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Circles problem:;

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by anuprajan5 » Tue Oct 23, 2012 9:48 pm
The answer is B

k is the hypotenuse for a triangle with sides r and r

Therefore k = root 2 *r

[spoiler]r = k/ root 2[/spoiler]
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by bobdylan » Wed Oct 24, 2012 6:36 am
I do not think the hypothenuse is the radius of the circle in question.
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by anuprajan5 » Wed Oct 24, 2012 6:39 am
It isn't. If you draw a triangle using the centre as one of the vertices, you will find that the sides of the triangle are r in length.

k is the distance from O to C which makes it the hypotenuse of that triangle.
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by abhi131 » Wed Oct 31, 2012 6:29 pm
Draw a line through C parallel to the y-axis intersecting the x-axis at D, and then join C and O.
You will see that COD is an isosceles right angled triangle, with the two sides as r. And the hypotenuse as k. For a right angled isosceles, k = r*sqrt(2).
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by das.ashmita » Wed Oct 31, 2012 10:17 pm
Hi bobdylan

For your reference.


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