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9th Class Mathematics Triangles Question Bank Triangle

  • question_answer
                      In a \[\Delta \mathbf{ABC},\overline{\mathbf{AD}},\overline{\mathbf{BE}}\], and \[\overline{\mathbf{CF}}\] are three medians. Then the ratio \[\left( {{\overline{\mathbf{AD}}}^{\mathbf{2}}}\mathbf{+}{{\overline{\mathbf{BE}}}^{\mathbf{2}}}\mathbf{+}{{\overline{\mathbf{CF}}}^{\mathbf{2}}} \right)\mathbf{:}\]\[\left( {{\overline{\mathbf{AB}}}^{\mathbf{2}}}\mathbf{+}{{\overline{\mathbf{AC}}}^{\mathbf{2}}}\mathbf{+}{{\overline{\mathbf{BC}}}^{\mathbf{2}}} \right)\] is

    A)  Equal to \[\frac{3}{4}\]    

    B)  Equal to 1

    C)  Greater than \[\frac{3}{4}\]        

    D)  Equal to \[\frac{1}{2}\]

    Correct Answer: A

    Solution :

    (a): In any triangle. The sum of the squares of any two sides is equal to twice the square of half of the third side together with twice the square of the median bisecting it. \[\therefore \]\[A{{B}^{2}}+A{{C}^{2}}=2\left( A{{D}^{2}}+B{{D}^{2}} \right)\] \[\Rightarrow \]\[A{{B}^{2}}+A{{C}^{2}}=2\left( A{{D}^{2}}+\frac{B{{C}^{2}}}{4} \right)\] \[\Rightarrow \]\[2\left( A{{B}^{2}}+A{{C}^{2}} \right)=4A{{D}^{2}}+B{{C}^{2}}\] Similarly, \[\Rightarrow \]\[2\left( A{{B}^{2}}+B{{C}^{2}} \right)=4B{{E}^{2}}+A{{C}^{2}}\] \[\Rightarrow \]\[2\left( A{{C}^{2}}+B{{C}^{2}} \right)=4C{{F}^{2}}+A{{B}^{2}}\] On adding all three \[4\left( A{{B}^{2}}+B{{C}^{2}}+A{{C}^{2}} \right)\] \[=4(A{{D}^{2}}+B{{E}^{2}}+C{{F}^{2}})+B{{C}^{2}}+A{{C}^{2}}+A{{B}^{2}}\] \[\Rightarrow \]\[3\left( A{{B}^{2}}+B{{C}^{2}}+A{{C}^{2}} \right)\] =\[4(A{{D}^{2}}+B{{E}^{2}}+C{{F}^{2}})\] \[\Rightarrow \]\[\frac{A{{D}^{2}}+B{{E}^{2}}+C{{F}^{2}}}{A{{B}^{2}}+B{{C}^{2}}+C{{A}^{2}}}=\frac{3}{4}\]           


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