First, let's find LK by using Power of a Point theorem on point L:
LK * LJ = LA * LB
LK * (LK + KJ) = 18 * (18 + 15)
LK^2 + LK * KJ = 18 * 33
LK^2 + LK * KJ = 594
Next, let's use Power of a Point theorem on point K:
LK * KJ = KC * KB
LK * KJ = 6 * (6 + LK)
Substituting this into the equation we found earlier, we get:
LK^2 + 6 * (6 + LK) = 594
LK^2 + 36 + 6LK = 594
LK^2 + 6LK - 558 = 0
(LK - 18)(LK + 24) = 0
LK = 18 (since length cannot be negative)
So, JL = 18 + 24 = 42
Perimeter of triangle JKL = JA + AL + LK + KJ + JL = 15 + 18 + 18 + 6 + 42 = 99
Therefore, the perimeter of triangle JKL is 99.
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