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Re: What is the area of circle O above? [#permalink]
25 Aug 2013, 23:04

1

This post received KUDOS

In the Quadilateral(OABC), since Ang(O),Ang(B), and Ang(C) are Right angles, Ang(A) also must be right angle. (Since sum of interior angles of a Quadilateral = (n-2) * 180 where n = number of sides)

Hence we could say, Quadilateral(OABC) is a Rectangle.

Re: What is the area of circle O above? [#permalink]
26 Aug 2013, 00:13

Smallwonder wrote:

In the Quadilateral(OABC), since Ang(O),Ang(B), and Ang(C) are Right angles, Ang(A) also must be right angle. (Since sum of interior angles of a Quadilateral = (n-2) * 180 where n = number of sides)

Hence we could say, Quadilateral(OABC) is a Rectangle.