Chapter 12.2, Problem 1E

a.

To determine

To find: Area of given parallelogram.

Answer to Problem 1E

Area of given parallelogram $=53m^2$

Explanation of Solution

Given information:

  $\begin{array}{l}\theta ={45}^0\\ a=10m\\ b=15m\end{array}$

Concept:

Area of parallelogram $=\frac{1}{2}ab\sin \theta$

Given

  $\begin{array}{l}\theta ={45}^0\\ a=10m\\ b=15m\end{array}$

We know

Area of parallelogram $=\frac{1}{2}ab\sin \theta$

On substituting values

We get

  $\begin{array}{l}\Rightarrow Area=\frac{1}{2}\left(10\right)\left(15\right)\sin \left({45}^0\right)\\ \Rightarrow Area=\frac{1}{2}\left(10\right)\left(15\right)\left(0.707\right)\\ \Rightarrow Area=\frac{106.1}{2}\\ \Rightarrow Area=53m^2\end{array}$

Therefore,

Area of given parallelogram $=53m^2$

b.

To determine

To find: Area of given parallelogram.

Answer to Problem 1E

Area of given parallelogram $=28.7m^2$

Explanation of Solution

Given information:

  $\begin{array}{l}\theta =\frac{{45}^0}{2}={22.5}^0\\ a=10m\\ b=15m\end{array}$

Concept:

Area of parallelogram $=\frac{1}{2}ab\sin \theta$

Given

  $\begin{array}{l}\theta =\frac{{45}^0}{2}={22.5}^0\\ a=10m\\ b=15m\end{array}$

We know

Area of parallelogram $=\frac{1}{2}ab\sin \theta$

On substituting values

We get

  $\begin{array}{l}\Rightarrow Area=\frac{1}{2}\left(10\right)\left(15\right)\sin \left({22.5}^0\right)\\ \Rightarrow Area=\frac{1}{2}\left(10\right)\left(15\right)\left(0.383\right)\\ \Rightarrow Area=\frac{57.4}{2}\\ \Rightarrow Area=28.7m^2\end{array}$

Therefore,

Area of given parallelogram $=28.7m^2$

c.

To determine

To find: Area of given parallelogram.

Answer to Problem 1E

Area of given parallelogram $=75m^2$

Explanation of Solution

Given information:

  $\begin{array}{l}\theta =2\left({45}^0\right)={90}^0\\ a=10m\\ b=15m\end{array}$

Concept:

Area of parallelogram $=\frac{1}{2}ab\sin \theta$

Given

  $\begin{array}{l}\theta =2\left({45}^0\right)={90}^0\\ a=10m\\ b=15m\end{array}$

We know

Area of parallelogram $=\frac{1}{2}ab\sin \theta$

On substituting values

We get

  $\begin{array}{l}\Rightarrow Area=\frac{1}{2}\left(10\right)\left(15\right)\sin \left({90}^0\right)\\ \Rightarrow Area=\frac{1}{2}\left(10\right)\left(15\right)\left(1\right)\\ \Rightarrow Area=\frac{150}{2}\\ \Rightarrow Area=75m^2\end{array}$

Therefore,

Area of given parallelogram $=75m^2$