8th Class Mathematics Practical Geometry Question Bank Geometry

  • question_answer
    In the given figure, PQ is a diameter of a circle with centre O and PQRS is a cyclic quadrilateral. SQ is joined. If\[{{240}^{o}}\], find\[{{60}^{o}}\].

    A)  \[{{120}^{o}}\]                              

    B)  \[{{180}^{o}}\]

    C)  \[\angle R={{138}^{o}}\]                

    D)  \[\angle PQS\]

    Correct Answer: C

    Solution :

    \[\angle 1+\angle 2+\angle 3+\angle 4>\angle 5+\angle 6+\angle 7+\angle 8\] PQRS is a cyclic quadrilateral. \[AB=2x+5,\]  \[CD=y+1,\] \[AD=y+5\]  \[BC=3x-4\] \[\angle BOC={{90}^{o}}\] \[\angle AOB\] Angle in a semicircle is a right angle. So      \[{{10}^{o}}\] In    \[{{40}^{o}}\] \[{{50}^{o}}\] \[{{25}^{o}}\] \[{{25}^{o}}\] \[{{40}^{o}}\]   \[{{50}^{o}}\]                


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