Jason T. answered • 12/19/12
Master's degree in math education
You are not given whether triangle HIJ is acute, right, or obtuse, so you must consider all possibilities. With this particular problem it is easy rush to a couple of "nice" solutions instead of thinking a little bit more to consider all possible solutions.
HI is congruent to JI. Solving for x gives x=5, which leads to lengths of HI and JI are 17. A bisector from the vertex to the base of an isosceles triangle is perpendicular to the base, and smaller triangles HIK and JIK are congruent. The only restriction on the length of HJ, the 3rd side of the triangle is that it be < 34.
Since angles JKI and HKI are right angles, the Pythagorean Theorem says that:
(IK)^2 + (KJ)^2 = 17^2
(IK)^2 + (KJ)^2 = 289
If L is the midpoint of segment IK, then the length of KL= (IK)/2
Using the equations above, solve for IK to get (IK)=sqrt(289 - (KJ)^2). We know KJ < 17 because HJ < 34.
KL= (IK)/2 so KL = ( sqrt(289 - (KJ)^2) ) / 2
If you now create segment JL, there is another right triangle JKL. Use the Pythagorean Theorem again, with JL as the hypotenuse and KL and KJ as the legs.
(KL)^2 + (KJ)^2 = (JL)^2 Substitute from earlier equations.
(KL)^2 = (289 - (KJ)^2) / 4
(289 - (KJ)^2) / 4 + (KJ)^2 = (JL)^2
(289/4) + (3(KJ)^2)/4 = (JL)^2
But we want JL, not (JL)^2, so take the square root of both sides.
sqrt(289 + 3(KJ)^2)/2 = JL
Now look at small triangle JKI. If the three sides are only integer lengths, they have to be 8, 15, and 17. But there is no such restriction in your problem and we have solved for all possible cases. We do know that KJ<17 and that is our only restriction to guarantee that our results make physical sense.
If it turns out that your lengths are only integers, or that your intial triangle HIJ was supposed to be limited to acute, or to obtuse, or to right, then you can substitute the appropriate value of KJ and calculate JL.
