91Ó°ÊÓ

Given that the triangle has vertices:$D\left(0,0\right),E\left(3,1\right),\text{and}F\left(−2,6\right)$

We have to show that the triangle with vertices$D\left(0,0\right),E\left(3,1\right),\text{and}F\left(−2,6\right)$ is a right triangle. Then we find its area.

We will find the lengths DE, DF and EFusing the distance formula:

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

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

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

For a right triangle the lengths must follow the Pythagorean theorem. Here, hypotenuse = longest side = EF.

Two legs = DE and DF.

So,

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

EF² = 50 which is equal to .

It means, the triangle is a right triangle.

Hence, the area of the right triangle is:

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

So, the triangle is a right triangle and the area is 10 square unit.