Like suggested in the comments, you should reflect in the x-axis. Then you will obtain a closed loop whose length is $\tilde{L}=2L$ and enclosed area is $\tilde{A}=2A$. Now apply the equality case of the isoperimetric theorem (< which states that $$ \tilde{L}^2/\tilde{A} = 4 \pi $$ if and only if the curve is a circle. Plugging in $L$ and $A$, you do get equality (by your hypothesis). Therefore you have a circle and the curve you start with is a semi-circle.