Question

The base of an equilateral triangle with side 2$a$ lies along the $y$ -axis such that the mid-point of the base is at the origin. Find vertices of the triangle.

Answer 1

Given: The midpoint of the base of the triangle is at origin, and the length of the base is $2a$ (on y-axis).

Hence we can say that the vertices of the base are $(0,a)$ and $(0,-a)$

Height of an equilateral triangle $=$ side$\times \frac{\sqrt{3}}{2}$

$h=2a\times \frac{\sqrt{3}}{2}$

$=a\sqrt{3}$

Let us mark $a\sqrt{3}$ units perpendicular to y-axis from the origin, i.e., on the x-axis.

$\therefore$ The third vertex is $(a\sqrt{3},0)$ or $(-a\sqrt{3},0)$

Hence there are two triangles possible.

Vertices of first triangle $\Rightarrow (0,a)(0,-a),(a\sqrt{3},0)$

Vertices of second triangle $\Rightarrow (0,a)(0,-a),(-a\sqrt{3},0)$