10th Class Mathematics Solved Paper - Mathematics-2015 Term-I

  • question_answer In triangle \[ABC\], if \[AP\bot BC\] and \[A{{C}^{2}}=B{{C}^{2}}-A{{B}^{2}}\], then prove that \[P{{A}^{2}}=PB\times CP\].

    Answer:

    \[A{{C}^{2}}=B{{C}^{2}}-A{{B}^{2}}\]    [Given]
    \[A{{C}^{2}}+A{{B}^{2}}=B{{C}^{2}}\]  
                \[\therefore \angle BAC=90{}^\circ \]
    [By converse of Pythagoras? theorem]
    \[\Delta \,APB\sim \Delta \,CPA\]
                   
    \[\Rightarrow \frac{AP}{CP}=\frac{PB}{PA}\]
                    [In similar triangle, corresponding sides are proportional]
    \[\Rightarrow P{{A}^{2}}=PB.CP\]                   Hence Proved.


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