Answer
$R^2\sqrt{3}$ where $R$ is the radio of the semicircle
Work Step by Step
Let the trapezoid be an isosceles trapezoid, as it is the only type of trapezoid that can be inscribed in a circle with its larger base as the diameter. This is due to the fact that an angle in one base is supplementary to the opposite angle in the other base. This is a key property to inscribe any quadrilateral in a circle. Let $a$ be the smaller base of the trapezoid and $b$ the bigger base, which is as long as the diameter of the semicircle. Let $h$ be the height of the trapezoid measured from $a$ to $b$. Let $R$ be the radius of the semicircle.
Having stated this, we can relate the formulas with the equation for the area of a trapezoid:
$$A = \frac{(a+b)h}{2}$$
We want to maximize $A$.
First, we must simplify the expression that uses a single variable. To do this, let $\theta$ be the angle formed between the rightmost vertex in $b$ and the rightmost vertex in $a$. Thus, we can express $a$, $b$, $h$ in terms of $R$ and $\theta$.
$$b = 2R$$
$$h = R \sin \theta$$
Now, $R$, $h$ and a segment of $b$ form a right triangle thanks to the perpendicular construction of the height $h$. As the height is taken from the vertex of $a$, we can assume that the segment corresponds to a segment of equal lenght from the vertex of $a$ to the center of $a$. Additionally, as the trapezoid is isosceles, we can deduce that the same applies at the other side of the trapezoid, in such a way that the two equal segments have a combined lenght of $a$. In mathematical terms,
$$\frac{1}{2}a = R \cos \theta \rightarrow a = 2R \cos \theta$$
It is important to note that $R$ is a constant, thus, there is no need to "eliminate" it from the equation.
Having done this, the area can be expressed as:
$$A = \frac {(2R \cos \theta + 2R)(R \sin \theta)}{2} = \frac {2R^2(\cos \theta + 1)\sin \theta}{2} = R^2(\cos \theta + 1)\sin \theta$$
$$\rightarrow A = A(\theta) = R^2(\cos \theta + 1)\sin \theta$$
We can now differentiate in order to find the $\theta$ at which $A$ is maximum.
$$\frac {dA}{d\theta} = \frac {d}{d\theta} R^2(\cos \theta + 1)\sin \theta$$
$$ = R^2 \frac {d}{d\theta}(\cos \theta + 1)\sin \theta$$
$$ = R^2 [\sin \theta\frac {d}{d\theta}(\cos \theta + 1) + (\cos \theta + 1)\frac {d}{d\theta}\sin \theta]$$
$$ = R^2 [\sin \theta(-\sin \theta) + (\cos \theta + 1)(\cos \theta)]$$
$$ = R^2 [-\sin^2 \theta + \cos^2 \theta + \cos \theta]$$
We can now take advantage of the trigonometric identity $\sin^2 \theta + \cos^2 \theta = 1$ to solve the brackets:
$$ \frac {dA}{d\theta} = R^2 [-(1-cos^2 \theta) + \cos^2 \theta + \cos \theta]$$
$$ = R^2 [-1 + 2\cos^2 \theta + \cos \theta]$$
Next, we have to find the critical point at which $\frac {dA}{d\theta} = 0$.
$$ \frac {dA}{d\theta} = 0 \rightarrow R^2 [2\cos^2 \theta + \cos \theta-1] = 0$$
$R$ cannot be $0$ due to the construction of the trapezoid. Thus, the brackets must equal $0$. Furthermore, this can be solved as a quadratic equation based on $\cos \theta$.
$$\cos \theta = \frac {-(1) \pm \sqrt{1^2 - 4(2)(-1)} } {2(2)} = \frac {-1 \pm 3}{4} = -1, \frac{1}{2}$$
$$\rightarrow \theta = \cos^-1 -1, \cos^-1 \frac{1}{2} = \pi, \frac{\pi}{3}$$.
Due to the construction of the angle, $\theta \neq \pi$. Thus, $\theta = \frac{\pi}{3}$. Finally, we know that this critical point represents a maximum, as the quadratic equation opens upwards (coefficient of the squared term is positive).
We can now find $a$, $b$, $h$ and, ultimately, $A_{max}$
$$a = 2R \cos \frac{\pi}{3}= 2R * \frac{1}{2} = R$$
$$b = 2R$$
$$h = R \sin \frac{\pi}{3} = R \frac{\sqrt{3}}{2}$$
$$\therefore A_{max} = A(\frac{\pi}{3}) = \frac {3R^2 * \sqrt{3}/2}{2} = \frac{3}{4}R^2\sqrt{3}$$
