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9th Class Mathematics Quadrilaterals Question Bank Quadrilateral

  • question_answer
    In the adjoining figure, AP and BP are angle bisectors of \[\angle \mathbf{A}\] and \[\angle B\] which meets At P in the parallelogram ABCD. Then\[\mathbf{2}\angle \mathbf{APB=?}\]                                

    A)  \[\angle C+\angle D\]    

    B)  \[\angle A+\angle C\]

    C)  \[\angle B+\angle D\]                

    D)  \[2\angle C\]

    Correct Answer: A

    Solution :

    (a)\[\angle A+\angle B+\angle C+\angle D={{360}^{{}^\circ }}\] \[\frac{\angle A}{2}\text{+}\frac{\angle B}{2}\text{+}\frac{\angle C}{2}\text{+}\frac{\angle D}{2}={{180}^{{}^\circ }}~~\] \[{{180}^{{}^\circ }}-\angle APB+\frac{\angle C}{2}+\frac{\angle D}{2}={{180}^{{}^\circ }}\] \[\Rightarrow \]\[\angle APB=\frac{\angle C}{2}+\frac{\angle D}{2}\]    \[2\angle APB=\angle C+\angle D.\]        


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