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Can you really equate the sides whose lengths are 6x and 2x+1?
Error: The student incorrectly assumed which legs were congruent.
Explanation: See solution.
Let's take a look at the steps taken by the student and identify the error. Then, we will find the value of x that makes the triangles congruent.
The student supposed that JK ≅ JM, and concluded that x= 14. If this was true, when substituted into the expressions that represent the length of the other two legs, x= 14 would produce the same value for 4x+4 and 3x-1. Let's see if this is the case.
| Side | Expression | x= 1/4 | Side Length |
|---|---|---|---|
| KL | 4x+4 | 4( 1/4)+4 | 5 |
| ML | 3x-1 | 3( 1/4)-1 | - 0.25 |
As we can see, KL ≠ML when x = 14. Therefore, the sides KL and ML are not congruent when x= 14.
If two triangles are congruent, then they have the same perimeter. JK+KL+JL=JM+ML+JL ⇕ JK+KL=JM+ML We can substitute the given expressions into the above equation and solve for x.
Substitute expressions
Add terms
LHS-6x=RHS-6x
LHS+1=RHS+1
.LHS /3.=.RHS /3.
Rearrange equation
Let's now find the length of the legs of each right triangle.
| Side | Expression | x= 2 | Side Length |
|---|---|---|---|
| KL | 4x+4 | 4( 2)+4 | 12 |
| KJ | 2x+1 | 2( 2)+1 | 5 |
| ML | 3x-1 | 3( 2)-1 | 5 |
| MJ | 6x | 6( 2) | 12 |
We can now label and mark congruent sides.
We see that △ JKL and △ LMJ have two pairs of congruent sides and congruent included angles. Therefore, by the SAS Congruence Theorem, △ JKL ≅ △ LMJ.