smc

The area of $T$ is $\frac{1}{2}\cdot 8\cdot 3=12$ and the perimeter is 18. The area of $T^{\prime }$ is $\frac{1}{2}b\sqrt{a^2-(\frac{b}{2}{)}^2}$ and the perimeter is $2a+b$. Thus $2a+b=18$, so $2a=18-b$. Thus $12=\frac{1}{2}b\sqrt{a^2-(\frac{b}{2}{)}^2}$, so $48=b\sqrt{4a^2-b^2}=b\sqrt{(18-b{)}^2-b^2}=b\sqrt{324-36b}$. We square and divide 36 from both sides to obtain $64=b^2(9-b)$, so $b^3-9b^2+64=0$. Since we know $b=8$ is a solution, we divide by $b-8$ to get the other solution. Thus, $b^2-b-8=0$, so $b=\frac{1+\sqrt{33}}{2}<\frac{1+6}{2}=\mathrm{3.5.}$ The answer is $\boxed{3}$.