A) \[\frac{2\,\,Ab}{\sqrt{{{b}^{4}}+4{{A}^{2}}}}\]
B) \[\frac{2\,\,{{A}^{2}}b}{\sqrt{{{b}^{2}}+4{{A}^{2}}}}\]
C) \[\frac{2A{{b}^{2}}}{\sqrt{{{b}^{2}}+4{{A}^{2}}}}\]
D) \[\frac{2{{A}^{2}}{{b}^{2}}}{\sqrt{{{b}^{4}}+{{A}^{2}}}}\]
Correct Answer: A
Solution :
In \[\Delta ABC,\] |
\[A=\frac{1}{2}\times \]base \[\times \]altitude \[=\frac{1}{2}\times b\times AC\] |
\[AC=\frac{2A}{b}\] |
Using Pythagorus theorem, |
\[A{{C}^{2}}+A{{B}^{2}}=B{{C}^{2}}\] |
\[\Rightarrow \] \[BC=\sqrt{\frac{4{{A}^{2}}}{{{b}^{2}}}+{{b}^{2}}}\] |
Again in \[\Delta ABC,\]\[A=\frac{1}{2}\times BC\times AD\] |
\[\Rightarrow \] \[AD=\frac{2A}{\sqrt{\frac{4{{A}^{2}}+{{b}^{2}}}{{{b}^{2}}}}}=\frac{2Ab}{\sqrt{4{{A}^{2}}+{{b}^{4}}}}\] |
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