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∆PQR is an equilateral triangle with side 18 cm. A circle is drawn on segment QR as diameter. Find the length of the arc of this circle within the triangle.

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Let ‘O’ be the centre of the circle drawn on QR as a diameter. 

Let the circle intersect seg PQ and seg PR at points M and N respectively. 

Since l(OQ) = l(OM), 

m∠OM Q = m∠OQM = 60°

m∠MOQ = 60° 

Similarly, m∠NOR = 60° 

Given, QR =18 cm. 

r = 9 cm

θ = 60° = (60 x π/180)c 

= (π/3)c

l(arc MN) = S = rθ = 9 x π/3 = 3π cm.

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