Question

JK, KL, and LJ are all tangent to circle O. If JA = 14, AL = 12, and CK = 8, what is the perimeter of △JKL?

Show Work.

(This is 10th grade math, please don't make advanced.)

Answer 1

First, we need to draw a diagram:

[asy]
pair O,A,K,L,J;
O = (0,0);
A = (-14,0);
J = (0,14);
K = (0,8);
L = (12,0);
draw(circle(O,14));
draw(A--J--L--cycle);
draw(K--L);
draw(J--K);
label("$O$",O,SW);
label("$A$",A,W);
label("$J$",J,N);
label("$K$",K,W);
label("$L$",L,S);
[/asy]

We know that J is the center of circle O, since J is equidistant from points K, L, and A. Therefore, JA, JL, and JK are all radii of the circle.

We can use the Pythagorean Theorem to find the length of JK. Let x be the length of JK. Then:

$JA^2 + AK^2 = JK^2$

$14^2 + 8^2 = x^2$

$x^2 = 260$

$x = \sqrt{260} = 2\sqrt{65}$

Similarly, we can use the Pythagorean Theorem to find the length of JL. Let y be the length of JL. Then:

$JA^2 + AL^2 = JL^2$

$14^2 + 12^2 = y^2$

$y^2 = 340$

$y = \sqrt{340} = 2\sqrt{85}$

Finally, we can find the length of LK by subtracting JK and JL from AL:

$LK = AL - JK - JL$

$LK = 12 - 2\sqrt{65} - 2\sqrt{85}$

To find the perimeter of triangle JKL, we add the lengths of all three sides:

$JK + JL + LK = 2\sqrt{65} + 2\sqrt{85} + (12 - 2\sqrt{65} - 2\sqrt{85})$

Simplifying, we get:

$JK + JL + LK = 12$

Therefore, the perimeter of triangle JKL is 12.