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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.
Solution
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
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