ABC is a triangle. AB = 5 cm, AC = $$\sqrt {41} $$  cm and BC = 8 cm. AD is perpendicular to BC. What is the area (in cm²) of triangle ABD?

$$\eqalign{ & {\text{Let }}\angle ACB = \alpha \cr & \angle DAC = {90^ \circ } - \alpha \cr & \cos \alpha = \frac{{{{\left( {\sqrt {41} } \right)}^2} + {{\left( 8 \right)}^2} - {{\left( 5 \right)}^2}}}{{2\sqrt {41} \times 8}} = \frac{{DC}}{{\sqrt {41} }} \cr & \frac{{41 + 64 - 25}}{{16\sqrt {41} }} = \frac{{DC}}{{\sqrt {41} }} \cr & \frac{{80}}{{16}} = DC \cr & DC = 5{\text{ cm}} \cr & BD = 8 - 5 = 3{\text{ cm}} \cr} $$

$${\text{Area of }}\Delta ABD = \frac{1}{2} \times 4 \times 3 = 6{\text{ c}}{{\text{m}}^2}$$