A) \[1/{{p}^{2}}\]
B) \[2/{{p}^{2}}\]
C) \[{{p}^{2}}\]
D) \[2{{p}^{2}}\]
Correct Answer: A
Solution :
Area of \[\Delta ABC=\frac{1}{2}\times c\times p=\frac{1}{2}\times b\times a\] \[\Rightarrow \] \[cp=ba\,\,\Rightarrow \,\,\,\frac{1}{{{p}^{2}}}=\frac{{{c}^{2}}}{{{a}^{2}}{{b}^{2}}}\] But, \[{{c}^{2}}={{a}^{2}}+{{b}^{2}}\][By Pythagoras theorem] \[\therefore \] \[\frac{1}{{{p}^{2}}}=\frac{{{a}^{2}}+{{b}^{2}}}{{{a}^{2}}{{b}^{2}}}\,\,\Rightarrow \,\,\frac{1}{{{a}^{2}}}+\frac{1}{{{b}^{2}}}=\frac{1}{{{p}^{2}}}\]
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