find length of PB

[Albert]

please find the length of PB

 

[caffeinemachine]

please find the length of PB

View attachment 1178

HINT:

Rotating the square by 90 degrees in the clockwise direction, the new position of $P$ is collinear with $P$ and $O$.

 

[Opalg]

HINT:

Rotating the square by 90 degrees in the clockwise direction, the new position of $P$ is collinear with $P$ and $O$.

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.

 

[Albert]

It is easy to see that four points A,B,O,P are
located on the same circle
let AP=5x ,BP=14X
$AP^2+BP^2=221x^2=AB^2=1989$
$\therefore x=3cm$
$BP=14x=42cm$