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

  • question_answer
    In Fig. 9, is shown a sector OAP of a circle with centre O, containing \[\angle \theta .\text{ }AB\] is perpendicular to the radius OA and meets OP produced at B. Prove that the perimeter of shaded region is \[r\left[ \tan \,\theta +\sec \,\theta +\frac{\pi \theta }{180}-1 \right]\]


    Given, the radius of circle with centre O is r.
    \[\angle POA=\theta \].
    then, length of the arc \[\overset\frown{PA}=\frac{2\pi r\theta }{360{}^\circ }=\frac{\pi r\theta }{180{}^\circ }\]
    And                              \[\tan \theta =\frac{AB}{r}\]
                                        \[AB=r\,\tan \theta \]
    And                              \[\sec \theta =\frac{OB}{r}\]
                                        \[OB=r\,\sec \theta \]
    Now,                             \[PB=OB-OP\]
                                        \[=r\,\sec \,\theta -r\]
    \[\therefore \] Perimeter of shaded region
                          \[=r\,\tan \,\theta +r\,\sec \,\theta -r+\frac{\pi r\theta }{180{}^\circ }\]
                          \[=r\left[ \tan \theta +\sec \theta +\frac{\pi \theta }{180}-1 \right]\]                           Hence Proved.


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