Question

If an equilateral triangle ABC is drawn as shown below with A(0,0) and a side of 8 units then find the other two vertices of the triangle.

Answer 1

In equilateral triangle ABC draw $CD\perp$ to AB, then D will be the mid-point of AB.

$\because B$ is on X-axis and AB =8 $\therefore B=(8,0)$

$\because$ Given AB =8

$\Rightarrow AD=\frac{1}{2}\times AB$

=4 units.

$\therefore$ x coordinate of C is 4.

Now ADC makes right angled triangle with $\angle ADC={90}^{\circ }$

$\therefore A{C}^2=A{D}^2+D{C}^2$

$\Rightarrow 8^2=4^2+D{C}^2$

$\Rightarrow D{C}^2=64-16$

=48

$\Rightarrow DC=4\sqrt{3}$

$\therefore$ Coordinates of $C=(4,4\sqrt{3})$

$\therefore$ Remaining vertices are B(8,0) and $C=(4,4\sqrt{3})$