10th Class Mathematics Solved Paper - Mathematics-2016 Delhi Term-II Set-II

  • question_answer The perimeter of a right triangle is 60 cm. Its hypotenuse is 25 cm. Find the area of the triangle.

    Answer:

    Given, the perimeter of right triangle \[=60\text{ }cm\]
    and hypotenuse \[=25\text{ }cm\]
    \[\therefore AB+BC+CA=60\text{ }cm\]
                            \[AB+BC+25=60\]
    \[\therefore AB+BC=35\]                                                            ?(i)
    Now, by pythagoras theorem,
                      \[{{(AC)}^{2}}={{(AB)}^{2}}+{{(BC)}^{2}}\]
                       \[{{(25)}^{2}}={{(AB)}^{2}}+{{(BC)}^{2}}\]
    \[\therefore A{{B}^{2}}+B{{C}^{2}}=625\]                                                   ?(ii)
    we, know that, \[{{(a+b)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab\]
    then,               \[{{(AB+BC)}^{2}}\text{=(}AB{{)}^{2}}+{{(BC)}^{2}}+2AB.BC\]
                                    \[{{(35)}^{2}}=625+2AB.BC\]
    \[\therefore 2AB.BC=1225-625\]
                              \[2AB.BC=600\]
    \[\therefore AB.BC=300\]
    \[\therefore \]      Area of \[\Delta \text{ }ABC=\frac{1}{2}~\times AB\times BC\]
                            \[=\frac{1}{2}\times 300=150\,\,c{{m}^{2}}\]


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