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 32 32 78 78 39 39 64

Answer 1

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

Since JK, KL, and LJ are tangent to circle O, they are all radii of the circle. Let's label the radii as follows: OJ = x, OK = y, OL = z.

We know that JA = 15 and AL = 18. By adding these two lengths, we get JL, which equals 15 + 18 = 33. Since JL is the sum of two radii (OJ + OL), we can write the equation:

x + z = 33

We also know that CK = 6. Since CK is a radius of the circle, it is also equal to OJ. Therefore, OJ = 6, which means that x = 6.

Now, we can substitute x = 6 into the equation above to find z:

6 + z = 33
z = 27

Next, we need to find y. Since KL is the sum of two radii (OK + OL), we can write:

y + 27 = 18
y = -9

However, since lengths cannot be negative, this suggests that point K lies beyond point L in the diagram. Therefore, we can use the absolute value of y: y = 9.

Now we have the lengths of all sides of triangle JKL:

JK = x = 6
KL = y = 9
LJ = z = 27

The perimeter of triangle JKL is the sum of these lengths:

Perimeter = 6 + 9 + 27 = 42

Therefore, the correct answer is 42.