Ok Let us solve with property of triangles 30-60-90..
look at the attached figure..
Take triangle RPO....
It is a right angled triangle with sides RO:PO:RP in the ratio of r/2:PO:r or 1:PO:2.
This is nothing but a 30-60-90 triangle with ratio of sides as 1:\(\sqrt{3}\):2
so Angle PRO is 60 and similarly Angle QRO is also 60

Thus the central angle is 120 and the arc made by it will be \(\frac{120}{360}=\frac{1}{3}\)

B

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Any other way ?
Also, why is this not a square. Diagnols bisect at 90 in a square as well


edited: answer to myself rhombus also has bisectors at 90 degrees

Posted from my mobile device

Please have a reference of the above figure posted by GMATinsight for the solution i am posting right now.

We know that radius of circle is OB and OC both.
Pick triangles OCS and SCB , they are congruent by SAS (CS=CS Angle = 90 and OS = SB given)
---> OC= CB
or OC = OB = CB ---> Equilateral triangle . Hence angle COA is 60 + 60 = 120 (Tirangle OAB is also equilateral by same reeasoning as above).

Now (120/360*2pi*OC)/2piOC = required answer = 1/3.

Please reply, if you did not get this LT2018