Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter - Section 11.4 - Applications with Acute Triangles - Exercises - Page 529: 41

Answer

The area of the parallelogram is $2\times \frac{1}{2}a\times b \times sin\gamma$

Work Step by Step

In order to calculate the area of the parallelogram, we have to calculate the area of the two triangles $\triangle MNQ$ and $\triangle PQN$. $MN=QP=b$, $MQ=NP=a$ and $\angle QMN =\angle NPQ = \gamma$ because $MNQP$ is a parallelogram. So, the area of both triangles can be calculated by $\frac{1}{2}a\times b \times sin\gamma$. The area of the parallelogram is $2\times \frac{1}{2}a\times b \times sin\gamma$ which is exactly the given formula.
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