How to express a circumference of half circle with this angle θ?

Question

So this is my homework:

$PQ=2$ and $PQ$ is also the radius of the half-circle, $QR$ is parallel to $PS$, $P$ is the origin,

How do we express the circumference in terms of $\theta$? I still can't imagine how the size $\theta$ affects the circumference. The question asks to prove that $K = 2 + 6 \cos(\theta) + 2 \sin (\theta)$ where $K$ is the circumference of this half circle.

Thank you very much!

Answer 1

The circumference, or the perimeter of $PQRS$ in terms of the given radius $|PQ|=r=2$ and the value of angle $\theta$ can be found as follows:

\begin{align} K&=|PQ|+|QR|+|RS|+|PS| \tag{1}\label{1} ,\\ |QR|&=2\,|QT| \tag{2}\label{2} ,\\ |QT|&=|PS|=r\cos\theta \tag{3}\label{3} ,\\ |RS|&=r\,\sin\theta \tag{4}\label{4} , \end{align}

so, indeed, we have \begin{align} K&=r+2r\cos\theta+r\,\sin\theta+r\cos\theta =r+3r\cos\theta+r\,\sin\theta \\ &=2+6\cos\theta+2\,\sin\theta \tag{5}\label{5} . \end{align}