Geometry: Common Core (15th Edition)

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

Chapter 7 - Similarity - 7-3 Providing Triangles Similar - Lesson Check - Page 455: 1

Answer

Yes, the triangles are similar. $\triangle AZE$ ~ $\triangle RBE$ using the AA Similarity Postulate.

Work Step by Step

We are given the measures of several angles in the two triangles. Let's see if the Angle-Angle Similarity Postulate can be applied here: $\angle REB ≅ \angle AEZ$, so if $m \angle REB = 35^{\circ}$, then $m \angle AEZ = 35^{\circ}$. Now we have the measures of $\angle REB$ and $\angle AEZ$. We need to find the measure of another pair of corresponding angles to be able to use the AA Similarity Postulate to compare the two triangles. We will now use the triangle sum theorem to find $m \angle Z$ in $\triangle AZE$, which should be congruent to its corresponding angle, $\angle B$ in $\triangle RBE$: $\angle Z = 180 - (100 + 35)$ Evaluate parentheses first: $\angle Z = 180 - (135)$ Subtract to solve: $\angle Z = 45^{\circ}$ $m \angle Z ≅ m \angle B$, so we have two congruent angles. Thus, $\triangle AZE$ ~ $\triangle RBE$ using the AA Similarity Postulate.
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