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shubhamsrg
 3 years ago
now the question which follows up is :
If all the sides and area of a triangle were rational numbers then show that the
triangle is got by `pasting' two rightangled triangles having the same property
shubhamsrg
 3 years ago
now the question which follows up is : If all the sides and area of a triangle were rational numbers then show that the triangle is got by `pasting' two rightangled triangles having the same property

This Question is Closed

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0both triangles should have rational sides.

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1345129143913:dw if ABC has rational sides, we have to show AMB and MBC also has rational sides.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0AC.MB is rational because area is rational and AC is rational so MB is rational ...now u just proveAM and MC are rational

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I think it will be enough to show just one side Am or MC rational

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0beause if one is rational than the other must be rational

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0if we say tanA = BC/AC = rational also tanA = MB/AM is it good enough to say here that AM will be rational ?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0sorry,,tanA=BC/Ab ,,but still rational..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ABC is right triangle !!? work on \[\sqrt{AB^2MB^2}+\sqrt{BC^2MB^2}=AM+MC\]

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i thought its not right

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0hmm,,well,,how will sum of sides help ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sorry that will not help..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0AM^2 and MC^2 will be rational

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0yep..thats where even i was wondering

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0this ques is a subpart of my previous ques,,do both have any relation ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0i mean ABC is a right triangle?

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0aha...thats what i had assumed!! so its not a right triangle!! hmm..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0sauravshakya mentiond a good point \[(AM+MC)^2=AC^2=AM^2+2AM.MC+MC^2\]so \(AM.MC\) is rational

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0now our problem is like this if \(a+b\) and \(ab\) are rational prove that both \(a\) and \(b\) are rational

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0well why were AM^2 and MC^2 rational ?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Take right triangle AMB and triangle BMC

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0ohh leave it got it..

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0if ab and and a+b are rational,,its not necessary both are rational,, eg. sqrt(2) +1 and sqrt(2) + 1

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0has to be an other way around..

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0yeah i was thinkin of that

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0guys i gotta go...but i love to know what is the answer

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1345131550151:dw here,,we have AC/sin(a+b) = BC/sinx now sin(a+b) = sina cosb + cosa sinb = AM/AB . BM/BC + BM/AB . MC/BC = (AM+BM)/AB + (BM +MC)/BC also, sin x =BM/AB on substituing, AC . BM/AB = BC . ( (AM+BM)/AB + (BM +MC)/BC ) AC.BM = (AM+BM)BC + (MB +MC)AB AC BM = AM BC + BM BC + MB AB + MC AB I DONT KNOW IF IT HELPS OR NOT!! lol.. :P

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1hasn't this been solved yet?

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1i lost track of it ... it's fuzzy ... well what's the Q?

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0dw:1345132117021:dwif ABC has rational sides, we have to show AMB and MBC also has rational sides.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0@experimentX thats the question

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1dw:1345138702333:dw a^2 + b^2 = c^2 b^2 = rational AM + MC = rational AM^2 + MC^2 = rational. AM*MC = rational.

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1Let me finish it man!! AM + MC = rational < either both are irrational or rational) AM*MC = rational < either both are irrational or rational Let's assume both to be rational, AM ( rational  AM) = rational AM (rational)  AM^2 = rational irrational  rational = rational (contradiction) hence both must be rational

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1typo: Let's assume both to be irrational,

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0here is my proofdw:1345140385316:dw

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0\(x+y=c\) is rational Area=\(ch/2\)=rational so \(h\) is rational since \(c\) is rational we have\[x^2+h^2=a^2\]\[y^2+h^2=b^2\]\[(xy)(x+y)=(xy).c=a^2b^2\]so we obtain\[xy=\frac{a^2b^2}{c}\]\[x+y=c\]it gives\[x=\frac{a^2+c^2b^2}{2c}\]\[y=\frac{b^2+c^2a^2}{2c}\]\(x\) and \(y\) both are rational

experimentX
 3 years ago
Best ResponseYou've already chosen the best response.1lol ... your method is more obvious!!

shubhamsrg
 3 years ago
Best ResponseYou've already chosen the best response.0aha..i see and understand both methods! thanks again @mukushla and @experimentX
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