10th Class Mathematics Solved Paper - Mathematics 2017 Outside Delhi Set-III

  • question_answer 8)
    In the given figure, O is the centre of the circle with \[AC=24\text{ }cm,\,\,AB=7\text{ }cm\] and \[\angle BOD=90{}^\circ \]. Find the area of the shaded region.

    Answer:

    Given, C (O, OB) with \[AC=24\text{ }cm\text{ }AB=7\text{ }cm\] and \[\angle BOD=90{}^\circ \]
    \[\angle CAB=90{}^\circ \] (Angle in semi-circle)
    Using pythagoras theorem in \[\Delta \text{ }CAB\]
    \[B{{C}^{2}}=A{{C}^{2}}+A{{B}^{2}}\]
                \[={{(24)}^{2}}+{{(7)}^{2}}\]
                \[=576+49\]
                \[=625\]
        \[BC=25\,\,cm\]
    Radius of circle \[=OB=OD=OC=\frac{25}{2}cm\]
    Area of shaded region
    = Area of semi-circle with diameter BC\[\]Area of \[\Delta \text{ }CAB\]+Area of sector \[BOD\]
                            \[=\frac{1}{2}\pi {{\left( \frac{25}{2} \right)}^{2}}-\frac{1}{2}\times 24\times 7+\frac{90}{360}\pi {{\left( \frac{25}{2} \right)}^{2}}\]
                            \[=\frac{3}{4}\times \frac{22}{7}\times \frac{25}{2}\times \frac{25}{2}-84\]
                            \[=\frac{20625}{56}-84\]
                            \[=\frac{20625-4704}{56}\]
                            \[=\frac{15921}{56}=284.3\,\,c{{m}^{2}}\]     (approx..)


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