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

  • question_answer
    In the given figure, \[AD=3cm,\,\,AE=5cm,\,\,BD=4cm,\,\,CE=4cm,\,\,CF=2cm,\,\,BF=2.5cm\], then find the pair of parallel lines and hence their lengths.

    Answer:

    \[\frac{EC}{EA}=\frac{CF}{FB}\]  and \[\frac{CF}{FB}=\frac{2}{2.5}=\frac{4}{5}\]
    \[\Rightarrow \frac{EC}{EA}=\frac{CF}{FB}\]
    In \[\Delta \text{ }ABC,\text{ }EF\parallel AB\]
    [Converse of Thales' theorem]
                Also,     \[\frac{CE}{CA}=\frac{4}{4+5}=\frac{4}{9}\]
                            \[\frac{CE}{CB}=\frac{2}{2+2.5}=\frac{2}{4.5}=\frac{4}{9}\]
                            \[\frac{EC}{EA}=\frac{CF}{CB}\]
                            \[\angle ECF=\angle ACB\]                                  [Common]
                        \[\Delta \text{ }CFE\tilde{\ }\Delta \text{ }CBA\]                   [SAS similarity]
                \[\Rightarrow \frac{EF}{AB}=\frac{CE}{CA}\]
                                        [In similar \[\Delta 's\], corresponding sides are proportional]
                \[\Rightarrow \frac{EF}{7}=\frac{4}{9}\]
                                       \[[\because ~\,AB=3+4=7\text{ }cm]\]
    \[\therefore EF=\frac{28}{9}cm\] and \[AB=7\,cm\]


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