A community for students. Sign up today!
Here's the question you clicked on:
 0 viewing
 2 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
 2 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

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2same property means?

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

shubhamsrg
 2 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.

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2AC.MB is rational because area is rational and AC is rational so MB is rational ...now u just proveAM and MC are rational

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2I think it will be enough to show just one side Am or MC rational

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2beause if one is rational than the other must be rational

shubhamsrg
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0sorry,,tanA=BC/Ab ,,but still rational..

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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2i thought its not right

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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2sorry that will not help..

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

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

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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2i mean ABC is a right triangle?

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

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

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2now our problem is like this if \(a+b\) and \(ab\) are rational prove that both \(a\) and \(b\) are rational

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

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2Take right triangle AMB and triangle BMC

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

shubhamsrg
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.0has to be an other way around..

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2yeah i was thinkin of that

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2guys i gotta go...but i love to know what is the answer

shubhamsrg
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1hasn't this been solved yet?

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

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2dw:1345132117021:dwif ABC has rational sides, we have to show AMB and MBC also has rational sides.

sauravshakya
 2 years ago
Best ResponseYou've already chosen the best response.2@experimentX thats the question

experimentX
 2 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
 2 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
 2 years ago
Best ResponseYou've already chosen the best response.1typo: Let's assume both to be irrational,

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2here is my proofdw:1345140385316:dw

mukushla
 2 years ago
Best ResponseYou've already chosen the best response.2\(x+y=c\) 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
 2 years ago
Best ResponseYou've already chosen the best response.1lol ... your method is more obvious!!

shubhamsrg
 2 years ago
Best ResponseYou've already chosen the best response.0aha..i see and understand both methods! thanks again @mukushla and @experimentX
Ask your own question
Ask a QuestionFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.