Question

In the figure below, S is the center of the circle. Suppose that JK= 2x-2,LM=12, SN=5 and SP=5. Find the followingx=PL=

Answer 1

1) Looking at that circle, we have two right triangles within. And Since SN is perpendicular to JK we can state that SN bisects JK. Therefore we have a

Δ JNS

2) By the same principle, we have SP bisecting, the Line segment LM into two equal segments therefore PL = 6

3) As we have 4 similar triangles SAS: Δ SPL, ΔSPM, ΔSNJ, ΔSNK we can write this equation:

JK≅LM

2x-2=12

2x=12+2

2x =14

x=7