K, 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

78
78

39

32

64

1 answer

To find the perimeter of triangle \( JKL \), we need to look at the properties of the tangents and how they relate to each other. Here are the key points:

1. \( K, L, \) and \( J \) are points where tangents touch the circle.
2. Tangents from a common external point to a circle are equal in length.

Given the following distances:
- \( JA = 15 \)
- \( AL = 18 \)
- \( CK = 6 \)

We can infer:
- Since \( J \) is tangent to the circle at point \( A \) and these distances are from point \( A \), we know \( JA \) (15) is a segment from \( J \).
- \( AL \) (18) suggests the tangential lengths related to \( A \) and \( L \).

Let’s find lengths of the tangents originating from each points:
- Tangents from point \( J \) to circle are \( JA = 15 \) and another part let's denote the point where the other tangent from \( J \) touches the circle as \( JX \). Since we don't have full data on \( J \), this tangent part may not be directly needed here.

- Tangents from point \( L \) to circle will be the lengths \( AL = 18 \).

- Length given \( CK = 6 \) indicates one part of the length \( C \).

To make sense of how these tangents combine, think of \( J, L, K \) tangents pairing tangential points around the circle. To complete \( JKL \), satisfying integer specifics:
- Calculate how remaining part or summation: connect \( J-L \), \( L-K \), \( K-J \).

Considering full engagement to perimeter calculations,
assuming summation proper along tangents uniformly covering points: Perimeter \( JKL = JA + AL + CK + remaining distances to complete the loop: - Rough Estimate to Total intersection Closure: 1 loop multiply by simple covered tangents.

Finally:
Perimeter of triangle \( JKL \) reliably estimated by points closure along any remaining divided among:
\[ 15+18+6 = 39.0, has no extra loop cycle disparities. \]

Thus:
Final Perimeter Closure \( = 39\).

Thus, Perimeter \( JKL = 39.0\).

Correct Answer Registry aligns as **39**.