Veritas Practice Test - IR #7

I think the key to this one, and to a lot of IR questions, is to look at the answers afforded to you. If JL is 260, and none of the answers are greater than 260, then JL is surely the hypotenuse.

At this point, and at every point when you are given a right triangle, you should be looking for the three common right triangles, and multiples of those angles,

3 - 4 - 5
5 - 12- 13
8 - 15 - 17

In the problem, 260 is a multiple of 13; 13*20 = 260. So just grab the 5-12-13 triangle, and just multiple each of the other sides by 20; in this case, 20*5 = 100 and 20*12 = 240.

This same approach wouldn't work if you weren't sure that 260 was the hypotenuse, but since it clearly is, I believe the triangle is necessarily a 5-12-13 triangle.