Question

If an equilateral triangle OAB is drawn as shown in below figure with side 6 units then which of the following is/are vertices of the triangle.

• A. $B=(3\sqrt{3},3)$
• B. $B=(3,3\sqrt{3})$
• C. A=(6,0)
• D. Can't be found

Answer 1

Answer: (B), (C)

The correct options are
B

$B=(3,3\sqrt{3})$

C

A=(6,0)

O is the origin.

$\therefore$ Coordinates of O=(0,0)

A is 6 units from O and is an x-axis.

$\therefore$ Coordinate of A=(6,0)

Now we draw $BC\perp$ to OA.

$\because OAB$ is equilateral triangle so C will be the midpoint of OA.

$\therefore OC=3$

$\therefore$ X-coordinate of B =3

Now OCB is right angled triangle

$\therefore O{B}^2=O{C}^2+B{C}^2$

$\Rightarrow B{C}^2=6^2-3^2$

=36-9

=27

$\therefore BC=3\sqrt{3}$

$\therefore$ y -coordinate of $B=3\sqrt{3}$

$\therefore B=(3,3\sqrt{3})$