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: 32

Answer

$a = 8$ $h = 30$ $k = 120$

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 $a$: $a + 15 = 23$ Subtract $15$ from both sides of the equation to solve for $a$: $a = 8$ In parallelograms, consecutive angles are supplementary, so let's set the sum of the expressions for the two consecutive angles equal to $180$: $4h + 2h = 180$ Combine like terms: $6h = 180$ Divide both sides of the equation by $6$ to solve for $h$: $h = 30$ Again, we shall use the fact that consecutive angles in a parallelogram are supplementary to solve for $k$. Let's set up that equation: $2h + k = 180$ $2(30) + k = 180$ Multiply first, according to order of operations: $60 + k = 180$ Subtract $60$ from each side of the equation to solve for $k$: $k = 120$
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