10th Class Mathematics Solved Paper - Mathematics-2018

  • question_answer
    In an equilateral \[\Delta \text{ }ABC,\text{ }D\]is a point on side BC such that \[BD=\frac{1}{3}BC\]. Prove that \[9{{(AD)}^{2}}\] \[=7{{(AB)}^{2}}\].
    Prove that, in a right triangle, the square on the hypotenuse is equal to the sum of the squares on the other two sides.


    Given, ABC is an equilateral triangle and D is a point on BC such that \[BD=\frac{1}{3}BC\].
    To prove:
    Construction: Draw \[AE\bot BC\]
    Proof:      \[BD=\frac{1}{3}BC\]                                                            ?(i) (Given)
                   \[AE\bot BC\]
    We know that perpendicular from a vertex of equilateral triangle to the base divides base in two equal parts.
    \[\therefore \]      \[BE=EC=\frac{1}{2}BC\]                                             ?(ii)
    In \[\Delta \,AEB\],
                                        (Pythagoras theorem)
    Or         \[A{{E}^{2}}=A{{D}^{2}}-D{{E}^{2}}\]                                        ?(iii)
    Similarly, In \[\Delta \,AEB\],
    \[=A{{D}^{2}}-D{{E}^{2}}+{{\left( \frac{1}{2}BC \right)}^{2}}\] [From (ii) and (iii)]
    \[=A{{D}^{2}}-{{\left( \frac{1}{2}BC \right)}^{2}}-{{\left( \frac{1}{3}BC \right)}^{2}}\]\[+2.\frac{1}{2}BC.\frac{1}{3}BC+\frac{1}{4}B{{C}^{2}}\]
    \[\Rightarrow \]   \[A{{B}^{2}}=A{{D}^{2}}+\frac{2}{9}B{{C}^{2}}\]
    \[\Rightarrow \]   \[A{{B}^{2}}=A{{D}^{2}}+\frac{2}{9}A{{B}^{2}}\]          (\[\because \] \[BC=AB\])
    \[\Rightarrow \]   \[A{{B}^{2}}-\frac{2}{9}A{{B}^{2}}=A{{D}^{2}}\]
    \[\Rightarrow \]   \[\frac{7}{9}A{{B}^{2}}=A{{D}^{2}}\]
    \[\Rightarrow \]   \[7A{{B}^{2}}=9A{{D}^{2}}\]
    Or         \[9{{(AD)}^{2}}=7{{(AB)}^{2}}\]                              Hence Proved.
    Given: \[\Delta \text{ }ABC\]is a right angle triangle, right angled at A.
    To prove:         \[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\]
    Construction: Draw \[AD\bot BC\].
    Proof: In \[\Delta \text{ }ADB\]and \[\Delta \,BAC\],
    \[\angle B=\angle B\]                        (Common)
    \[\angle ADB=\angle BAC\]                          (Each \[90{}^\circ \])
    \[\therefore \]      \[\Delta \text{ }ADB\tilde{\ }\Delta \text{ }BAC\]
               (By AA similarity axiom)
    \[\therefore \]      \[\frac{AB}{BC}=\frac{BD}{AB}\]                                                         (CPCT)
                \[A{{B}^{2}}=BC\times BD\]                                                   ?(i)
    \[\Delta \text{ }ADC\tilde{\ }\Delta \text{ }CAB\]
                            \[A{{C}^{2}}=BC\times DC\]                           ?(ii)
    On adding eq. (i) and (ii)
    \[A{{B}^{2}}+A{{C}^{2}}=BC\times BD+BC\times CD\]
         \[=BC\text{ (}BD+CD)\]
         \[=BC\times BC\]
    \[\Rightarrow \]   \[B{{C}^{2}}=A{{B}^{2}}+A{{C}^{2}}\]                                       Hence Proved.


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