Geometry: Common Core (15th Edition)

Published by Prentice Hall
ISBN 10: 0133281159
ISBN 13: 978-0-13328-115-6

Chapter 6 - Polygons and Quadrilaterals - 6-5 Conditions for Rhombuses, Rectangles, and Squares - Practice and Problem-Solving Exercises - Page 388: 42

Answer

The two sides that are marked congruent both measure $18$. Another side measures $16.4$. The last side is already given; it measures $4.8$.

Work Step by Step

We can set the first set of congruent sides equal to one another to solve for $b$: $4y + 6 = 7y - 3$ Subtract $6$ from each side of the equation to move constants to the left side of the equation: $4y = 7y - 9$ Subtract $7y$ from each side of the equation to move the variable to the left side of the equation: $-3y = -9$ Divide each side of the equation by $-3$ to solve for $y$: $y = 3$ Let's plug in $3$ for $y$ to find the length of one of the sides: length of side = $4(3) + 6$ Multiply first, according to order of operations: length of side = $12 + 6$ Add to solve: length of side = $18$ Let's substitute $3$ for $y$ in another expression: length of side = $5(3) + 1.4$ Multiply first, according to order of operations: length of side = $15 + 1.4$ Add to solve: length of side = $16.4$ Two of the sides are congruent, so they both measure $18$. Another side measures $16.4$. The last side is already given; it measures $4.8$.
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