JK¯¯¯¯¯¯¯¯, KL¯¯¯¯¯¯¯¯, LJ¯¯¯¯¯¯¯
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
32
32
64
64
78
78
39
To find the perimeter of triangle JKL, we need to find the lengths of JL, JK, and KL.
First, let's find the length of JL. Since JA is tangent to circle O, by the tangent-secant theorem, JA^2 = JL * JM. Given that JA = 15, we can determine JL by substituting the values:
15^2 = JL * (JL + LJ)
225 = JL^2 + LJ * JL
Similarly, CK is also tangent to circle O, so CK^2 = KL * KM. Given CK = 6, we can determine KL by substituting the values:
6^2 = KL * (KL + LJ)
36 = KL^2 + LJ * KL
We also know that AL = JL - AJ and that JA + JK = KL, so by adding the equations and substituting the values, we can solve for KL:
18 = JL - 15 (AL = JL - JA)
18 + 15 = KL (KL = JA + JK)
33 = KL
Now we can substitute KL = 33 into the equation involving JL and solve for JL:
225 = JL^2 + LJ * JL
225 = JL^2 + LJ * 33
225 = JL^2 + 33LJ
Substitute LJ = 33 - JL from the equation LJ = KL - JK:
225 = JL^2 + 33(33 - JL)
225 = JL^2 + 33(33) - 33JL
225 = JL^2 + 1089 - 33JL
JL^2 - 33JL + 864 = 0
(JL - 27)(JL - 36) = 0
JL = 27 or JL = 36
Since JL cannot be 27 (as it would make JA longer than AL), JL = 36 and LJ = 33. Therefore, JK = 33 - 15 = 18.
The perimeter of triangle JKL is JL + JK + KL = 36 + 18 + 33 = 87.
Therefore, the correct answer is not given in the options provided.