Plane Geometry Question: Distinction

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  • #1
aek
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PQR is an equilateral triangle, and A is a point on QR such that RA = 2 QA. Prove that PA^2 = 7QA^2

If someone can help, i'd really appreciate it. Thanks
 

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  • #2
aek said:
PQR is an equilateral triangle, and A is a point on QR such that RA = 2 QA. Prove that PA^2 = 7QA^2

If someone can help, i'd really appreciate it. Thanks
How have you attempted to prove this statement ?
 
  • #3
Curious3141
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aek said:
PQR is an equilateral triangle, and A is a point on QR such that RA = 2 QA. Prove that PA^2 = 7QA^2

If someone can help, i'd really appreciate it. Thanks
HINT : Apply cosine rule on triangle PQA.
 
  • #4
aek
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Thanks. . .

Why are you answering my question with a question? Just abit ironic thats all. Thanks for the hint hyper, it was helpful.
 
  • #5
Curious3141
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aek said:
Why are you answering my question with a question? Just abit ironic thats all. Thanks for the hint hyper, it was helpful.
I can't even tell if that's sarcasm. In any case, it was I that gave the hint, not hyper.

The policy of this forum is not to do homework or just solve the problem for the poster unless it is evident some effort or at least some serious thinking has been put into the solution by the poster. So I can understand why hyper posted what he did.

Have you solved the problem yet ? If you're still finding difficulty, apply the Cosine Rule as I suggested and see what you get. If stuck, post here with working, then you'll get help.
 
  • #6
Icebreaker
You can choose any arbitrary equilateral triangle PQR and let a point A on QR such that RA = 2QA. Show arithmatically (with numbers) that PA^2 = 7QA^2. Then proceed to argue that all triangles PQA are congruent, and therefore any proportions wrt its lengths (PA^2 = 7QA^2) must remain constant.
 
  • #7
aek
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yeah, it was sarcasm

yep i got the point,
but i too know that you have to use the cosine rule and i've tried and couldn't find a relevant answer. If you want proof that i tried, i could scan my work but you being so trustful that wouldn't be necassary :P
 
  • #8
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Here is an alternative but longer way to do it that avoids trig. Let M be the midpoint of RQ. Since PQR is equilateral, you know (it's a theorem) that PM must be perpendicular to RQ. Then you can use Pythagoras's theorem a couple of times to find how the different segments relate to QA.
 
  • #9
Curious3141
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aek said:
yep i got the point,
but i too know that you have to use the cosine rule and i've tried and couldn't find a relevant answer. If you want proof that i tried, i could scan my work but you being so trustful that wouldn't be necassary :P
[tex]PQ = QR = 3QA[/tex]

[tex]PA^2 = PQ^2 + QA^2 - 2(PQ)(QA)\cos 60^o[/tex]

[tex]PA^2 = 9QA^2 + QA^2 - 2(3QA)(QA)(\frac{1}{2})[/tex]

[tex]PA^2 = 7QA^2[/tex]

Perhaps now the sarcasm can end.
 
  • #10
aek
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perhaps it may. well thanks for the help, even though it took some time but truly, i really do appreciate it.
 

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