Big Ideas Math Integrated I, 2016
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Big Ideas Math Integrated I, 2016 View details
5. Proving Triangle Congruence by SSS
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Exercise 20 Page 623

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.

Practice makes perfect

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.

Error

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.

Correct the Error

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.

JK+KL=JM+ML
2x+1+ 4x+4= 6x+3x-1
â–¼
Solve for x
6x+5=9x-1
5=3x-1
6=3x
2=x
x=2

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.