Question

triangle JKL with vertices J(-6,7), K(-4,6), and L(-3,9) is drawn inside a rectangle. what is the area in square units of triangle JKL

Answer 1

One way to find the area of ​​the triangle inscribed in the square is

$A_T=A_S-(A_1+A_2+A_3)$

Where

The area of ​​a triangle is given by the formula

$\begin{gathered} A=\frac{b\cdot h}{2} \\ \text{ Where b is the base and} \\ h\text{ is the height of the triangle} \end{gathered}$

So, you have

A1

$\begin{gathered} A_1=\frac{b\cdot h}{2} \\ A_1=\frac{2units\cdot3\text{ units}}{2} \\ A_1=3\text{ units}^2 \end{gathered}$

A2

$\begin{gathered} A_2=\frac{b\cdot h}{2} \\ A_2=\frac{1unit\cdot3\text{ units}}{2} \\ A_2=\frac{3}{2}\text{ units}^2 \\ \text{ or} \\ A_2=1.5\text{ units}^2 \end{gathered}$

A3

$\begin{gathered} A_3=\frac{b\cdot h}{2} \\ A_3=\frac{2units\cdot1\text{ units}}{2} \\ A_3=1\text{ unit}^2 \end{gathered}$

The area of ​​a square is given by the formula

$A=b\cdot h$

AS

$\begin{gathered} A_S=3\text{ units}\cdot3\text{ units} \\ A_S=9\text{ units}^2 \end{gathered}$

Then,

$\begin{gathered} A_T=A_S-(A_1+A_2+A_3) \\ \text{ Replacing} \\ A_T=9units^2-(3units^2+1.5units^2_{}+1unit^2) \\ A_T=9units^2-5.5units^2 \\ A_T=5.5\text{ units}^2 \end{gathered}$

Therefore, the area of the triangle JKL is 5.5 square units.