146k views
2 votes
In the figure, JKLM is a rectangle inscribed in circle O . JK=6 and KL=14 . Find OK in simplest radical form.

In the figure, JKLM is a rectangle inscribed in circle O . JK=6 and KL=14 . Find OK-example-1
User Babon
by
8.7k points

1 Answer

6 votes

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.

User Ben Usman
by
8.0k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories