One way to find the area of the triangle inscribed in the square is
[tex]A_T=A_S-(A_1+A_2+A_3)[/tex]
Where
The area of a triangle is given by the formula
[tex]\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}[/tex]
So, you have
A1
[tex]\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}[/tex]
A2
[tex]\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}[/tex]
A3
[tex]\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}[/tex]
The area of a square is given by the formula
[tex]A=b\cdot h[/tex]
AS
[tex]\begin{gathered} A_S=3\text{ units}\cdot3\text{ units} \\ A_S=9\text{ units}^2 \end{gathered}[/tex]
Then,
[tex]\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}[/tex]
Therefore, the area of the triangle JKL is 5.5 square units.
