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Solution : Let AB = p; BC = b; AC = h By Pythagoras theorem, p^{2} + b^{2} = h^{2} Area of circle with diameter' d' = \frac{{πd}^{2}}{8} So, Area of semicircle with AB as diameter = \frac{{πp}^{2}}{8} Area of semicircle with BC as diameter = \frac{{πb}^{2}}{8} Area of semicircle with AC as diameter = \frac{{πh}^{2}}{8} Area of semicircle with AB as diameter + Area of semicircle with BC as diameter = \frac{{πp}^{2}}{8} + \frac{{πb}^{2}}{8} = \frac{π (p^{2} + b^{2})}{8} = \frac{{πh}^{2}}{8} = Area of semicircle with AC as diameter

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