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 - Lesson Check - Page 379: 4

Answer

$x = 4$ $MK = 4$ $JL = 4$

Work Step by Step

In the rectangle described, both $\overline{JL}$ and $\overline{MK}$ are diagonals of the rectangle. According to Theorem 6-15, diagonals of a rectangle are congruent to one another. Therefore, we can set the two expressions that are given for the two diagonals equal to one another: $JL = MK$ Let's plug in our expressions for each diagonal: $4x - 12 = x$ Add $12$ to each side of the equation to move constants to the right of the equation: $4x = x + 12$ Subtract $x$ from each side to move variables to the left side of the equation: $3x = 12$ Divide each side by $3$ to solve for $x$: $x = 4$ Therefore, $MK = 4$. Let's plug in $4$ for $x$ into the expression for $JL$: $JL = 4(4) - 12$ Multiply first, according to order of operations: $JL = 16 - 12$ Subtract to solve: $JL = 4$
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