![]() 05/24/2019 at 11:30 • Filed to: None | ![]() | ![]() |
[EDIT]: Thank you.
A 16-foot chord is inscribed on a circle with a radius of 767.81 feet. What is the altitude of the bisector of the chord to the circle? +2 if you can tell me how you come up with the answer.
![]() 05/23/2019 at 20:38 |
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I vote 16’, since the two extremes of the function are the points where x=r and thus the length of the chord equals r, and x=0 when the length of the chord equals 0.
I’m sure there’s calculus to prove me wrong, but FAKE NEWS.
![]() 05/23/2019 at 20:40 |
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![]() 05/23/2019 at 20:41 |
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https://www.mathopenref.com/sagitta.html
- it explains it and provides a calculator - the bisector in your problem is the “sagitta”
![]() 05/23/2019 at 20:42 |
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.04 ft
![]() 05/23/2019 at 20:42 |
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![]() 05/23/2019 at 20:43 |
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As a Sagittarius I’m more than a little disappointed I didn’t know that term.
![]() 05/23/2019 at 20:44 |
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that’s the MathNerd ™ term, it’s pretty obscure
![]() 05/23/2019 at 20:44 |
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If the chord connects on either end to the circle, then you have a right triangle with hypotenuse with the same length as your radius. Using half the length of the chord (8') and R (767.81'), you can solve for the third leg using Pythagoreum’s theorem, which would be equal to (R - x).
So then (R)^2 = (8)^2 + (R-x)^2, solve for x.
x = .0417'
ed: damn too slow
![]() 05/23/2019 at 20:46 |
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hmm
![]() 05/23/2019 at 20:49 |
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Not all paths between two points are a straight line. In fact, to a very good approximation, none of them are. REAL NUMBERS
![]() 05/23/2019 at 20:55 |
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Id just draw it in CAD and measure it but that would mean id have to get up
![]() 05/23/2019 at 21:07 |
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Gotta love constraints and the measure tool
![]() 05/23/2019 at 23:06 |
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Yes! Indeed.
![]() 05/23/2019 at 23:09 |
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I did that in Algebra once. Or was it Calculus? The math is familiar to me, but it's been many years. Thank you!
![]() 05/23/2019 at 23:31 |
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![]() 05/23/2019 at 23:46 |
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!!! UNKNOWN CONTENT TYPE !!!
Maybe no t in your hyperspace.
![]() 05/24/2019 at 00:37 |
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![]() 05/24/2019 at 01:07 |
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It takes an engineer! They can’t get ride of us so easily.
![]() 05/24/2019 at 09:44 |
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Pythagoras strikes again!
![]() 05/24/2019 at 11:33 |
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Was that a practical problem or an intellectual problem?
![]() 05/24/2019 at 12:10 |
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![]() 05/24/2019 at 12:26 |
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Practical. My BIL is building something for his job, probably a bridge, and he’ll be looking for how high the crest is from either side. Probably ordering some ready-made fabricated steel structure. But Oppo came through and showed me the geometry, and it’s really quite accessible.
![]() 05/24/2019 at 16:14 |
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I’m just trying to figure out how you measure music in feet
![]() 05/24/2019 at 16:18 |
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Yes, an interesting notion how that word shows up in both contexts. I’ll ask Ttyymmnn about that one...