PQR is a triangle, P and Q r the centres of 2 intersecting circles, prove that PQ=QR=PR

- anonymous

PQR is a triangle, P and Q r the centres of 2 intersecting circles, prove that PQ=QR=PR

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- schrodinger

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- anonymous

DO you have a picture?

- nincompoop

we need to draw the picture based on the description and prove

- anonymous

yep i have one but how to insert it
|dw:1377587687555:dw|

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## More answers

- nincompoop

that's the figure you were given?

- oOKawaiiOo

Nincompoop, make sure P and Q are the at the middle of the circle. Maybe that would help?

- anonymous

yes this is the fig but it is not neat

- nincompoop

|dw:1377588053427:dw|

- oOKawaiiOo

|dw:1377588160226:dw|

- nincompoop

that's what I was trying to draw, so I was a little confused when you put up your own figure… well it's intersecting… LOL |dw:1377588177376:dw|

- anonymous

nincompoop's is wrong

- nincompoop

|dw:1377588261886:dw|

- anonymous

ya like the second one

- oOKawaiiOo

Yeah, the red circles should be right

- anonymous

kawaii's is right

- oOKawaiiOo

Thats what I thought........

- nincompoop

if the circle has the same area or perimeter then all sides of the triangle are equal. that's the only condition

- anonymous

look the twelve question

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- nincompoop

I don't understand anything besides Greek Alphabet and English Alphabet, man
Don't show me sanskrit-looking scribbles LOL

- anonymous

no under that it is written in english

- anonymous

it is written in two languages

- oOKawaiiOo

She got that terrorist homework.

- nincompoop

ohhhh I see! LOL kawaii you're messed up

- oOKawaiiOo

NO WE WILL NOT HELP YOU DEVELOP THE NEXT BOMB!

- nincompoop

well to prove that PQ=RQ=PR
like I said that in that figure we assume that the circles are identical
and any distance from the center of a circle represents the radius, and all segments from the center (radius) are equidistant. since each segment (radius) is represented by P, Q and R we have the following demonstration:
PR = r
RQ = r
PQ = r
r = r = r
∴PQ=RQ=PR
I don't know the jargons yet. but this should suffice
if you want you can wait until I finish my analytical geom to finis this up.

- nincompoop

Circle got this I think… LOL it's written in his/her name

- anonymous

Actually, the question is not really complete, because it also needs to state that both circles have equal radii. Assuming the radius of both circles is RAD, then you know that the distance from the center of such a circle to the circle itself is by definition RAD. All lengths, PQ, QR and PR are either on centers or on the circles itself so all these distances need to be equal to RAD and hence have the same length. Hope this helps...

- anonymous

@nincmpoop....seems you've said the same thing, my apologies for duplicating

- anonymous

ya thank yuuuuuuuuuuuuuuu

- nincompoop

http://www.regentsprep.org/Regents/math/geometry/math-GEOMETRY.htm
http://info.tetonscience.org/JSR/High%20School/Core%20Classes/Math/IB%20Math%20HL%20(McClennen)/Mathematics%20HL%20OPTIONS%20(D)/contents/euclidian_geom.pdf

- anonymous

so PQ AND QP R THE RADII OF THE CIRCLES Q AND P. EVEN PR AND QR ARE THE RADII OF THE CIRCLES THUS IT IS PROVED

- nincompoop

NO. PQ does not represent a circle. Just the centers of an intersecting identical circles which make up a diameter PQ

- anonymous

K

- nincompoop

since diameter = 2r
the rest follows

- anonymous

THANKS

- nincompoop

don't use short language like "r" and "u" when referring to "are" and "you" they are troublesome to read specially when dealing with geometry

- anonymous

K

- nincompoop

"okay"

- anonymous

OKAY

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