- Thread starter
- #1
Albert
Well-known member
- Jan 25, 2013
- 1,225
HINT:
Good idea! But what I would do is to rotate the square by 90 degrees in an anticlockwise direction. The new position (call it $P'$) of $P$ will be on the line $PB$. The lines $PB$, $P'C$ will be perpendicular, and the triangles $APB$, $BP'C$ will be congruent. Therefore $\angle APB$ is a right angle. If the lengths of $AP$, $BP$ are $5x$ and $14x$ then by Pythagoras $AB$ will be $\sqrt{221}x$ and the area of the square is $221x^2$. Since $1989 = 9\times 221$ it follows that $x=3$ and so $PB = 42$cm.HINT:
Rotating the square by 90 degrees in the clockwise direction, the new position of $P$ is collinear with $P$ and $O$.