, 2014

Solution 1

Consider the following diagram.

[[IMAGE0]]

The outer square has an area of 9 cm$^2$, so the sides of this outer square have length 3 cm (since $3\times3=9$), and thus $PN=3$ cm.

The inner square has an area of 1 cm$^2$, so the sides of this inner square have length 1 cm (since $1\times1=1$), and thus $MR=1$ cm.

Since $PN=3$ cm, then $PS+SN=3$ cm and so $QR+SN=3$ cm (since $QR=PS$).

But $QR=QM+MR$, so then $QM+MR+SN=3$ cm or $QM+1+SN=3$ cm (since $MR=1$ cm).

From this last equation we get $QM+SN=2$ cm. Since each of $QM$ and $SN$ is the width of an identical rectangle, then $QM=SN=1$ cm.

Using $PS+SN=3$ cm, we get $PS+1=3$ cm and so $PS=2$ cm.

Since the rectangles are identical, then $SN=PQ=1$ cm.
The perimeter of rectangle $PQRS$ is $2\times(PS+PQ)=2\times(2+1)=2\times3=6$ cm.

Solution 2

Consider the following diagram.

[[IMAGE1]]

The outer square has an area of 9 cm$^2$, so the sides of this outer square have length 3 cm (since $3\times3=9$), and thus $PN=3$ cm.

Since $PN=3$ cm, then $PS+SN=3$ cm.

Since each of $PQ$ and $SN$ is the width of an identical rectangle, then $PQ=SN$ and so $PS+SN=PS+PQ=3$ cm.
The perimeter of $PQRS$ is $2\times(PS+PQ)=2\times3=6$ cm.