1. **Problem statement:** We have a right triangle with one leg labeled $x$, the other leg labeled $7$, and the hypotenuse is a side of an equilateral triangle. We need to find $x$ in simplest radical form with a rational denominator. 2. **Key fact:** In an equilateral triangle, all sides are equal. The hypotenuse of the right triangle is one side of the equilateral triangle, so its length is equal to the side length of the equilateral triangle. 3. **Using the Pythagorean theorem:** For the right triangle, $$x^2 + 7^2 = \text{(hypotenuse)}^2$$ Let the hypotenuse be $s$, the side length of the equilateral triangle. 4. **Equilateral triangle property:** The altitude of an equilateral triangle of side $s$ is $$h = \frac{\sqrt{3}}{2} s$$ 5. **Relating the right triangle to the equilateral triangle:** The side labeled $7$ is the altitude of the right triangle, so $$7 = \frac{\sqrt{3}}{2} s$$ 6. **Solve for $s$:** $$s = \frac{7}{\frac{\sqrt{3}}{2}} = 7 \times \frac{2}{\sqrt{3}} = \frac{14}{\sqrt{3}}$$ 7. **Rationalize the denominator:** $$s = \frac{14}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{14 \sqrt{3}}{3}$$ 8. **Use Pythagorean theorem to find $x$:** $$x^2 + 7^2 = s^2$$ $$x^2 + 49 = \left(\frac{14 \sqrt{3}}{3}\right)^2 = \frac{196 \times 3}{9} = \frac{588}{9} = \frac{196}{3}$$ 9. **Simplify for $x^2$:** $$x^2 = \frac{196}{3} - 49 = \frac{196}{3} - \frac{147}{3} = \frac{49}{3}$$ 10. **Find $x$:** $$x = \sqrt{\frac{49}{3}} = \frac{7}{\sqrt{3}}$$ 11. **Rationalize denominator for $x$:** $$x = \frac{7}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{7 \sqrt{3}}{3}$$ **Final answer:** $$x = \frac{7 \sqrt{3}}{3}$$