Answer:
Step-by-step explanation:
If BD is altitude then AC is the base.
The length of AC is:
The length of BD is:
The area is:
Calculate distances
[tex]\boxed{\sf \sqrt{(x_2-x_1)^2+(y_2-y_1)^2}}[/tex]
AC:-
[tex]\\ \sf\longmapsto \sqrt{(0+6)^2+(6-0)^2}[/tex]
[tex]\\ \sf\longmapsto \sqrt{6^2+6^2}[/tex]
[tex]\\ \sf\longmapsto \sqrt{36+36}[/tex]
[tex]\\ \sf\longmapsto \sqrt{72}[/tex]
[tex]\\ \sf\longmapsto 6\sqrt{2}[/tex]
[tex]\\ \sf\longmapsto \sqrt{(-3-0)^2+(3-0)^2}[/tex]
[tex]\\ \sf\longmapsto \sqrt{(-3)^2+(3)^2}[/tex]
[tex]\\ \sf\longmapsto \sqrt{9+9}[/tex]
[tex]\\ \sf\longmapsto \sqrt{18}[/tex]
[tex]\\ \sf\longmapsto 3\sqrt{2}[/tex]
BD is height and AC is base.
[tex]\\ \sf\longmapsto Area=\dfrac{1}{2}Base\times Height[/tex]
[tex]\\ \sf\longmapsto Area=\dfrac{1}{2}(6\sqrt{2})(3\sqrt{2})[/tex]
[tex]\\ \sf\longmapsto Area=18units^2[/tex]
