Arc of inscribed circle : Quant Question Archive [LOCKED]
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# Arc of inscribed circle

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Director
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24 Nov 2007, 10:13
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

Here's a MGMT problem. Can some please explain how they have obtained the value of the arc. I was able to understand the explanation. Details would be helpful. Thanks

Thanks for you help in advance.
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arc of the inscribed circle.doc [188.5 KiB]

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24 Nov 2007, 10:37
All you need to know is that the arc degree faceing x is 2x (the whole circle is 360 degrees round) , that way if you solve for x=30 then 2x = 60 and since the perimeter of the circale is 2*pi*5 = 10*pi, and you need (60+60)/360 = 1/3 of that perimeter. Then the arc is 10/3*pi.

Did you understood the values for the sides (i.e. 10*sqrt(3) ?

Director
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24 Nov 2007, 11:09
KillerSquirrel says :

All you need to know is that the arc degree faceing x is 2x (the whole circle is 360 degrees round) , that way if you solve for x=30 then 2x = 60 and since the perimeter of the circale is 2*pi*5 = 10*pi, and you need (60+60)/360 = 1/3 of that perimeter. Then the arc is 10/3*pi.

Alimad : Asking a dumb question, is this a a hard and fast rule that the arc's degree would be twice its opposite angle?

KillerSquirrel says :
Did you understood the values for the sides (i.e. 10*sqrt(3) ?

Alimad : Sort of, I thought when you draw a straight line from C to A, BAC is the right angle instead of BCA, or don't you think we should draw a straight line from C to E, that would make BAC and BAE right angles ? Thanks for all you help
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24 Nov 2007, 11:33
1. Yes - its a rule see here ---> http://www.gmatclub.com/forum/t53871

2. See post below attachment

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untitled.GIF [ 6.49 KiB | Viewed 798 times ]

Last edited by KillerSquirrel on 25 Nov 2007, 13:39, edited 3 times in total.
Director
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24 Nov 2007, 11:38
Keeping your attachment in mind, the hypotenese we get the 30:60:90 rule or x : xRoot3 : 2x . We know that 2x = 10, so x = 5

Do you agree?
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25 Nov 2007, 13:36
In order to solve the sides of the triangle you need to know that an inscribed triangle in a circle will always have one side at 90 degrees.

So if you draw a line from C to A you get an perfect right triangle with one side of 10 ---> x:x*sqrt(3):2*x = 5:5*sqer(3):10

Since you need two sides then 2*5*sqrt(3) = 10*sqrt(3)

:)
25 Nov 2007, 13:36
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