A) \[({{a}^{2}}+{{b}^{2}}){{p}^{2}}={{a}^{2}}{{b}^{2}}\]
B) \[{{a}^{2}}+{{b}^{2}}={{a}^{2}}{{b}^{2}}{{p}^{2}}\]
C) \[{{p}^{2}}={{a}^{2}}+{{b}^{2}}\]
D) \[{{p}^{2}}={{a}^{2}}-{{b}^{2}}\]
Correct Answer: A
Solution :
In right \[\Delta ABC,\] Area \[=\frac{1}{2}\times a\times b\] Again, in right \[\Delta ABC,\] Area \[=\frac{1}{2}\times AB\times DC\] \[\Rightarrow \] \[\frac{1}{2}ab=\frac{1}{2}\times c\times p\] \[\Rightarrow \] \[ab=p(\sqrt{{{a}^{2}}+{{b}^{2}})}\]\[(\because {{c}^{2}}={{a}^{2}}+{{b}^{2}})\] \[\Rightarrow \] \[{{a}^{2}}{{b}^{2}}={{p}^{2}}({{a}^{2}}+{{b}^{2}})\]You need to login to perform this action.
You will be redirected in
3 sec