To find the perimeter of triangle JKL, we first notice that triangle JKL is similar to triangle JAO by AA similarity.
Since lines JA, KL, and LJ are tangent to circle O, we know that triangle JAO is a right triangle with AO being the radius of the circle. This means that triangle JKL is also a right triangle.
By proportions, we have:
JK/AK = LJ/AL
JK/(AK - 15) = LJ/18
JK = (AK - 15)LJ / 18
Also, since triangle JKL is similar to triangle JAO, we have:
JK / AO = KJ / JA
JK / r = 15 / 18
Substitute JK = (AK - 15)LJ / 18 into the second equation:
(AK - 15)LJ / 18 = r * 15 / 18
AK - 15 = r
AA - AK = 15
3r = 3AK, r = AK
Since AK = 15 + KL, KL = 18
perimeter = 15 + 18 + 18 = 51
Therefore, the perimeter of triangle JKL is 51.
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?
39
78
32
64
1 answer