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: 34

Answer

$x = 28$

Work Step by Step

The kite pictured in this exercise has been divided into two triangles by one of its diagonals, with two sides marked congruent and one side that is shared by both triangles. It 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. We now turn our attention to one of the smaller triangles within the kite. In one triangle, we already have expressions for two of the angles. We also have a value for the third angle, which is $90^{\circ}$, because it is an angle that is formed by the diagonals. Diagonals of a kite are perpendicular to one another. $(x + 6) + (2x) + (90) = 180$ Combine like terms: $3x + 96 = 180$ Subtract $96$ from each side to move constants to the right side of the equation: $3x = 84$ Divide each side by $3$ to solve for $x$: $x = 28$
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