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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