JK¯¯¯¯¯¯¯¯, KL¯¯¯¯¯¯¯¯, and LJ¯¯¯¯¯¯¯ are all tangent to O (not drawn to scale). JA = 12, AL =15 and CK = 16. Find the perimeter of △JKL (1 point) Responses 42 42 84 84 86 86 43

Answer 1

To find the perimeter of triangle JKL, we need to find the lengths of JK, KL, and JL.

Since JK and KL are tangent to O, we know that JK = JL (tangents from a point to a circle are equal in length).

Therefore, JK = JL = x (let's call this length x).

Now, we can use the Pythagorean theorem to find the length of KL.

JK^2 + JL^2 = KL^2
x^2 + x^2 = 15^2
2x^2 = 225
x^2 = 112.5
x ≈ 10.61 (rounded to 2 decimal places)

Now we can find the perimeter of triangle JKL:

JK + KL + JL = x + 16 + x
= 2x + 16
= 2(10.61) + 16
= 21.22 + 16
= 37.22

So, the perimeter of triangle JKL is approximately 37.22.

Therefore, the closest response is 42.