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calculate a diagonal of the rectangular havind sides as 40.6, 40.2, 35.2 and 40

Guest May 18, 2014

#12**+5 **

Give Francis my thanx for the reduction....that was vey "mule"-if-i-cent of him.

And as for those "flakes,' I believe he can find plenty of those on the forum....and he won't have to search very hard, either!!!

Now...you'll have to excuse me....I've got to get back to my game of Donkey Kong with Sisyphus. At this point, I think I have him "up a hill."

CPhill May 19, 2014

#2**0 **

It's not a "rectangle," but just a quadrilateral........I call it an "oddtangle"

Here's what you asked for:

CPhill May 18, 2014

#4**0 **

If I could get this stupid thing to display properly, I'm going to try to provide that, if I can!!

No promises, though!!

CPhill May 18, 2014

#6**0 **

Well ...IF you could see the whole thing (I'm tired of messing with it!!) The sides AB and BC (BC is off the chart, so to speak) form a right angle

So sqrt(40.6^2 + 40.2^2) =57.1 and that's the length of AC

I haven't actually figured out a way to calculate the other diagonal...if there is one?? I might look at it later and see if anything "pops" into my head. Maybe we could set it up as a question in the "puzzles" section??

Besisdes, the questioner only asked for "a" diagonal.......he didn't atually specify * WHICH* one !!

I claim I've actually done my duty!!!

Time for lunch!!

CPhill May 18, 2014

#7**+5 **

To get the length of the other diagonal, DB, we'll use this "formula" to find the measure of angles DAC and CAB :

Cos θ = [u(dot)v] /( llull * llvll) Where u and v are vectors and

[u(dot)v] - the dot product of u and v and

llull = length of u and

llvll = length of v

So, taking AD and AC as vectors

AD =<5.4, 39.7> AC = <40.6, 40.2> AD(dot)AC = 1815.18

llADll = 40 llACll = 57.1

cos^{-1} (1815.18)/(40*57.1) = m< DAC =37.37

And again, taking AB and AC as vectors

AB = <40.6 ,0> AC <40.6, 42.7> AB(dot)AC = 1648.36

llABll = 40.6 llACll = 57.1

cos^{-1} (1648.36)/(40.6*57.1) = m< CAB = 43.8

m< DAB = m< DAC + m< CAB = 37.37 + 44.7 = 82.07

And using the Law of Cosines, DB = SQRT (40^2 + 40.6^2 - 2(40)(40.6)cos(82.07)) = 52.92 ≈ 53

And that's close enough for me!!

CPhill May 18, 2014

#8**+5 **

Impressive calculations, but the quadrilateral isn't unique - see illustration below (drawn to scale in Geogebra). Compare the blue and red quadrilaterals, both on the same black baseline.

Alan May 18, 2014

#9**+5 **

Yeah, Alan....I realized that I could have "manipulated" the Geogebra generated object to produce any number of quads with these sides....this one just had a right angle which was fairly easy to deal with.

....But, at the moment.....I'm still looking for a missing zero along with my pal Sisyphus.....no more time for this "four-sided" silliness!!

CPhill May 19, 2014

#10**+5 **

Okay,

So what Alan has demonstrated is that there is no specific answer because many different quadrilaterals can have these side lengths.

-----------------------------------------------------------------------------

Hi Alan,

I am impressed with your use of GeoGebra. I have not found it very intuitive and have not progressed passed the real basics.

I believe you can also convert your diagram to Latex code and implant it that way!

I want an easy way to make and use a number line. I did it last time with GeoGebra just by cutting out the y axis but it wasn't as good as I would have liked. Do you know if this sort of thing can be done properly with GeoGebra?

Maybe Heureka knows if you do not ?

Melody May 19, 2014

#11**0 **

Francis reduced the image for you, CPhill. His fee is two flakes of hay, and 1.5 quarts of sweet feed, --Remember, “imperial quarts”. Frances says he’s very imperious, and she’d really like crown him sometimes.

Frances and Co.

PS IT didn’t work. It resizes to large after publishing. Francis says he’ll work on it.

Guest May 19, 2014

#12**+5 **

Best Answer

Give Francis my thanx for the reduction....that was vey "mule"-if-i-cent of him.

And as for those "flakes,' I believe he can find plenty of those on the forum....and he won't have to search very hard, either!!!

Now...you'll have to excuse me....I've got to get back to my game of Donkey Kong with Sisyphus. At this point, I think I have him "up a hill."

CPhill May 19, 2014