Chapter 11.6, Problem 28PPS

To determine

To Find: The area of triangle with given vertices.

Answer to Problem 28PPS

  $A=12.375sq.unit$

Explanation of Solution

Given information:

The given vertices of triangle are $E\left(-2.5,2\right),F\left(3,-2.5\right)\&G\left(3,2\right)$

Concept and Formula Used:

Area of triangle $A=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$

Calculation:

The given vertices of triangle are $E\left(-2.5,2\right),F\left(3,-2.5\right)\&G\left(3,2\right)$

Using the formula

  $\begin{array}{l}A=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|\\ \text{where }\left(x_1,y_1\right)=\left(-2.5,2\right),\left(x_2,y_2\right)=\left(3,-2.5\right)\&\left(x_3,y_3\right)=\left(3,2\right)\\ A=\frac{1}{2}\left|\left(-2.5\right)\left(-2.5-2\right)+3\left(2-2\right)+3\left(2-\left(-2.5\right)\right)\right|\\ A=\frac{1}{2}\left|\left(-2.5\right)\left(-4.5\right)+3\left(0\right)+3\left(2+2.5\right)\right|\\ A=\frac{1}{2}\left|11.25+0+13.5\right|\\ A=\frac{1}{2}\left|24.75\right|\\ A=12.375sq.unit\end{array}$

Conclusion:

  $A=12.375sq.unit$