Question

Show that the points O(0, 0), `A(3, sqrt(3))` and `B(3, -sqrt(3))` are the vertices of an equilateral triangle. Find the area of this triangle.

Answer 1

The given points are O(0,0), A`( 3,sqrt(3)) and B (3,-sqrt(3))` .

`OA = sqrt((3-0)^2 +{ (sqrt(3)) -0}^2) = sqrt((3)^2 +(sqrt(3))^2) = sqrt(9+3) = sqrt(12) = 2sqrt(3) `units

`AB = sqrt((3-3)^2 +(-sqrt(3)-sqrt(3))^2) = sqrt((0) + (2 sqrt(3))^2) = sqrt(4(3)) = sqrt(12) =2sqrt(3)` units

`OB = sqrt((3-0)^2 + (-sqrt(3) -0)^2) = sqrt((3)^2 +(sqrt(3)^2) = sqrt(9+3) = sqrt(12) = 2 sqrt(3)` units 

Therefore, `OA =AB = OB = 2 sqrt(3) `  units

Thus, the points O(0,0), A`( 3,sqrt(3)) and B (3,-sqrt(3))`  are the vertices of an equilateral triangle Also, the area of the triangle `OAB = sqrt(3)/4 xx (" side")^2`

`=sqrt(3)/4 xx(2 sqrt(3) )^2`

`= sqrt(3)/4 xx 12`

`= 3 sqrt(3) `   square units