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Re: 90 < x < 180 [#permalink]
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The question doesn't state that O is the center. How can we assume that this is the case?
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Re: 90 < x < 180 [#permalink]
ImRoma wrote:
The question doesn't state that O is the center. How can we assume that this is the case?

You are right, Carcass should fix the question. It should have been expressed that O stood for the center of the circle, otherwise, no way to assume it is an isosceles!
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Re: 90 < x < 180 [#permalink]
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The fact that O is the center of the circle or not is irrelevant to find the perimeter

OB is 5.

So, regardless the shape of the triangle, drawn or not the scale, similarly OA must be 5 alike

\(x:x:x\sqrt{2}\)

\(5*\sqrt{2}\)=7.07

The perimeter is 17.07

A is the answer
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Re: 90 < x < 180 [#permalink]
Carcass wrote:
The fact that O is the center of the circle or not is irrelevant to find the perimeter

OB is 5.

So, regardless the shape of the triangle, drawn or not the scale, similarly OA must be 5 alike

\(x:x:x\sqrt{2}\)

\(5*\sqrt{2}\)=7.07

The perimeter is 17.07

A is the answer


How do you know OA must be same as OB if there is no need to know whether O is the center or not. OA can be assumed as 5 only when it is explicitly specified in the question root that OB and OA stand for the radius of the circle, center of which is O.

We have no information indicating the existence of isosceles... Only Center O provides this info.
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Re: 90 < x < 180 [#permalink]
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