Question

If an equilateral triangle ABC is drawn as shown below with A(0,0) and a side of 10 units then the other two vertices of the triangle are___

• A. $B=(5,5\sqrt{3})$
  C =(10,0)
• B. B =(5,5)
  C =(10,0)
• C. C =(5,5)
  B =(10,0)
• D. $C=(5,5\sqrt{3})$
  B =(10,0)

Answer 1

The correct option is D

$C=(5,5\sqrt{3})$
B =(10,0)

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 =10 $\therefore B=(10,0)$

$\because$ Given AB =10

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

=5 units.

$\therefore$ x coordinate of C is 5.

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

$\therefore A{C}^2=A{D}^2+D{C}^2$ ( Since its an equilateral triangle, AB = BC = AC = 10 )

$\Rightarrow {10}^2=5^2+D{C}^2$

$\Rightarrow D{C}^2=100-25$

=75

$\Rightarrow DC=5\sqrt{3}$

$\therefore$ Coordinates of $C=(5,5\sqrt{3})$

$\therefore$ Remaining vertices are B(10,0) and $C=(5,5\sqrt{3})$