(a) \(\frac{2Ab}{\sqrt{4A^2+b^4}}\)
Let ABC be the given right angled triangle, right angled at B. Let BC = b.
Then, Area of ΔABC = \(\frac12\) x BC x AB
⇒ A = \(\frac12\) x b x AB ⇒ AB = \(\frac{2A}{b}\)
In ΔABC, AC2 = AB2 + BC2
AC2 = \(\frac{4A^2}{b^2}+b^2\)
⇒ AC = \(\sqrt{\frac{4A^2}{b^2}+b^2}\)
Again in ΔABC,
A = \(\frac12\) x AC x BD
⇒ A = \(\frac12\) x \(\sqrt{\frac{4A^2}{b^2}+b^2}\) x BD
⇒ BD = \(\frac{2Ab}{\sqrt{4A^2+b^4}}.\)