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
39
39
78
78
32
32
64
To find the perimeter of triangle JKL, we need to find the lengths of JK, KL, and LJ.
Since JK¯¯¯¯¯¯¯¯ is tangent to circle O, we know that JK is perpendicular to the radius at the point of tangency. This means that triangle JKO is a right triangle. By the Pythagorean theorem, we can find the length of KO:
JA^2 = KO^2 + JA*OA
15^2 = KO^2 + 15*OA
225 = KO^2 + 15*OA
Similarly, since KL¯¯¯¯¯¯¯¯ is tangent to circle O, triangle CKO is a right triangle. We can find the length of KO using the Pythagorean theorem for this triangle as well:
CK^2 = KO^2 + CK*OK
6^2 = KO^2 + 6*OA
36 = KO^2 + 6*OA
Now we have two equations for the values of KO^2 and OA, where KO is the same value for both equations:
225 = KO^2 + 15*OA
36 = KO^2 + 6*OA
Subtracting the second equation from the first equation:
189 = 9*OA
OA = 21
Now that we have found OA, we can substitute it back into one of the equations to find KO:
225 = KO^2 + 15*21
225 = KO^2 + 315
KO^2 = -90
KO = 3√10
Now we can find the lengths of JK, KL, and LJ:
JK = JA + AK = 15 + 21 = 36
KL = KC + LC = 21 + 6 = 27
LJ = LA - AJ = 21 - 15 = 6
The perimeter of triangle JKL is therefore:
36 + 27 + 6 = 69
Therefore, the closest answer choice to the perimeter of triangle JKL is 64.