Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 4 - Congruent Triangles - 4-5 Isosceles and Equilateral Triangles - Practice and Problem-Solving Exercises - Page 256: 38


$LM = 3$

Work Step by Step

In congruent polygons, vertices are named in the exact order of their matching angles and sides. If $\triangle LMN$ is congruent to $\triangle PQR$, then vertex $L$ corresponds to vertex $P$, vertex $M$ corresponds to vertex $Q$, and vertex $N$ corresponds to vertex $R$. Therefore, $\overline{LM}$ would be congruent to $\overline{PQ}$. In $\triangle PQR$, $PQ$ is $x$, so in $\triangle LMN$, $LM$ would also be $x$. We also know that $\overline{LN}$ is congruent to $\overline{PR}$, so we can set those lines equal to one another to solve for $x$, and we shall have the length of $LM$. Let's set up the equation: $LN = PR$ Let's plug in the values we are given: $2x + 4 = 10$ Subtract $4$ from each side of the equation to isolate the $x$ term: $2x = 6$ Divide each side of the equation by $2$ to solve for $x$: $x = 3$ Since $x = 3$, $LM = 3$.
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