Geometry: Common Core (15th Edition)

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

Chapter 6 - Polygons and Quadrilaterals - 6-3 Proving That a Quadrilateral Is a Parallelogram - Practice and Problem-Solving Exercises - Page 374: 34

Answer

$f = 11$ $e = 13$ $c = 204$

Work Step by Step

In parallelograms, opposite sides are congruent, so let's set the expressions for opposite sides equal to one another to solve for $f$: $f - 3 = 8$ Add $3$ to both sides of the equation to solve for $f$: $f = 11$ In parallelograms, consecutive angles are supplementary, so let's set the sum of the expressions for the two consecutive angles equal to $180$: $6e + 102 = 180$ Subtract $102$ from each side of the equation to isolate the constants on the right side of the equation: $6e = 78$ Divide both sides of the equation by $6$ to solve for $e$: $e = 13$ Again, we shall use the fact that consecutive angles in a parallelogram are supplementary to solve for $c$. Let's set up that equation: $6e + \frac{c}{2} = 180$ $6(13) + \frac{c}{2} = 180$ Multiply first, according to order of operations: $78 + \frac{c}{2} = 180$ $\frac{c}{2} = 102$ Multiply both sides by $2$ to solve for $c$: $c = 204$
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