$|FE| = 7 cm$
$|AB| = 2 cm$
OAEF is a rectangle. ABCD is a square.
O is the center of the quarter circle.
$|ED| = x =?$
$|FE| = 7 cm$
$|AB| = 2 cm$
OAEF is a rectangle. ABCD is a square.
O is the center of the quarter circle.
$|ED| = x =?$
Copyright © 2021 JogjaFile Inc.

Note that $OC,OE $ are radii of circle
By Pythagoras Theorem
$$(OB)^2+(BC)^2=r^2$$ where $r$ is radius of circle SO, $$r=\sqrt{85}$$
Similarly, $$(OF)^2+(FE)^2=r^2$$
$$(x+2)^2+49=r^2$$
which gives $x=4$