To find the perimeter of triangle JKL, we need to first find the lengths of the sides JK, KL, and LJ.
Since JK, KL, and LJ are all tangent to circle O, we can use the fact that the length of a tangent from an external point to a circle is equal to the length of the segment connecting the external point to the circle's center.
Let's consider triangle JKL. The length of JA is given as 15, AL is given as 18, and CK is given as 6.
Since JA is the radius of circle O and AL is tangent to the circle, we can say that AL is perpendicular to JA. Similarly, since CK is the radius of circle O and KL is tangent to the circle, we can say that CK is perpendicular to KL.
Given this, we can see that triangle ALJ is a right triangle. Using the Pythagorean theorem, we can find the length of LJ:
(15)^2 = (AL)^2 + (LJ)^2
225 = 324 + (LJ)^2
(LJ)^2 = 225 - 324
(LJ)^2 = 99
LJ = √99 = 9√11
Similarly, we can find the length of JK:
(6)^2 = (CK)^2 + (JK)^2
36 = 36 + (JK)^2
(JK)^2 = 0
JK = 0
Since JK is 0, the triangle JKL degenerates into a line segment KL.
Therefore, the perimeter of triangle JKL (line segment KL) is KL + LJ = 0 + 9√11 = 9√11.
The correct answer is 39.
JK¯¯¯¯¯¯¯¯, KL¯¯¯¯¯¯¯¯, LJ¯¯¯¯¯¯¯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 64 64 32
1 answer