Question

The quadrilateral RSTV has vertices $R\left(-4,6\right),S\left(4,5\right),T\left(6,3\right)\text{and}V\left(5,-8\right)$. Find the perimeter of quadrilateral.

Short answer

The perimeter of quadrilateral is approximately$38.6\text{units}$.

Step-by-step solution

Step 1. Write down the given information.

The given quadrilateral RSTV has vertices $R\left(-4,6\right),S\left(4,5\right),T\left(6,3\right)\text{and}V\left(5,-8\right)$.

Step 2. Calculation.

The distance$\left(d\right)$ between the pair of points with given coordinates $R\left(-4,6\right),S\left(4,5\right),T\left(6,3\right)\text{and}V\left(5,-8\right)$ is evaluated using the distance formula.

$\begin{array}{c}\left|d_{RS}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(4+4\right)}^2+{\left(5-6\right)}^2}\\ =\sqrt{65}\text{units}\end{array}$$\begin{array}{c}\left|d_{ST}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(6-4\right)}^2+{\left(3-5\right)}^2}\\ =2\sqrt{2}\text{units}\end{array}$$\begin{array}{c}\left|d_{TV}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(5-6\right)}^2+{\left(-8-3\right)}^2}\\ =\sqrt{122}\text{units}\end{array}$$\begin{array}{c}\left|d_{RV}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(5+4\right)}^2+{\left(-8-6\right)}^2}\\ =\sqrt{277}\text{units}\end{array}$

Now the perimeter of quadrilateral is the sum of its four sides. Therefore perimeter (P) of quadrilateral (RSTV) is evaluated as:

$\begin{array}{c}P=\left|d_{RS}\right|+\left|d_{ST}\right|+\left|d_{TV}\right|+\left|d_{RV}\right|\\ =\sqrt{65}+2\sqrt{2}+\sqrt{122}+\sqrt{277}\text{units}\\ =38.6\text{\hspace{0.17em}units}\left(\text{Approx.}\right)\end{array}$

Step 3. Conclusion.

The perimeter of quadrilateral (RSTV) is approximately $38.6 \text{units}$.