Curve enclosing the maximum area the curve of fixed length $l$ that joins the points $(0,0)$ and $(1,0)$ lies above the $x-axis$ and encloses the maximum area between itself and the $x-axis$, is a segment of 


A straight line
A parabola
An ellipse
A circle


I don't know exactly how to solve it but it seems Circle is the right Answer
Here l is the fixed perimeter of plane curve passing through two given points (0,0) and (1,0). Let S be the area enclosed by the plane curve and x-axis 

Then, we are maximize $S=\int _0^1 y\;dx$ with boundary conditions y(0)=y(1)=0 subject to constraint $\int_0^1(1+y'^2)^\frac{1}{2}\;dx=l$

Let $F(x,y,y')=y+ \lambda(1+y'^2)^\frac{1}{2} $

Where $\lambda$ is the lagrange multiplier . Then the required extremal satiesfied the Euler's equation 

$\frac{\partial F}{\partial y}-\frac{d}{dx}(\frac{\partial F}{\partial y'})=0\\
\Rightarrow 1-\frac{d}{dx}\{\frac{\lambda y'}{(1+y'^2)^\frac{1}{2}}\}=0\\
Integrating \;w\;r\;to\;x, x-\{\frac{\lambda y'}{(1+y'^2)^\frac{1}{2}}\}=a\\
\Rightarrow (x-a)^2=\{\frac{\lambda^2 y'^2}{(1+y'^2)}\}\\
\Rightarrow \frac{1+y'^2}{y'^2}=\frac{\lambda^2}{(x-a)^2}\\
\Rightarrow y'^2=\frac{(x-a)^2}{\lambda^2-(x-a)^2}\\
\frac{dy}{dx}=\pm \frac{(x-a)}{(\lambda^2-(x-a)^2)^\frac{1}{2}}\\
Integrating, \; y=b\pm\{\lambda^2-(x-a)^2\}^\frac{1}{2}\\
\Rightarrow (y-b)^2=\lambda^2-(x-a)^2\\
\Rightarrow (x-a)^2+(y-b)^2=\lambda^2$
this is equation of a circle with radius $\lambda$
By reflecting the curve across the $x$-axis we can make it in to a closed loop, which doubles both the length and the area. Thus this is equivalent to the usual isoperimetric problem (maximize area inside a loop of fixed length) with the constraint of reflection symmetry. Since the circle is the unconstrained optimum and also has the reflection symmetry, it provides the solution to your problem: the curve is a semicircle.