Got Homework?
Connect with other students for help. It's a free community.
Here's the question you clicked on:
 0 viewing
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
 one year ago
 one year 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
 one year ago
 one year ago

This Question is Closed

sauravshakyaBest ResponseYou've already chosen the best response.2
same property means?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
both triangles should have rational sides.
 one year ago

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

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

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

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

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

shubhamsrgBest ResponseYou've already chosen the best response.0
sorry,,tanA=BC/Ab ,,but still rational..
 one year ago

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

mukushlaBest ResponseYou've already chosen the best response.2
i thought its not right
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
hmm,,well,,how will sum of sides help ?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
sorry that will not help..
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
AM^2 and MC^2 will be rational
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
yep..thats where even i was wondering
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
this ques is a subpart of my previous ques,,do both have any relation ?
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
i mean ABC is a right triangle?
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
aha...thats what i had assumed!! so its not a right triangle!! hmm..
 one year ago

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

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

shubhamsrgBest ResponseYou've already chosen the best response.0
well why were AM^2 and MC^2 rational ?
 one year ago

sauravshakyaBest ResponseYou've already chosen the best response.2
Take right triangle AMB and triangle BMC
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
ohh leave it got it..
 one year ago

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

shubhamsrgBest ResponseYou've already chosen the best response.0
has to be an other way around..
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
yeah i was thinkin of that
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
guys i gotta go...but i love to know what is the answer
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
dw: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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
hasn't this been solved yet?
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
i lost track of it ... it's fuzzy ... well what's the Q?
 one year ago

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

sauravshakyaBest ResponseYou've already chosen the best response.2
@experimentX thats the question
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
dw:1345138702333:dw a^2 + b^2 = c^2 b^2 = rational AM + MC = rational AM^2 + MC^2 = rational. AM*MC = rational.
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
Let 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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
typo: Let's assume both to be irrational,
 one year ago

mukushlaBest ResponseYou've already chosen the best response.2
here is my proofdw:1345140385316:dw
 one year ago

mukushlaBest 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
 one year ago

experimentXBest ResponseYou've already chosen the best response.1
lol ... your method is more obvious!!
 one year ago

shubhamsrgBest ResponseYou've already chosen the best response.0
aha..i see and understand both methods! thanks again @mukushla and @experimentX
 one year ago
See more questions >>>
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.