How do you find the values for right triangles that are congruent by HL one is x + x = 2 the other is 3y + y +4 how do you solve for x and y

Question

Please explain how to use the HL therom

Answer 1

Karen,

The question isn't exactly clear, so you probably aren't going to get a useful answer.

Are x, x, and 2 sides of a right triangle (which would then be isosceles)? If so, 3y, y, and 4 *can't* be sides of a congruent right triangle. Perhaps you mean, you're given two sides of the second triangle as 3y, and (y+4)? In this case, you have two possibilities: Either y+4 is the hypotenuse, or it's the second leg.

In the first case, then the other leg is also 3y (both triangles are isosceles), and x=(2)^0.5 (i.e. squareroot of 2), (3y)^2 + (3y)^2 = 4^2; 18y^2 = 16, y=(8/9)^0.5

In the second case, 3y=y+4; y=2; then the hypotenuse = ((6^2)+(6^2))^0.5 = 6*(2)^0.5

In neither case is HL (hypotenuse leg) what you need, I think? But what HL does allow you to do, is to set up ratios in each of the triangles, once you have the following: (A) at least both of one factor (either H or L) is specified for each triangle, and one of the other factor is specified for one triangle, AND you have a scale factor between the triangles, or (B) you have otherwise enough information to figure out which numbers must correspond to H and L in each triangle (you may need to use areas, other legs and the Pythagorean Theorem to figure this out).