Answer
$$h=\frac{2}{5}$$
Work Step by Step
First off, let $radius\:OP$ have a length that is equivalent to $z$, which is the altitude of the triangle that's added to $\frac{h}{2}$. If that is the case, then this would be a general equality:
$$z=1-\frac{h}{2}$$
Based on the previous equality, the following would be the general area of a triangle:
$$A=\frac{1}{2}bz=\frac{1}{2}b(1-\frac{h}{2})$$
The following would be the general area of a rectangle: $$A=bh$$
Thanks to both of the being equal, since they are both areas, the following would be the solved height:
$$\begin{matrix}
\frac{1}{2}bz=bh\\
\frac{1}{2}b(1-\frac{h}{2})=bh\\
\frac{1}{2}b-\frac{bh}{4}=bh\\
\frac{1}{2}b\cdot 4-\frac{bh}{4}\cdot 4=bh\cdot 4\\
2b-bh=4bh\\
2b-bh-2b=4bh-2b\\
-bh=4bh-2b\\
-bh-4bh=4bh-2b-4bh\\
-5bh=-2b\\
\frac{-5bh}{-5b}=\frac{-2b}{-5b}\\
\mathbf{h=\frac{2}{5}}
\end{matrix}$$
