Question

Find the perimeter and area of triangle shown below.

Short answer

The perimeter of triangle is$7\sqrt{2}+\sqrt{58}\text{units}$.The area of given triangle is$10\text{square units}$.

Step-by-step solution

Step 1. Write down the given information.

The given triangle is shown below having coordinates $A\left(-3,-2\right)\text{,B}\left(-1,-4\right)\text{and}C\left(4,1\right)$.

Step 2. Concept used.

If a line segment has end-points at $\left(x_1,y_1\right)\text{and}\left(x_2,y_2\right)$ then the distance $\left(d\right)$ between these points is given as:

$\left|d\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}....\left(1\right)$

Step 3. Calculation.

The distance $\left(d\right)$ between the pair of points with given coordinates $A\left(-3,-2\right)\text{,B}\left(-1,-4\right)\text{and}C\left(4,1\right)$ is evaluated using the distance formula.

$\begin{array}{c}\left|d_{AB}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(-1+3\right)}^2+{\left(-4+2\right)}^2}\\ =\sqrt{8}\\ =2\sqrt{2}\text{units}\end{array}$$\begin{array}{c}\left|d_{BC}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(4+1\right)}^2+{\left(1+4\right)}^2}\\ =\sqrt{50}\\ =5\sqrt{2}\text{units}\end{array}$$\begin{array}{c}\left|d_{BC}\right|=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2}\\ =\sqrt{{\left(4+1\right)}^2+{\left(1+4\right)}^2}\\ =\sqrt{50}\\ =5\sqrt{2}\text{units}\end{array}$

Now the perimeter of triangle is the sum of its three sides. Therefore perimeter (P) of triangle (ABC) is evaluated as:

$\begin{array}{c}P=\left|d_{AB}\right|+\left|d_{BC}\right|+\left|d_{CA}\right|\\ =2\sqrt{2}+5\sqrt{2}+\sqrt{58}\\ =7\sqrt{2}+\sqrt{58}\text{units}\end{array}$

And area of triangle with the given vertices $A\left(-3,-2\right)\text{,B}\left(-1,-4\right)\text{and}C\left(4,1\right)$ is evaluated as:

$\begin{array}{c}\text{Area}\left(A\right)=\frac{1}{2}\left[x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right]\\ =\frac{1}{2}\left[-3\left(-4-1\right)-1\left(1+2\right)+4\left(-2+4\right)\right]\\ =\frac{1}{2}\left[15-3+8\right]\\ =10\text{squareunits}\end{array}$

Step 4. Conclusion.

The perimeter of triangle is $7\sqrt{2}+\sqrt{58}\text{units}$. The area of given triangle is $10\text{square units}$.