Question

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.)
(The perimeter should be 68 units.)

Answer 1

Since JA and AL are tangent to circle O, we know that OJ is perpendicular to JA and OL is perpendicular to AL. Similarly, we know that OK is perpendicular to CK and OL is perpendicular to KL.

Let x be the length of JK. Then, KJ = x and LJ = x.

Using the Pythagorean Theorem, we can find the length of OJ and OL:

OJ^2 = JA^2 - AO^2 = JA^2 - r^2

OL^2 = AL^2 - AO^2 = AL^2 - r^2

where r is the radius of the circle.

Since OJ = OL, we can set the two expressions equal to each other:

JA^2 - r^2 = AL^2 - r^2

Simplifying this equation gives:

JA^2 = AL^2

Substituting the given values, we get:

14^2 = 12^2 + r^2

r^2 = 196 - 144 = 52

r = sqrt(52)

Now we can find the lengths of JK, KL, and LJ:

KJ^2 = r^2 + OK^2 = 52 + 8^2 = 116

LJ^2 = r^2 + OL^2 = 52 + 12^2 = 196

KJ = LJ = sqrt(116) = sqrt(4*29) = 2sqrt(29)

JKL is an isosceles triangle with two sides of length 2sqrt(29) and one side of length 14.

The perimeter is therefore:

P = 2*(2sqrt(29)) + 14 = 4sqrt(29) + 14

Simplifying this expression gives:

P = 2(2sqrt(29) + 7) = 4sqrt(29) + 14

P ≈ 68 (rounded to the nearest whole number)