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March 16th, 2018, 12:44 PM   #1
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The "height" of Pythagoras Theorem

Determine h = ? in the figure.
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March 16th, 2018, 01:41 PM   #2
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reference the attached diagram.

similar triangles ...

$\dfrac{h}{c + \frac{ab}{c}} = \dfrac{b}{a}$

$h = \dfrac{b}{a}\left(c + \dfrac{ab}{c}\right) = \dfrac{b(c^2+ab)}{ac}$
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March 16th, 2018, 03:41 PM   #3
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Your starting point in the drawing is not correct.
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March 16th, 2018, 04:50 PM   #4
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Quote:
Originally Posted by avniu View Post
Your starting point in the drawing is not correct.
... complete the problem yourself, then.
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March 16th, 2018, 05:19 PM   #5
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There are two possibilities:

1. h is perpendicular to the two long lines and the two long lines are parallel. If this is the case then Skeeter's solution is perfect.

2. h was not marked perpendicular and the two long lines were not marked parallel. If this is the case then the OP is a clown.
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March 17th, 2018, 01:27 AM   #6
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This is an elegant geometry puzzle for geometry fans. A clown is the one that can not solve this puzzle.
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March 17th, 2018, 05:19 AM   #7
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Unfortunately, skeeter slipped up and got an incorrect answer, and avniu correctly pointed out that skeeter's diagram is the source of that slip.

Assuming that the various apparently perpendicular and apparently parallel lines in the original diagram are perpendicular and parallel respectively, there arises the question of how the diagram should be drawn if a > b.

I therefore suggest the additional assumption that a $\small\leqslant$ b.

With all the above assumptions, one gets h = (b² + 3ab)/c, but I have yet to find an elegant way to obtain this.
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March 17th, 2018, 08:38 AM   #8
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Many thanks Skipjack for the constructive and correct answer, a math forum needs people like you.
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