Question

If an equilateral triangle ABC is drawn as shown below with A(0,0) and a side of 4 units then the other two vertices of the triangle are___

• A. $B=(2,2\sqrt{3})$
  C =(4,0)
• B. $C=(2,2)$
  B =(4,0)
• C. $C=(2,2\sqrt{3})$
  B =(4,0)
• D. $B=(2,2)$
  C =(4,0)

Answer 1

The correct option is C

$C=(2,2\sqrt{3})$
B =(4,0)

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 =4 $\therefore B=(4,0)$

$\because$ Given AB =4

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

=2 units.

$\therefore x$ coordinate of C is 4.

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

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

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

$\Rightarrow D{C}^2=16-4$

=12

$\Rightarrow DC=2\sqrt{3}$

$\therefore$ Coordinates of $C=(2,2\sqrt{3})$

$\therefore$ Remaining vertices are B(4,0) and $C=(2,2\sqrt{3})$