196k views
5 votes
PLEASE HELP!! S is the center of the circle. Suppose that JK=12, LK=12, NS=8, and PS= 2x - 6. Fine the following

PLEASE HELP!! S is the center of the circle. Suppose that JK=12, LK=12, NS=8, and-example-1

1 Answer

2 votes

Answer:

x = 7

LP = 6

Explanation:

JK = 12

LK = 12

NS = 8

PS = 2x - 6

radius bisects the chords JK and LK because it is perpendicular to them.

JK = JN + NK = 6 + 6 = 12

LK = LP + PK = 6 + 6 = 12

LP = 6

radius= root(6^2 + 8^2) = 10 then PS = 8 = 2x -6,

8 = 2x - 6

14 = 2x

x = 7

User Nrako
by
6.5k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.