A chord AB of length 24 cm is drawn in a circle of radius 13cm.Find the area of the shaded portion APB.
 In $$\triangle$$OAB   Â
                           Â
     OC is drawn perpendicular to ABÂ
      By symmetry AC=CB=AB/2 =24/2= 12Â
                OA=13
          $$OC^{2}=OA^{2}-AC^{2}$$
          $$OC^{2}=13^{2}-12^{2}$$=25
                 OC=5
So area of shaded region = Area of sector OAPB - Area of $$\triangle$$OAB
                   =$$\frac{x}{360}*\pi*(OA)^{2}$$-$$\frac{1}{2}*(AB)*(OC)$$
                   =$$\frac{x}{360}*\pi*(13)^{2}$$-$$\frac{1}{2}*(24)*(5)$$
                   =$$\frac{169\pi x}{360}-60cm^{2}$$
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