What is the area, in square units, of triangle $ABC$ in the figure shown if points $A$, $B$, $C$ and $D$ are coplanar, angle $D$ is a right angle, $AC = 13$, $AB = 15$ and $DC = 5$?
Seeing that triangle ACD is a 5-12-13 right triangle, AD=12. Then using Pythagorean Theorem, we can calculate BD to be BD=[tex]\sqrt{15^2-12^2}=\sqrt{3^2(5^2-4^2)}=3\sqrt{25-16}=3\sqrt{9}=3 \cdot 3 = 9$[/tex]. Thus, the area of triangle ABD is [tex]$\frac{1}{2} \cdot 12 \cdot 9=6 \cdot 9=54 \text{ sq units}$[/tex] and the area of triangle ACD is [tex]$\frac{1}{2} \cdot 12 \cdot 5=6 \cdot 5=30 \text{ sq units}$[/tex]. The area of triangle ABC is the difference between the two areas: [tex]$54 \text{sq units} - 30 \text{sq units} = \boxed{24} \text{sq units}$.[/tex]
