1. **State the problem:** We have a right triangle ABC with a right angle at C, sides AC = $2\sqrt{6}$, BC = $3\sqrt{3}$, and angle at C is 45°. We drop a height from A to BC, meeting BC at D, and want to find the length $x = AD$. 2. **Use the properties of right triangles and heights:** The height from the right angle vertex to the hypotenuse splits the triangle into two smaller right triangles similar to the original. 3. **Given ratios:** - $CD : 2\sqrt{6} = \sqrt{3} : 2$ - $2CD = 2\sqrt{6}$ - $CD = 3\sqrt{2}$ - $AD : 2\sqrt{6} = \sqrt{}$ (incomplete, but we will use the known values) 4. **Calculate $CD$ from the given:** From $2CD = 2\sqrt{6}$, dividing both sides by 2: $$ \cancel{2}CD = \cancel{2}\sqrt{6} \implies CD = \sqrt{6} $$ But the user also states $CD = 3\sqrt{2}$, so we clarify the consistent value is $CD = 3\sqrt{2}$. 5. **Use Pythagoras theorem in triangle ADC:** Since $AD$ is the height, triangle ADC is right angled at D. 6. **Calculate $AD$ using the relation:** Using the similarity or the Pythagorean theorem, and given $AC = 2\sqrt{6}$ and $CD = 3\sqrt{2}$, we find $AD$: Since $AD$ is the height from A to BC, and $CD$ is part of BC, we use the formula for height in right triangle: $$ AD = \frac{AC \times BC}{AB} $$ But $AB$ is the hypotenuse, which can be found by: $$ AB = \sqrt{AC^2 + BC^2} = \sqrt{(2\sqrt{6})^2 + (3\sqrt{3})^2} = \sqrt{4 \times 6 + 9 \times 3} = \sqrt{24 + 27} = \sqrt{51} $$ 7. **Calculate $AD$:** $$ AD = \frac{AC \times BC}{AB} = \frac{2\sqrt{6} \times 3\sqrt{3}}{\sqrt{51}} = \frac{6 \sqrt{18}}{\sqrt{51}} = \frac{6 \times 3\sqrt{2}}{\sqrt{51}} = \frac{18 \sqrt{2}}{\sqrt{51}} $$ 8. **Simplify the denominator:** $$ AD = \frac{18 \sqrt{2}}{\sqrt{51}} = \frac{18 \sqrt{2}}{\sqrt{51}} \times \frac{\sqrt{51}}{\sqrt{51}} = \frac{18 \sqrt{102}}{51} = \frac{6 \sqrt{102}}{17} $$ **Final answer:** $$ x = AD = \frac{6 \sqrt{102}}{17} $$