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Class 10 › Mathematics › Coordinate Geometry
CBSE2026Class Class 10 · Mathematics1 Marks · Mcq✅ Verified
In the given figure, $\triangle ABC$ is an equilateral triangle. $AD$ is a median of the triangle joining the points $A\left(0, \dfrac{5\sqrt{3}}{2}\right)$ and $D(0,0)$. Points $B$ and $C$ are (in same order) :
A. $(-5, 0),\ (5, 0)$
B. $\left(-\dfrac{5}{2}, 0\right),\ \left(\dfrac{5}{2}, 0\right)$
C. $(-10, 0),\ (10, 0)$
D. $(-5\sqrt{3}, 0),\ (5\sqrt{3}, 0)$
✅ Answer & Solution
✅ Correct Answer: B
Step 1: $AD$ is the median (also the altitude) of the equilateral triangle.
$$AD = \frac{5\sqrt{3}}{2}$$
Step 2: For an equilateral triangle of side $a$, altitude $= \dfrac{\sqrt{3}}{2}a$.
$$\frac{\sqrt{3}}{2}a = \frac{5\sqrt{3}}{2} \Rightarrow a = 5$$
Step 3: $D(0,0)$ is the midpoint of $BC$ and $BC$ lies along the $x$-axis, so
$$BD = DC = \frac{a}{2} = \frac{5}{2}$$
Step 4: Hence $B\left(-\dfrac{5}{2}, 0\right)$ and $C\left(\dfrac{5}{2}, 0\right)$.
