∆PQR is an equilateral triangle and M is the midpoint of Line QR. Hence, PM is the radius of the circle likewise Line PA and Line PB Using Pythagoras theorem to find Line PM : \( 18^2 = PM^2 + 9^2\) \( PM^2 = 324 - 81 = 243\) PM = \(\sqrt{243} = 9\sqrt{3}\) ∠QPR = \(\frac{180^0}{3} = 60^0\) ( each ∠ of an equilateral △ are equal) Area of △,PQR = \(\frac{1}{2}absin\theta\), but \theta = 60^\circ = \(\frac{1}{2}\times18\times18sin60 = 81\sqrt{3} = 140.30 cm^2\) Area of PAB = \(\frac{\theta}{360} \times\pi r^2\) = \(\frac{60}{360}\times \frac{22}{7} \times(9\sqrt{3})^2\) = \(\frac{1}{6} \times \frac{22}{7} \times243\) = \(127.29cm^2\) therefore, the Area of the shaded portion = area of △ PQR - area of sector PAB = 140. 30 - 127.29 = \(13.01cm^2\) (to 2 d.p)