Chapter 15, Problem 45CR

To determine

To find: The area of the given parallelogram.

Answer to Problem 45CR

  $34.64uni{t}^2$

Explanation of Solution

Given:

  $\begin{array}{l}\text{Measures of diagonals of parallelogram}=8\&10\\ \text{Angle of intersection of diagonals}={60}^o\end{array}$

Formula Used:

  $\text{Area of parallelogram}=\frac{1}{2}\left(d_1\times d_2\times \sin y\right)$

Calculation:

  $\begin{array}{l}\text{Measures of diagonals of parallelogram}=8\&10\\ let,\\ d_1=8\\ d_2=10\\ \text{Angle of intersection of diagonals}=y={60}^o\\ \text{Area of parallelogram}=\frac{1}{2}\left(d_1\times d_2\times \sin y\right)\\ =\frac{1}{2}\left(8\times 10\times \sin {60}^o\right)\\ =\frac{1}{2}\left(80\times \frac{\sqrt{3}}{2}\right)\\ =34.64uni{t}^2\end{array}$

Conclusion:

Thus the required area is $34.64uni{t}^2$ .