Equilateral Side 3E7Ad9

1. **Problem Statement:** We have an equilateral triangle with a perpendicular segment of length $5$ from the left vertex to the right side, meeting it at a right angle. The upper portion of the right side is labeled $x$. We need to find the length of side $x$ in simplest radical form with a rational denominator. 2. **Understanding the problem:** In an equilateral triangle, all sides are equal, and all angles are $60^\circ$. The perpendicular from a vertex to the opposite side bisects that side. 3. **Set up the problem:** Let the side length of the equilateral triangle be $s$. The perpendicular from the left vertex to the right side creates two segments on the right side: one of length $x$ (upper portion) and the other of length $s - x$ (lower portion). 4. **Use properties of equilateral triangles:** The perpendicular from a vertex to the opposite side also acts as an altitude. The altitude $h$ of an equilateral triangle with side length $s$ is given by: $$h = \frac{\sqrt{3}}{2} s$$ 5. **Analyze the right triangle formed:** The perpendicular segment of length $5$ is the altitude from the left vertex to the right side. Since the triangle is equilateral, the altitude also bisects the side it meets. Therefore, the right side is divided into two equal parts: $$x = \frac{s}{2}$$ 6. **Relate the altitude and side length:** Given the altitude $h = 5$, we have: $$5 = \frac{\sqrt{3}}{2} s$$ 7. **Solve for $s$:** Multiply both sides by $\frac{2}{\sqrt{3}}$: $$s = 5 \times \frac{2}{\sqrt{3}} = \frac{10}{\sqrt{3}}$$ 8. **Rationalize the denominator:** $$s = \frac{10}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}} = \frac{10 \sqrt{3}}{3}$$ 9. **Find $x$:** Since $x = \frac{s}{2}$: $$x = \frac{1}{2} \times \frac{10 \sqrt{3}}{3} = \frac{10 \sqrt{3}}{6} = \frac{5 \sqrt{3}}{3}$$ **Final answer:** $$x = \frac{5 \sqrt{3}}{3}$$