Q. 32 - Geometry

Question

 In Exercises 27-32 find and then compare lengths of segments.

 Show that the triangle with vertices D(0, 0), E(3, 1), and F(-2, 6) is a right triangle, then find the area of the triangle.

Step-by-Step Solution

Answer

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

Step-1 – Given

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

Step-2 – To determine

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.

Step-3 – Calculation

We will find the lengths DE, DF and EF using 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.