$A B C$ is a right-angled triangle in which $∠ A = 90^{\circ}$ $A B = 21 c m$ and $A C = 28 c m$ . Semi-circles are described on $A B, B C$ , and $A C$ as diameters. Find the area of the shaded region.

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Solution

Area of shaded region = ar. semi-circle AB + ar. semi-circle AC + ar. $△$ ABC - ar. semi-circle BC

$= \frac{π}{2} [r_{1}^{2} + r_{2}^{2}] + \frac{1}{2} \times A B \times A C - \frac{π}{2} \times r_{3}^{2}$

$= \frac{1}{2} \times \frac{22}{7} [{(\frac{21}{2})}^{2} + {(\frac{28}{2})}^{2}] + \frac{1}{2} \times 21 \times 28 - \frac{1}{2} \times \frac{22}{7} \times {(\frac{B C}{2})}^{2}$

$= \frac{11}{7} [\frac{441}{4} + \frac{784}{4}] + 294 - \frac{11}{7} \times [\frac{{A B}^{2} + {A C}^{2}}{4}]$

$= \frac{11}{7} \times \frac{1225}{4} + 294 - \frac{11}{7} \times \frac{1}{4} [21^{2} + 28^{2}] = 775.25 - \frac{11}{28} \times 1225$

$= 775.25 - 481.25 = 294 {c m}^{2}$

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