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: 52


$BD = 1$ $AC = 1$

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 $c$: $AC = BD$ Substitute with the expressions given for each diagonal: $\frac{3c}{9} = 4 - c$ Multiply both sides of the equation by $9$ to get rid of the fraction: $3c = 36 - 9c$ Add $9c$ to each side of the equation to move the variable to the left side of the equation: $12c = 36$ Divide each side of the equation by $12$ to solve for $c$: $c = 3$ Now that we have the value of $c$, we can plug $3$ in for $c$. We do not need to evaluate both expressions because we know that the lengths of the two diagonals are equal to one another: $BD = 4 - 3$ Subtract to solve: $BD = 1$ If $BD = 1$, then $AC = 1$ because diagonals of rectangles are congruent.
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