Chapter 13.1, Problem 32WE

To determine

To show that the triangle with vertices DEF is a right triangle

Explanation of Solution

Given information:

The triangle has vertices is

  $D\left(0,0\right),E\left(3,1\right),\text{ and }F\left(-2,6\right)$

Proof:

The triangle DEF is a right triangle can be show as

Use distance formula to show the triangle is a right triangle

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

First find DE, DF and EF

Therefore,

  $\begin{array}{l}DE=\sqrt{{\left(3-0\right)}^2+{\left(1-0\right)}^2}\left[\text{since }D\left(0,0\right)\text{ and }E\left(3,1\right)\right]\\ DE=\sqrt{{\left(3\right)}^2+{\left(1\right)}^2}\\ DE=\sqrt{9+1}\\ DE=\sqrt{10}\end{array}$

  $\begin{array}{l}DF=\sqrt{{\left(-2-0\right)}^2+{\left(6-0\right)}^2}\left[\text{since }D\left(0,0\right)\text{ and }F\left(-2,6\right)\right]\\ DF=\sqrt{{\left(-2\right)}^2+{\left(6\right)}^2}\\ DF=\sqrt{4+36}\\ DF=\sqrt{40}\end{array}$

  $\begin{array}{l}EF=\sqrt{{\left(-2-3\right)}^2+{\left(6-1\right)}^2}\left[\text{since }E\left(3,1\right)\text{ and }F\left(-2,6\right)\right]\\ EF=\sqrt{{\left(-5\right)}^2+{\left(5\right)}^2}\\ EF=\sqrt{25+25}\\ EF=\sqrt{50}\\ EF=5\sqrt{2}\end{array}$

The right triangle states that the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides

  $\begin{array}{l}{\left(EF\right)}^2={\left(DE\right)}^2+{\left(DF\right)}^2\\ {\left(EF\right)}^2={\left(\sqrt{10}\right)}^2+{\left(\sqrt{40}\right)}^2\\ {\left(EF\right)}^2=10+40\\ {\left(EF\right)}^2=50\end{array}$

  $EF=50$ which is equal to $\left(DE+DF\right)$

So, the triangle is a right triangle

Area of a right triangle is

  $\begin{array}{l}A=\frac{ab}{2}\left[\begin{array}{l}A=\text{area}\\ a=\text{base}\\ b=\text{height}\end{array}\right]\\ A=\frac{\left(DE\right)\left(DF\right)}{2}\left[\begin{array}{l}a=DE\\ b=DF\end{array}\right]\\ A=\frac{\left(\sqrt{10}\right)\left(\sqrt{40}\right)}{2}\\ A=\frac{\sqrt{400}}{2}\\ A=\frac{20}{2}\\ A=10\end{array}$

Hence,

The triangle is a right triangle and the area is 10.