Is this a thing?

Let $a,b$ be the legs of a right triangle and $c$ is the hypotenuse. So, we have, $\begin{aligned} \frac{ab}{2}&=11\implies ab=22 &&\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(1)\\ a+b+c&=16\implies a+b=16-c && \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(2) \\ a^2+b^2&=c^2 && \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;(3) \end{aligned}$ Squaring equation $(2)$ , using equations $(1)$ and $(3)$ , $\begin{aligned} (a+b)^2&=(16-c)^2 \\ a^2+b^2+2ab&=256-32c+c^2 \\ c^2+44&=256-32c+c^2 \\ 32c&=212 \\ \therefore \; c&=\frac{53}{8} \end{aligned}$ So equation $(2)$ transforms to, $\begin{aligned} a+b&=\frac{75}{8} \\ a+\frac{22}{a}&=\frac{75}{8} \\ 8a^2-75a+176&=0 \\ \end{aligned}$ In the above quadratic, the discriminant , $\Delta=(-75)^2-4\cdot 8\cdot 176=-7<0$ . Hence, the quadratic does not have real solutions and both $a$ , $b$ are complex numbers.

Therefore, there does not exist a right triangle such that its area is $11$ and perimeter is $16$ .

Assume that it is possible, and let $a$ and $b$ be the legs of the right triangle.

By the Pythagorean Theorem, the hypotenuse is $\sqrt{a^2 + b^2}$ , so the perimeter would be $P = a + b + \sqrt{a^2 + b^2} = 16$ .

Also, the area would be $A = \frac{1}{2}ab = 11$ , which rearranges to $b = \frac{22}{a}$ .

Substituting $b = \frac{22}{a}$ into $a + b + \sqrt{a^2 + b^2} = 16$ gives $a + \frac{22}{a} + \sqrt{a^2 + (\frac{22}{a})^2} = 16$ , which rearranges to $8a^2 - 75a + 176 = 0$ (for $a \neq 0$ ).

However, the discriminant of $8a^2 - 75a + 176 = 0$ is $75^2 - 4 \cdot 8 \cdot 176 = -7$ , so there is no real solution for $a$ .

The original assumption leads to a contradiction, so we can reject it and conclude that such a triangle is not possible .