geometry question.
@JoannaBlackwelder
Well, we have lots of isosceles triangles, but I'm not sure yet hoe to find the radius...
*how haha
Ok, yea, I saw that too. I used the secant tangent theorem and extended BD so it intercepts the circle at point F and FD=6 with that theorem. And I hope that word was just a typo
Wouldn't you like to know? And I get FB=6. I think that is what you mean, right?
Oh yes, my typo
If we extend the line segment AB, we could make a secant line that crosses the other secant line.
|dw:1392014850059:dw|
|dw:1392014941530:dw| this is what i found
|dw:1392014989272:dw| and is this what you want to do?
Yup. :)
Because the products of the lengths of the secants are equal.
And since A is the center, A to the circle is a radius.
So, 4*6=r*(4+r-4)
Make sense?
i dont get the r*(4+r-4) part
Wait, that was wrong...just a sec.
Let's call the radius of the circle r.
So the piece of the secant that passes through A is r+4.
And the other piece of the secant is r-4.
So the equation is 6*4=(r+4)(r-4)
ok. but its not really a secant since it passes through the center. its a diameter
It is a diameter, but a diameter is just a special type of secant, right?
I 'm sorry, wrong terminology. I mean chord, not secant.
ok. i guess your right
You don't sound convinced. Do you not think my method is sound?
i know im convinced. i just didnt know a diameter was also a chord, but now im convinced http://answers.yahoo.com/question/index?qid=20090221092517AAXylCx
ok. and when i solved it, i got 2 root 10
Yup, me too. :) Yeah, the diameter is the longest chord.
ok, so thats it?
Looks like it to me.
Is it supposed to be harder than that?
i dont think so. if only i knew a diameter was also a chord... but i did learned something new today. thank you so much!
Haha, no worries. :)
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