Chapter 5.7, Problem 16PSB

To determine

To find: Thearea of a parallelogram whose four vertices are given.

Answer to Problem 16PSB

The area of the parallelogram is 54.

Explanation of Solution

Given Information:

Four vertices of parallelogram are $\left(-5,9\right),\left(4,5\right),\left(4,-7\right)\text{and}\left(-5,-3\right)$

Formula used:

Distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ is given by

  $\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

Area of a parallelogram $\frac{1}{2}bh$

  $b\text{and}h$ are the base and height of a parallelogram

Calculation:

Let the four vertices of the quadrilateral be $A\left(-5,9\right),B\left(4,5\right),C\left(4,-7\right)\text{and}D\left(-5,-3\right)$

If we draw a perpendicular from point $B$ to the side $AD,$ then the point where it meets $AD$ will be $E\left(-5,5\right)$

This is because the line drawn will be parallel to x-axis.

So, if we take $AD$ as the base, then $BE$ becomes height of the parallelogram

Now we know that,

Distance between two points $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)=\sqrt{{\left(x_1-x_2\right)}^2+{\left(y_1-y_2\right)}^2}$

  $\therefore AD=\sqrt{{\left(-5+5\right)}^2+{\left(9+3\right)}^2}=12$

  $BE=\sqrt{{\left(4+5\right)}^2+{\left(5-5\right)}^2}=9$

Now, Area of a parallelogram $\left(A\right)=\frac{1}{2}bh$

  $b\text{and}h$ are the base and height of a parallelogram

  $\begin{array}{l}\therefore A=\frac{1}{2}\times AD\times BE\\ \therefore A=\frac{1}{2}\times 12\times 9\\ \Rightarrow A=54\end{array}$

Hence, the area of the parallelogram is 54.