2020/007) We know for a right angled triangle $a^2+b^2=c^2$ where a,b are shorter sides and c is hypotenuse. What is $\frac{1}{a^2}+ \frac{1}{b^2}$ for a right angled triangle

We have $a^2+b^2=c^2$

Hence deviding by $a^2b^2$ we get

$\frac{1}{a^2}+ \frac{1}{b^2} = \frac{c^2}{a^2b^2}\cdots(1)$

Now area of the triangle $A = \frac{1}{2}ab\cdots(2)$

If h is the altitude drawn from the right angle to the hypotenuse then area of
the triangle $A = \frac{1}{2}ch\cdots(3)$

So from (2) and (3) ab = ch and putting in (1)

$\frac{1}{a^2}+ \frac{1}{b^2} = \frac{c^2}{a^2b^2} = \frac{c^2}{c^2h^2}= \frac{1}{h^2}$

Or  $\frac{1}{a^2}+ \frac{1}{b^2} = \frac{1}{h^2}$

Where h is the altitude drawn from the right angle to the hypotenuse