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Make a rectangle ABCD where AB & CD r the longer sides (lengths) of the rectangle ABCD.Connect the diagonal AD.Extend BD upto E & connect AE making a right triangle ABE.Connect EC.Here if CD is given as an integer value then we will not get AE in integer values if ECis having integer values or if AE is having integer values then we will not get EC in integer values.
Can this be proved?

2007-01-24 23:57:36 · 8 answers · asked by rajesh bhowmick 2 in Science & Mathematics Mathematics

8 answers

i dont know whether we can prove that or not
but i can prove that you are very intelligant in maths

2007-01-25 00:06:59 · answer #1 · answered by ani 2 · 0 0

If E is on the line BD ouside of the rectangle, then ABE cannot be a right triangle. So I suggest you correct your question.

Actually all you are saying starts to make sense if you assume that the diagonals are AD and BC which is not the usual notation... You also have to require that AC and BD have integer values otherwise nothing prevents EC and AE to take the values 5 and 6 if CD =3.

2007-01-25 00:18:03 · answer #2 · answered by gianlino 7 · 0 0

you probably meant a square rectangle ABDC, where E is a point on the extension of BD outside the rectangle, than ABE and CDE are square sided triangles where B and D are the square side. then AE = SQRT(AB^2 +BE^2) and CE = SQRT(DE^2+CD^2).
but this allows for AE and CE to be integer in many cases.
Maybe you should rephrase your question

2007-01-25 00:50:05 · answer #3 · answered by Anonymous · 0 0

It can't be proved because it is not true, as you have stated it here.

CD is the same length as AB. So you're asking that in the triangle ABD, if AB is integer, then AE should not be and if AE is integer, AB should not be.

But it is perfectly possible for them both to be integer, for example, 3 and 5.

2007-01-25 00:53:23 · answer #4 · answered by Gnomon 6 · 0 1

If u make a rectangele ABCD with AB & CD as longest sides, in any way u will get AD as a breadth & not as a diagnol. So, the question cannot be continued & hence according to me, it cannot be be proved.

2007-01-25 00:04:57 · answer #5 · answered by shailendra s 3 · 0 2

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2016-12-12 19:53:02 · answer #6 · answered by degennaro 4 · 0 0

Yes, it may be incorrect when finished but it can be proved.

2007-01-25 00:06:02 · answer #7 · answered by Anonymous · 0 0

the answer is yes.. or maybe no

i hope this helps!

2007-01-25 00:07:54 · answer #8 · answered by Sarah W 2 · 0 1

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