Answer:
OK = 6/2 = 3√2
Explanation:
Since JKLM is a rectangle inscribed in circle O, its diagonals JL and KM are diameters of the circle. Therefore, the center of the circle, O, is the midpoint of JL and KM. Let N be the midpoint of JK, so JL and KM pass through N.
Since JKLM is a rectangle, we have JK=ML=6 and KL=JM=14. Let x be the length of OM.
Then, ON is the midpoint of JL, so JL=2ON. Similarly, KM=2OM.
Since JL+KM=18 (the diameter of the circle), we have:
2ON + 2OM = 18
Simplifying this equation, we get:
ON + OM = 9
Since ON=NM=6/2=3 and JK=6, we have JN = sqrt(JK^2 - MN^2) = sqrt(6^2 - 3^2) = sqrt(27).
Then using the Pythagorean theorem, we find:
OM^2 = ON^2 + NM^2
OM^2 = 3^2 + (sqrt(27))^2
OM^2 = 9 + 27
OM^2 = 36
OM = 6
Therefore, the length of OK, which is half of OM, is:
OK = 6/2 = 3√2
So, OK in simplest radical form is 3√2.