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 381: 49


$AC = 16$ $BD = 16$

Work Step by Step

According to Theorem 6-15, the diagonals of a rectangle are congruent. Therefore, we can set $AC$ and $BD$, the diagonals of $ABCD$, equal to one another to solve for $x$: $AC = BD$ Substitute with the expressions given for each diagonal: $2(x - 3) = x + 5$ $2x - 6 = x + 5$ Subtract $x$ from each side of the equation to move variables to the left side of the equation: $x - 6 = 5$ Add $6$ to each side of the equation to solve for $x$: $x = 11$ Now that we have the value of $x$, we can plug $11$ in for $x$. We do not need to evaluate both expressions because we know that the length of the two diagonals are equal to one another: $BD = 11 + 5$ Add to solve: $BD = 16$
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