Correct derivation: The height $ h $ of an equilateral triangle is $ rac{\sqrt{3}}{2}s $. The centroid (which coincides with the circumcenter) is at $ rac{2}{3} $ of the height from a vertex, so:

["Understanding the Correct Derivation of the Height and Centroid of an Equilateral Triangle", "The equilateral triangle is a fundamental shape in geometry, cherished for its symmetry and balanced proportions. Whether used in architecture, design, or advanced mathematics, understanding its core properties—especially the height and the location of its centroid—is essential. One commonly referenced fact is that the height $ h $ of an equilateral triangle with side length $ s $ is $ h = \dfrac{\sqrt{3}}{2}s $. But how is this formula correctly derived, and what role does the centroid play in this? Let’s explore the accurate geometric derivation step by step.", "### The Geometry of an Equilateral Triangle", "An equilateral triangle has all three sides equal and all interior angles measuring $ 60^\circ $. To derive the height, we begin by drawing a perpendicular line from one vertex to the opposite side—thisbisects both the vertex angle and the base side, splitting the equilateral triangle into two identical right-angled triangles.", "### Deriving the Height: Step-by-Step", "Consider triangle $ \ riangle ABC $ with side length $ s $, and let $ D $ be the foot of the perpendicular from vertex $ A $ to side $ BC $. Since $ \ riangle ABC $ is equilateral:", "- $ AB = AC = BC = s $\n- The altitude $ AD $ splits $ BC $ into two equal segments: $ BD = DC = \dfrac{s}{2} $", "Using the Pythagorean theorem in right triangle $ \ riangle ABD $:", "[\nAB^2 = AD^2 + BD^2\n]", "Substitute known values:", "[\ns^2 = h^2 + \left(\dfrac{s}{2}\right)^2\n]", "[\ns^2 = h^2 + \dfrac{s^2}{4}\n]", "Isolate $ h^2 $:", "[\nh^2 = s^2 - \dfrac{s^2}{4} = \dfrac{4s^2}{4} - \dfrac{s^2}{4} = \dfrac{3s^2}{4}\n]", "Take the positive square root (since height is a positive length):", "[\nh = \sqrt{\dfrac{3s^2}{4}} = \dfrac{\sqrt{3}}{2}s\n]", "Thus, the height of the equilateral triangle is correctly given by:", "[\n\boxed{h = \dfrac{\sqrt{3}}{2}s}\n]", "### The Centroid: Location and Significance", "The centroid of any triangle—defined as the point where the medians intersect—is also the center of mass if the triangle is treated as a uniform lamina. For the equilateral triangle, the centroid lies at $ \dfrac{2}{3} $ of the height from the vertex opposite the base.", "To locate the centroid $ G $ in the same triangle, consider that the medians each connect a vertex to the midpoint of the opposite side. The centroid divides each median in a $ 2:1 $ ratio. Since the height $ h $ extends from the vertex vertically to $ D $ on $ BC $, the distance from vertex $ A $ to centroid $ G $ is:", "[\nAG = \dfrac{2}{3}h = \dfrac{2}{3} \cdot \dfrac{\sqrt{3}}{2}s = \dfrac{\sqrt{3}}{3}s\n]", "The distance from the centroid $ G $ to the base $ BC $ is therefore:", "[\nGB = h - AG = \dfrac{\sqrt{3}}{2}s - \dfrac{\sqrt{3}}{3}s = \left( \dfrac{3 - 2}{6} \right)\sqrt{3}s = \dfrac{\sqrt{3}}{6}s\n]", "This confirms that $ G $ lies exactly $ \dfrac{2}{3} $ of the height from the vertex and $ \dfrac{1}{3} $ from the base—aligning with the geometric fact.", "### Summary", "- The height $ h $ of an equilateral triangle with side $ s $ is accurately derived using the Pythagorean theorem:\n [\n h = \dfrac{\sqrt{3}}{2}s\n ]\n- The centroid (also circumcenter) lies at $ \dfrac{2}{3} $ of the height from any vertex, and $ \dfrac{1}{3} $ from the base.\n- Understanding these derivations deepens geometric insight and supports applications in art, engineering, and trigonometry.", "Proper mathematical derivation ensures clarity, accuracy, and confidence when applying equilateral triangle properties—essential knowledge for students, educators, and professionals alike.", "---", "Keywords: equilateral triangle height formula, derivation height $ h = \frac{\sqrt{3}}{2}s $, centroid of equilateral triangle, geometric derivation, triangle geometry, centroid location, Pythagorean theorem, right triangle in equilateral triangle, triangle centroid properties."]