Chapter 11, Problem 11STP

To determine

To calculate: The area of the given parallelogram.

Answer to Problem 11STP

The area of the parallelogram is $106.038{\text{ft}}^{\text{2}}$

Explanation of Solution

Given information: The two sides of the parallelogram is $13\text{ft}\text{and}9\text{ft}$ and the one angle between the sides is ${65}^0$

Concept Used:

• The area $A$ of the parallelogram having base $b$ and the height $h$is

  $A=b\times h$

And the opposite sides of the parallelogram are equal.

• The value of the trigonometric function is

  $\sin \theta =\frac{\text{perpendicular}}{\text{hypotenuse}}$

Calculation:

The two sides of the parallelogram is $13\text{ft}\text{and}9\text{ft}$.

Base of the parallelogram is $13\text{ft}$

• For the height of the parallelogram use of

Trigonometry in the triangle inside the parallelogram.

  $\sin \theta =\frac{\text{perpendicular}}{\text{hypotenuse}}$

Opposite angles of the parallelogram are equal so,

  $\theta ={65}^0$

And the hypotenuse is $9\text{ft}$

Then,

  $\begin{array}{l}\sin {65}^0=\frac{\text{perpendicular}}{9}\\ \text{perperdicular }=9\times \sin {65}^0\\ =9\times 0.9063\\ =8.1567\text{ft}\end{array}$

So, the height of the parallelogram is

  $h=8.1567\text{ft}$

• Now, the area of the parallelogram is

  $\begin{array}{l}A=b\times h\\ A=13\times 8.1567\\ A=106.038{\text{ft}}^{\text{2}}\end{array}$

So, area of the parallelogram is $106.038{\text{ft}}^{\text{2}}$

Conclusion:

The area of the parallelogram is $106.038{\text{ft}}^{\text{2}}$