• question_answer 11) In fig. 4, O is the centre of a circle such that diameter $AB=13\text{ }cm$ and $AC=12\,cm.\,\,BC$ is joined Find the area of the shaded region. (Take$\pi =3.14$)

 Given, AB is a diameter of length $13\text{ }cm$. And $AC=12\text{ }cm$. Then, by pythagoras theorem, ${{(BC)}^{2}}={{(AB)}^{2}}-{{(AC)}^{2}}$ ${{(BC)}^{2}}={{(13)}^{2}}-{{(12)}^{2}}$ $BC=\sqrt{169-144}$ $BC=\sqrt{25}$ $\therefore$                  $BC=5\,cm$ Now, Area of shaded region = Area of semi circleArea of $\Delta \text{ }ABC$ $=\frac{\pi {{r}^{2}}}{2}-\frac{1}{2}\times BC\times AC$ $=\frac{1}{2}\times 3.14\times {{\left( \frac{13}{2} \right)}^{2}}-\frac{1}{2}\times 5\times 12$ $=\frac{1.57\times 169}{4}-30$ $=66.33-30$ $=36.33\,c{{m}^{2}}$ So, area of shaded region is $36.33\,\,c{{m}^{2}}$.