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: 50


$BD = 2$ $AC = 2$

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 $a$: $AC = BD$ Substitute with the expressions given for each diagonal: $2(5a + 1) = 2(a + 1)$ Use the distributive property: $10a + 2 = 2a + 2$ Subtract $2a$ from each side of the equation to move variables to the left side of the equation: $8a + 2 = 2$ Subtract $2$ from each side of the equation to move constants to the right side of the equation: $8a = 0$ Now that we have the value of $a$, we can plug $0$ in for $a$. We do not need to evaluate both expressions because we know that the lengths of the two diagonals are equal. $BD = 2(0 + 1)$ Evaluate parentheses first: $BD = 2(1)$ Multiply to solve: $BD = 2$ If $BD = 2$, then $AC = 2$ because diagonals of rectangles are congruent.
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