Problem 649

We begin by extending $PT$ to
point $U$ on $QR$, as shown.

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Since $PT$ is parallel to $SR$, then $TU$ is parallel to $SR$.

Similarly, since $TS$ is parallel to
$QR$, then $TS$ is parallel to $UR$.

Since $TU$ is parallel to $SR$ and $TS$ is parallel to $UR$, then $TURS$ is a parallelogram, and so $TU=SR$ and $UR=TS=6$ cm.

In $\triangle PQU$, the sum of the
measures of the three angles is $180\degree$, and so $\$\angle PUQ=180\degree-60\degree-60\degree=60\degree\$.$

Thus, $\triangle PQU$ is an
equilateral triangle, and so $QU=PU=PQ=10$ cm.

The perimeter of $PQRST$ is $$\begin{align*} PQ+QR+SR+TS+PT&=PQ+QU+UR+SR+TS+PT \ \ \text{(since }QR=QU+UR)\\ &=PQ+QU+UR+TU+TS+PT \ \ \text{(since }SR=TU)\\ &=PQ+QU+UR+PT+TU+TS \ \ \text{(reordering some sides)}\\ &=PQ+QU+UR+PU+TS \ \ \text{(since }PT+TU=PU)\\ &=10\text{ cm}+10\text{ cm}+6\text{ cm}+10\text{ cm}+6\text{ cm}\\ &=42\text{ cm}\end{align*}$$