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SSC Sample Paper Mock Test-13 SSC CGL Tear-II Paper-1

  • question_answer
    In the given figure, PQ is the diameter of a circle with centre at 0. 05 is perpendicular to PR. Then, OS is equal to

    A)  \[\frac{1}{4}QR\]

    B)  \[\frac{1}{3}QR\]

    C)  \[\frac{1}{2}QR\]

    D)  \[QR\]

    Correct Answer: C

    Solution :

    As, O is mid-point of PQ and \[\angle PRQ=90{}^\circ \] (angle is semicircle) So, OS || QR as both \[\bot PR:\frac{PO}{PQ}=\frac{OS}{QR}\] \[\frac{OS}{QR}=\frac{1}{2}\]\[\Rightarrow \]\[OS=\frac{1}{2}QR\]


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