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.
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
1 answer