Chapter 11.2, Problem 10PSA

a.

To determine

To find the area of a parallelogram ABCD with side AD as 10 and side AB as 20.

Answer to Problem 10PSA

The area of a parallelogram ABCD is $173.2$ .

Explanation of Solution

Given:

  $\Delta AED$ isa ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$ .

  $\begin{array}{l}AB=20,\\ AD=10.\end{array}$

Formula Used:

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

Calculation:

  

  $\Delta AED$ isa ${30}^{\circ }{60}^{\circ }{90}^{\circ }\Delta$ .

  $\begin{array}{l}sin\theta =\frac{oppositeside}{hypotenuseside}\\ sin60^o=\frac{AE}{AD}\\ \frac{\sqrt{3}}{2}=\frac{AE}{10}\\ AE=5\sqrt{3}\end{array}$

  $DC=AB=20$

  $\begin{array}{l}Areaofparallelogram=b\ast h\\ =DC*AE\\ =20*5\sqrt{3}\\ =100\sqrt{3}\\ \approx 173.2\end{array}$

b.

To determine

To calculate the area of a parallelogram ABCD with side DC as 15 and angle C as ${45}^{\circ }$ .

Answer to Problem 10PSA

The area of a parallelogram ABCD is $84.9$ .

Explanation of Solution

Given:

  $\Delta BEC$ isa ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

  $\begin{array}{l}BC=8,\\ DC=15.\end{array}$

Formula Used:

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

Calculation:

  

  $\Delta BEC$ isa ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

  $\begin{array}{l}sin\theta =\frac{oppositeside}{hypotenuseside}\\ sin45^o=\frac{BE}{BC}\\ \frac{1}{\sqrt{2}}=\frac{BE}{8}\\ BE=\frac{8}{\sqrt{2}}\\ BE=5.66\end{array}$

  $\begin{array}{l}Areaofparallelogram=b\ast h\\ =DC*BE\\ =15*5.66\\ \approx 84.9\end{array}$