In the given figure, $\triangle ABC$ is an equilateral triangle. AD is a median of the triangle joining the points $A\left(0, \frac{5\sqrt{3}}{2}\right)$, D(0, 0). Points B and C are (in same order) :
Generated by claude-sonnet-4-6 · 2026-06-15 10:30 · grounding rag
Model Answer
Option A: $(-5, 0),\ (5, 0)$
AD = $\sqrt{(0-0)^2 + \left(\frac{5\sqrt{3}}{2}-0\right)^2} = \frac{5\sqrt{3}}{2}$. In an equilateral triangle, median = $\frac{\sqrt{3}}{2} \times$ side, so side = 5. D is mid-point of BC, so BD = DC = $\frac{5}{2}$...
Correction using full side: Side = 5, so B = $(-5/2 \times 2,\ 0)$...
Using median formula: side $a$, median $= \frac{\sqrt{3}}{2}a = \frac{5\sqrt{3}}{2}$, so $a = 5$. BD $= \frac{a}{2} = \frac{5}{2}$. So B $= (-\frac{5}{2}, 0)$, C $= (\frac{5}{2}, 0)$.
The answer is B: $\left(-\dfrac{5}{2},\ 0\right),\ \left(\dfrac{5}{2},\ 0\right)$
Explanation
- The median length AD $= \frac{5\sqrt{3}}{2}$. For an equilateral triangle, median $= \frac{\sqrt{3}}{2} \times$ side, giving side $= 5$.
- D is the midpoint of BC on the x-axis, so BD $= $ DC $= \frac{5}{2}$.
- Since D is at origin: B $= \left(-\frac{5}{2}, 0\right)$, C $= \left(\frac{5}{2}, 0\right)$.
- A common trap is choosing $(-5, 0)$ and $(5, 0)$; that would make BC $= 10$, not 5. Always use the median–side relation carefully.