JK¯¯¯¯¯¯¯¯, KL¯¯¯¯¯¯¯¯, LJ¯¯¯¯¯¯¯

Question

are all tangent to circle O. The diagram is not drawn to scale. If JA = 15, AL = 18 and CK = 6, what is the perimeter of triangle JKL?

(1 point)
Responses

39
39

64
64

32
32

78

Answer 1

To find the perimeter of triangle JKL, we need to find the lengths of segments JK, KL, and LJ.

Given that JK, KL, and LJ are tangent to circle O, we can use the Tangent-Secant Theorem:
JA*AL = JK^2
15*18 = JK^2
270 = JK^2
JK = sqrt(270) = 3sqrt(30)

Using the same theorem for CK:
CK^2 = KL^2
6^2 = KL^2
36 = KL

Now we can find LJ:
LJ = JA + AL - JL
LJ = 15 + 18 - 6
LJ = 27

Now, we can find the perimeter of triangle JKL:
Perimeter = JK + KL + LJ
Perimeter = 3sqrt(30) + 36 + 27
Perimeter = 3sqrt(30) + 63

Therefore, the perimeter of triangle JKL is 3sqrt(30) + 63. This is approximately 78.67, so the answer closest to this is 78.