Let PQR be an equilateral triangle with side equal to x and let the intersection point of PS and QR be M. Clearly, the circle is the circumcircle of the triangle PQR. QR = x => QM = $$frac{x}{2}$$ because a perpendicular from the centre to any chord bisects the chord. Angle OQM = 30 degrees and QM is equal to $$frac{x}{2}$$ => OQ = $$frac{frac{x}{2}}{cos(30)}$$ = $$frac{x}{sqrt{3}}$$ Hence the radius of the circumcircle of an equilateral triangle is equal to $$frac{x}{sqrt{3}}$$. Angle PQS = 90 degrees as it is an angle in a semicircle. PS bisects angle QPR => angle QPS is 30 degrees. Hence QS subtends an angle of 30 degrees in the major arc => QS subtends an angle of 60 degrees at the centre because angle subtended by a chord at the centre is twice the angle subtended by the chord in the major arc. Angle QOS = 60 degrees => Triangle QOS is equilateral and hence QS is equal to radius of the circle => QS = $$frac{x}{sqrt{3}}$$ Given that radius is r => r = $$frac{x}{sqrt{3}}$$ => x = $$rsqrt{3}$$ => Perimeter of PQRS = PQ + QR + RS + SP = $$ rsqrt{3} + r + r + rsqrt{3}$$ = $$2r(1+sqrt{3})$$