10th Class Mathematics Solved Paper - Mathematics-2016 Outside Delhi Set-I

  • question_answer
    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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