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\[\pi =3.14\]) |
Answer:
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 circle\[\]Area 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 \[36.33\,\,c{{m}^{2}}\].
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