Geometry: Common Core (15th Edition)

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

Chapter 6 - Polygons and Quadrilaterals - 6-4 Properties of Rhombuses, Rectangles, and Squares - Practice and Problem-Solving Exercises - Page 380: 15


$m \angle 1 = 55^{\circ}$ $m \angle 2 = 35^{\circ}$ $m \angle 3 = 55^{\circ}$ $m \angle 4 = 90^{\circ}$

Work Step by Step

Diagonals of rhombuses cross each other at right angles, so $m \angle 4 = 90^{\circ}$. Let's look at one of the smaller triangles. We have one angle measuring $90^{\circ}$ and an angle that measures $35^{\circ}$; therefore, we only have to find $m \angle 1$. The interior angles of a triangle add up to $180^{\circ}$, so let's set up an equation where we can find the measure of one of the interior angles given the measures of the other two angles: $m \angle 1 = 180 - (90 + 35)$ Evaluate parentheses first, according to order of operations: $m \angle 1 = 180 - (125)$ Subtract to solve: $m \angle 1 = 55^{\circ}$ Now that we know $m \angle 1$, we also know $m \angle 3$ because the two angles are alternate interior angles, and alternate interior angles are congruent: $m \angle 3 = 55^{\circ}$ We can get $m \angle 2$ by using the triangle-sum theorem, which states that the sum of the interior angles of a triangle equal $180^{\circ}$. One angle is congruent to $\angle 1$ because they are the angles resulting from a diagonal bisecting one of the angles of a rhombus. This angle has a measure of $55^{\circ}$. The other angle is formed by the crossing of the two diagonals of a rhombus and is a right angle. Now that we have two angles, we can figure out the third angle: $m \angle 2 = 180 - (90 + 55)$ Evaluate what's in parentheses first, according to order of operations: $m \angle 2 = 180 - (145)$ Subtract to solve: $m \angle 2 = 35^{\circ}$
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