mentioned I tried that approach. I couldn't get anywhere with it. Blue, can you expand on the proportionality arguments? I tried using similar figures to get proportions, but I couldn't show any helpful ones. –  Jan 15 '14 at 07:37
  • @Josh: For the proportionality argument, suppose that $a$ is the shorter base. Draw a line through an endpoint of the $a$-base, parallel to the opp side; this cuts the trapezoid into a parallelogram (with upper and lower bases $a$), and a triangle (with (lower) base $b-a$). The drawn line separates the "middle" base, $c$, into a part within the parallelogram, and a part within the triangle; the first part has length $a$, while the second part has, by similar triangle proportions, length $\frac{k}{h}(b-a)$, where $h$ is the height of the full triangle, and $k$ is the height of the sub-triangle. – Blue Jan 15 '14 at 10:59
  • My answer here is relevant. – Will Orrick Mar 01 '19 at 00:42

1 Answers1

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Suppose that $a$ is the shorter (and "upper") base. Draw a line through an endpoint of the $a$-base, parallel to the opposite lateral side; this cuts the trapezoid into a parallelogram (with upper and lower bases $a$) and a triangle (with (lower) base $b-a$). The drawn line separates your "middle base", say of length $c$, into a part (of length $a$) within the parallelogram and a part (of length $c-a$) within the triangle.

enter image description here

If $h$ is the height of the parallelogram (and triangle), and $k$ is the height the upper sub-trapezoid (and upper sub-triangle), then, by similarity, $$\frac{k}{h}=\frac{c-a}{b-a} \tag{1}$$

Now, assuming the middle base bisects the area, so that the upper sub-trapezoid's area is half the whole, we have $$\frac{1}{2}k(a+c) = \frac12\cdot\frac12h(a+b) \quad\to\quad \frac{k}{h} = \frac12\frac{a+b}{a+c} \tag{2}$$

Setting $(1)$ and $(2)$ equal gives $$\frac{c-a}{b-a}=\frac12\frac{a+b}{a+c}\quad\to\quad 2(c^2-a^2) = b^2-a^2 \quad\to\quad 2c^2 = a^2+b^2 \tag{3}$$ which is the target relation. $\square$

Blue
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  • @AaronLiu: The "mysterious square" $\square$ is a common typographic device for marking the end of a proof. It's sometimes called the tombstone, and was introduced by Halmos. – Blue Jan 19 '23 at 03:43