Show that the flow area, $A$, and wetted perimeter, $P$, of a channel with a triangular bottom section are given by
$$
\begin{aligned}
A= & \frac{T_b}{2}\left(\frac{T_b}{2 m_1}\right) \\
& +\left[T_b+\left(y-\frac{T_b}{2 m_1}\right) m_2\right]\left(y-\frac{T_b}{2 m_1}\right) \\
P= & 2 \sqrt{1+m_1^2}\left(\frac{T_b}{2 m_1}\right) \\
& +2 \sqrt{1+m_2^2}\left(y-\frac{T_b}{2 m_1}\right)
\end{aligned}
$$
where $T_b$ is the top width of the bottom section, $m_1$ is the side slope of the bottom section, $y$ is the depth of flow, and $m_2$ is the side slope of the channel.