How to prove $ \sqrt { \frac {s(c-a-b)}{2∆} } $ is equal to i?

Question

Consider a right angles triangle. Let, c be it's hypotenuse and a,b it's other sides.

Then prove,

$ \sqrt { \frac {s(c-a-b)}{2∆} } $ is complex. Where ∆= area of the triangle and s is semi perimeter.

I am not able to tackle this questions second part which is the above fraction can be proven equal to i.

Proving it is complex is easy by the triangle equality. But how to prove that it is equal to i. Also, is there any other way to prove that the above fraction is complex.

Edit- Including the other method, there are basically 2 ways for proving it is complex. Is there any other way?

Answer 1

We have that

$$\frac {s(c-a-b)}{2∆}=\frac {\frac12(c+(a+b))(c-(a+b))}{2\frac12ab}=\frac12\frac{c^2-(a+b)^2}{ab}=\frac12\frac{-2ab}{ab}=-1 $$

Answer 2

Semi-perimeter $s=\frac{a+b+c}2, \Delta=\frac12ab\implies\sqrt{\frac{s(c-a-b)}{2∆}}=\sqrt{\frac{c^2-(a+b)^2}{2ab}}=\sqrt{-\frac{2ab}{2ab}}\because a^2+b^2=c^2$

Answer 3

The first thing to do is to replace $s$ with $(a+b+c)/2$ and $\Delta$ with $ab/2$. This means the expression becomes

$$\sqrt{\frac{(a+b+c)(c-a-b)}{2ab}}$$

now, simplify the numerator (either by multiplying everything out or first writing it as $(c+(a+b))(c-(a+b))$ and then using the fact that $(A+B)(A-B)=A^2-B^2$. Also, use the fact that $c^2=a^2+b^2$.