JK, KL, and LJ are all tangent to circle O. If JA = 14, AL = 12, and CK = 8, what is the perimeter of ΔJKL?
Show Work.
(This is 10th grade math, please don't make advanced.)
Since JK, KL, and LJ are all tangent to circle O, they are all perpendicular to the radii drawn to the points of tangency. This means that JK, KL, and JL bisect angles J, K, and L respectively.
Let x be the length of JK, y be the length of KL, and z be the length of LJ.
Since JK and LJ are perpendicular to radii, they are equal in length:
JK = LJ = x
By the angle bisector theorem, we can set up the following proportions:
JA/JL = LK/LC
14/x = y/(8+2x)
Multiplying both sides by (8+2x):
14(8+2x)/x = y
112/x + 28 = y
Similarly:
LC/LK = AJ/JK
(8+2x)/y = 12/x
Multiplying both sides by xy:
8x + 24y = xy
Using the earlier substitution for y:
8x + 24(112/x+28) = x(112/x+28)
Simplifying:
8x + 2688/x + 672 = 112 + 28x
Multiplying both sides by x:
8x^2 + 2688 + 672x = 112x + 28x^2
Simplifying:
4x^2 - 168x + 672 = 0
Dividing both sides by 4:
x^2 - 42x + 168 = 0
Factoring:
(x-6)(x-36) = 0
Since x cannot be 6 (it would make LJ of length 0), we have x = 36.
Using the earlier formulas for y and z:
y = 112/x + 28 = 112/36 + 28 = 7
z = 8 + 2x = 8 + 2(36) = 80
Therefore, the perimeter of triangle JKL is:
JK + KL + JL = 36 + 7 + 36 = 79
So the perimeter of ΔJKL is 79.