Geometry: Common Core (15th Edition)

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

Chapter 6 - Polygons and Quadrilaterals - 6-2 Properties of Parallelograms - Practice and Problem-Solving Exercises - Page 364: 14


$x = 6$ $y = 8$

Work Step by Step

Theorem 6-6 states that in a quadrilateral that is a parallelogram, its diagonals bisect one another. We can now deduce that for bisector $\overline{ab}$, $\overline{ab}$ is congruent to $\overline{ab}$. For $\overline{ab}$, $\overline{ab}$ is congruent to $\overline{ab}$. Let's set up the two equations reflecting this information: $PT = TR$ $QT = TS$ Let's substitute in what we know: $2x = y + 4$ $x + 2 = y$ We can solve for both $x$ and $y$ by setting the two equations up as a system of equations: $2x = y + 4$ $x + 2 = y$ Let's get all the variables on one side and the constants on the other: $2x - y = 4$ $x - y = -2$ Let's modify the second equation so that the $y$ variables are the same but differ only in sign. We can do this by multiplying the second equation by $-1$: $2x - y = 4$ $-x + y = 2$ Now, we add the two equations together: $x = 6$ Now that we have the value for $x$, we can plug in this value for $x$ into one of the original equations to find $y$: $6 - y = -2$ Subtract $6$ from each side of the equation to isolate constants on the right side of the equation: $-y = -8$ Divide each side by $-1$ to solve for $y$: $y = 8$
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