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: 22


$x = \frac{5}{3}$ $LN = \frac{29}{3}$ $MP = \frac{29}{3}$

Work Step by Step

According to Theorem 6-15, the diagonals of a rectangle are congruent. Therefore, we can set $LN$ and $MP$, the diagonals of $LMNP$, equal to one another to solve for $x$: $LN = MP$ Substitute with the expressions given for each diagonal: $7x - 2 = 4x + 3$ Subtract $4x$ from each side of the equation to move variables to the left side of the equation: $3x - 2 = 3$ Add $2$ to each side of the equation to move constants to the right side of the equation: $3x = 5$ Divide each side by $3$ to solve for $x$: $x = \frac{5}{3}$ Now that we have the value of $x$, we can plug $\frac{5}{3}$ in for $x$: $LN = 7(\frac{5}{3}) - 2$ $LN = \frac{35}{3} - 2$ $LN = \frac{35}{3} - \frac{6}{3}$ $LN = \frac{29}{3}$ Now let's find $MP$: $MP = 4(\frac{5}{3}) + 3$ Multiply first, according to order of operations: $MP = \frac{20}{3} + 3$ Convert $3$ into a fraction with $3$ as its denominator: $MP = \frac{20}{3} + \frac{9}{3}$ Add to solve: $MP = \frac{29}{3}$
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