Chapter 11.2, Problem 21PSB

a.

To determine

To find the area of parallelogram ABCD with side BC as $14$ and DC as $6$ .

Answer to Problem 21PSB

The area of parallelogram ABCDis $72.7$ .

Explanation of Solution

Given:

  $\begin{array}{l}BC=AD=14,\\ AB=CD=6.\end{array}$

Formula Used:

  $sin\theta =\frac{oppositeside}{hypotenuseside}$

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

Calculation:

Draw altitude $\overline{AE}$ .

  $DC=6$

  $\begin{array}{l}\angle C={120}^{\circ }\\ \therefore \angle B={60}^{\circ }\\ \angle E={90}^{\circ }\\ \therefore \angle A={30}^{\circ }\end{array}$

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

  

In right angled $\Delta AEB$ , we get

  $\begin{array}{l}sin\theta =\frac{oppositeside}{hypotenuseside}\\ sin60^o=\frac{AE}{AB}\\ \frac{\sqrt{3}}{2}=\frac{AE}{6}\\ AE=3\sqrt{3}\end{array}$

  $\begin{array}{l}Areaofparallelogram=b\ast h\\ =BC*AE\\ =14*3\sqrt{3}\\ \approx 72.7\end{array}$

b.

To determine

To calculate the area of parallelogram ABCD with angle A as ${135}^{\circ }$ .

Answer to Problem 21PSB

The area of parallelogram ABCD is $120.2$ .

Explanation of Solution

Given:

  $\begin{array}{l}AB=CD=10,\\ BC=AD=17.\end{array}$

Formula Used:

  $\sin \theta =\frac{oppositeside}{hypotenuseside}$

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

Calculation:

Draw altitude $\overline{CE}$ .

  $DC=6$

  $\begin{array}{l}\angle A={135}^{\circ }\\ \therefore \angle D={45}^{\circ }\\ \angle E={90}^{\circ }\\ \therefore \angle C={45}^{\circ }\end{array}$

  $\Delta CED$ is a ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

  

In right angled $\Delta CED$ , we get

  $\begin{array}{l}sin\theta =\frac{oppositeside}{hypotenuseside}\\ sin60^o=\frac{AE}{AB}\\ \frac{1}{\sqrt{2}}=\frac{CE}{CD}\\ \frac{1}{\sqrt{2}}=\frac{CE}{10}\\ CE=\frac{10}{\sqrt{2}}\end{array}$

  $\begin{array}{l}Areaofparallelogram=b\ast h\\ =AD*CE\\ =17*\frac{10}{\sqrt{2}}\\ =17*7.07\\ \approx 120.2\end{array}$

c.

To determine

To compute the area of parallelogram ABCD .

Answer to Problem 21PSB

The area of parallelogram ABCD is $120$ .

Explanation of Solution

Given:

  $\begin{array}{l}AB=CD=8,\\ BC=AD=17.\end{array}$

Formula Used:

The below theorem is used:

Pythagoras theorem states that “In a right angled triangle, the square of the hypotenuse side is equal to the sum of squares of the other two sides”.

  $\begin{array}{l}Areaoftriangle=\frac{1}{2}\ast b\ast h\\ b=Base,\\ h=Height\end{array}$

Calculation:

Draw altitude $\overline{CE}$ .

  $DC=6$

  $\begin{array}{l}\angle B={45}^{\circ }\\ \therefore \angle D={45}^{\circ }\end{array}$

  $\Delta CAB$ is a ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

  $\Delta ACD$ is a ${45}^{\circ }{45}^{\circ }{90}^{\circ }\Delta$ .

  

In right angled $\Delta CAB$ , we get

  ${\left(CB\right)}^2={\left(AC\right)}^2+{\left(AB\right)}^2$

  $\begin{array}{l}{\left(AC\right)}^2={\left(17\right)}^2-{\left(8\right)}^2\\ {\left(AC\right)}^2=289-64\\ {\left(AC\right)}^2=225\\ AC=\sqrt{225}=15\end{array}$

  $\begin{array}{l}A(\Delta CAB)=\frac{1}{2}\ast AB\ast AC\\ =\frac{1}{2}\ast 8\ast 15\\ =60\\ A(ACD)=\frac{1}{2}\ast CD\ast AC\\ =\frac{1}{2}\ast 8\ast 15\\ =60\end{array}$

  $\begin{array}{l}Areaof(▱ABCD)=A(ACD)+A(\Delta CAB)\\ =60+60\\ =120\end{array}$