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 78 78 32 32 64 64 39

Answer 1

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

Since JA and AL are tangent to circle O, we know that JA^2 = JK * JL and AL^2 = KL * JL.

Given that JA = 15 and AL = 18, we can plug in these values to get:

15^2 = JK * JL
18^2 = KL * JL

225 = JK * JL
324 = KL * JL

Now we need to solve for JK, KL, and JL. To do this, we will multiply the two equations we have:

225 * 324 = JK * JL * KL * JL
72900 = JK * KL * JL^2

Next, we will take the square root of both sides to isolate JL:

JL = sqrt(72900 / (JK * KL))

Substitute the values for JK, KL, and CK:

JL = sqrt(72900 / (225 * 324))
JL = sqrt(72900 / 72900)
JL = 1

Now that we have found JL, let's plug it back into one of the original equations to solve for JK and KL:

225 = JK * 1
225 = JK
JK = 225

324 = KL * 1
324 = KL
KL = 324

Now that we have found the lengths of each side of triangle JKL, we can calculate the perimeter:

Perimeter = JK + KL + JL
Perimeter = 225 + 324 + 1
Perimeter = 550

Therefore, the perimeter of triangle JKL is 550.