Hero

Given the dimensions of this triangle, we can consider the unit circle with a radius of $1$ and points the circumference with coordinate values of $(\cos{\theta}, \sin{\theta})$ . Here, we are clearly dealing with a right triangle with a hypotenuse of $1$ .

When $\tan{\theta} = \dfrac{3}{4}$ for $0<\theta<\dfrac{\pi}{2}$ , $\sin{\theta} = \dfrac{3}{\sqrt{3^2+4^2}} = \dfrac{3}{5}$ and $\cos{\theta} = \dfrac{4}{\sqrt{3^2+4^2}} = \dfrac{4}{5}$ . Thus, the area of such a triangle is $\dfrac{1}{2} \cdot \dfrac{3}{5} \cdot \dfrac{4}{5} = \dfrac{6}{25}$

$\therefore a+b = 6+25 = \boxed{31.}$