Problem 548

Let $r$ be the radius of each of the six circles. Then $TY=TU=UV=YX=XW=VW=2r$ and $PQ=SR=6r$ and $PS=QR=4r$: Since each circle has radius $r$, then each circle has diameter $2r$, and so can be enclosed in a square with side length $2r$ whose sides are parallel to the sides of rectangle $PQRS$. Each circle touches each of the four sides of its enclosing square. Because each of the circles touches one or two sides of rectangle $PQRS$ and each of the circles touches one or two of the other circles, then these six squares will fit together without overlapping to completely cover rectangle $PQRS$. [[IMAGE0]] Therefore, $PQ=SR=6r$ and $PS=QR=4r$ since rectangle $PQRS$ is three squares wide and two squares tall. Finally, since the centre of each circle is the centre of its square, then the distance between the centres of each pair of horizontally or vertically neighbouring squares is $2r$ (which is two times half the side length of one of the squares). Therefore, the perimeter of rectangle $TVWY$ is $$TV+TY+YW+VW=2r+4r+4r+2r=12r$$ Since the perimeter of rectangle $TVWY$ is 60, then $12r=60$ or $r=5$. Since $r=5$, then in the larger rectangle, we have $PQ=SR=30$ and $PS=QR=20$. Therefore, the area of rectangle $PQRS$ is $PQ\cdot PS = 30\cdot 20 = 600$.