Chapter 11.2, Problem 29PSC

To determine

To calculate thearea of parallelogram $ABCD$ .

Answer to Problem 29PSC

The area of parallelogram $ABCD$ is $50$ .

Explanation of Solution

Given:

  $\begin{array}{l}AB=CD=5,\\ AD=BC=20.\end{array}$

Formula Used:

  $\begin{array}{l}Areaofparallelogram=b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

  

Draw AE perpendicular to BC .

  $\Delta BAE$ is a ${30}^{\circ }{60}^{\circ }{90}^{\circ }△$ .

  $\begin{array}{l}sin\theta =\frac{oppositeside}{hypotenuseside}\\ sin30^o=\frac{AE}{AB}\\ \frac{1}{2}=\frac{AE}{5}\\ AE=\frac{5}{2}\\ AE=2.5\end{array}$

  $\begin{array}{l}A(\square ABCD)=BC*AE\\ =20\ast 2.5\\ A(\square ABCD)=50\end{array}$