The trapezoid of largest area with three equal sides

Area. The average of the bases is $\frac{1+(1+2\cos\theta)}2=1+\cos\theta$, so
$$A(\theta)=\sin\theta\,(1+\cos\theta).$$

Derivative.
$$A'(\theta)=\cos\theta(1+\cos\theta)-\sin^2\theta=2\cos^2\theta+\cos\theta-1=(2\cos\theta-1)(\cos\theta+1).$$
On $\left(0,\frac\pi2\right)$, $\cos\theta+1>0$ and $2\cos\theta-1>0\iff\theta<\frac\pi3$. So $A$ increases until $\theta=\frac\pi3$, then decreases.

Answer. $\theta=\dfrac\pi3$, and
$$A_{\max}=\frac{\sqrt3}2\cdot\frac32=\frac{3\sqrt3}4\approx1.30.$$

Remark. At $\theta=\frac\pi3$ the trapezoid is half of a regular hexagon of side $1$: the best shape is the natural one. A regular hexagon has area $\frac{3\sqrt3}2$, twice the answer.