Is this a Possible Triangle?

The answer is $\boxed{\text{Yes}}$ , with one example being $(a,b,c) = \left(\dfrac{3}{2}, \dfrac{20}{3}, \dfrac{41}{6}\right)$ , which is a scaled-down $(9,40,41)$ Pythagorean triple. The idea is to find a Pythagorean triple with an area of the form $5n^{2}$ for some integer $n$ and then scaling down the sides of this triple by a factor $n$ . In the case of this example, the area of a $(9,40,41)$ triangle is $180 = 5 \times 6^{2}$ , so we just scale down its sides by a factor of $6$ to obtain the desired result.

I'm not sure if there are any other primitive Pythagorean triples we can scale down like this.....

first thing first

$r \in \mathbb{Q} \Rightarrow r^n \in \mathbb{Q} \\$

And

$\begin{aligned} \frac{a \times b} {2}&= 5 \\ a \times b&= 10 \\ a&= \frac{10}{b} \end{aligned}$

So

$a\in \mathbb{Q} \iff b \in \mathbb{Q}$

we have

$\begin{aligned} \text {let's spouse } c \in \mathbb{Q} \\ a^2+b^2&=c^2 \\ \left ( \frac{10}{b} \right )^2 + b^2&=c^2 \\ \frac{10^2}{b^2}+b^2&=c^2 \\ \frac{10^2+b^4}{b^2}&=c^2 \\ c&=\frac{\sqrt{10^2+b^4}}{\sqrt{b^2}}\\ c&=\frac{\sqrt{10^2+b^4}}{b} \end{aligned} \iff \boxed{b\in\mathbb{Q}} \\$

so for any $c\in\mathbb{Q}$ we have:

$\boxed{\left ( a,b,c \right ) \in \mathbb{Q}}$