Geometry: Common Core (15th Edition)

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

Chapter 6 - Polygons and Quadrilaterals - 6-6 Trapezoids and Kites - Practice and Problem-Solving Exercises - Page 395: 35

Answer

$x = 35$ $y = 30$

Work Step by Step

The kite pictured in this exercise has been divided into two triangles by its diagonal, with two sides marked congruent and one side that is shared by both triangles and, thus, represents another congruent side for the two triangles. Therefore, the two triangles are congruent by the SSS postulate. Corresponding parts of congruent triangles are congruent, so let's set up an equation to set two corresponding angles equal to one another: $3x + 5 = 4x - 30$ $3x = 4x - 35$ Subtract $4x$ from each side of the equation to move variable terms to the left side of the equation: $-x = -35$ Divide both sides by $-1$ to solve for $x$: $x = 35$ We now turn our attention to the angles within each triangle. In one triangle, we already have expressions for two of the angles. We can find the third angle by using the triangle sum theorem, which states that the sum of the interior angles of a triangle equals $180^{\circ}$. $(4x - 30) + (2y - 20) + (y) = 180$ Let's get rid of the parentheses: $4x - 30 + 2y - 20 + y = 180$ Group like terms on the left side of the equation: $4x + (2y + y) + (-30 - 20) = 180$ Combine like terms: $4x + 3y - 50 = 180$ Let's substitute $35$ for $x$: $4(35) + 3y - 50 = 180$ Multiply first: $140 + 3y - 50 = 180$ Move constants to the other side of the equation and combine them: $3y = 90$ Divide each side of the equation by $3$ to solve for $y$: $y = 30$
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