Finding Length DB in a Complex Triangle Configuration: 6-Unit Height Problem

This problem involves two connected right triangles. First, use triangle ABC to find CB, then use triangle CDB to find DB. Each triangle requires its own application of $a^2 + b^2 = c^2$.

The hypotenuse is always the longest side and sits opposite the right angle. In the diagram, look for the right angle symbols (small squares) to identify which side is opposite to them.

You don't need to convert to decimals! Keep exact values like $2\sqrt{11}$ throughout your work. When you square it: $(2\sqrt{11})^2 = 4 \times 11 = 44$.

Great observation! $\sqrt{8} = \sqrt{4 \times 2} = 2\sqrt{2}$, but for this problem, leaving it as $\sqrt{8}$ works perfectly fine since we square it in the next step.

Substitute back into triangle CDB: $CD^2 + DB^2 = CB^2$ becomes $2^2 + 2^2 = (\sqrt{8})^2$, which gives $4 + 4 = 8$ ✓