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Q. No 2 :
The base of an equilateral triangle with side 2a lies along the y-axis such that the mid point of the base is at origin.Find the vertices of triangle.
Sol :
BC is the base of ∆ABC such that
BO=OC=a
AB = AC = 2a
Now AO2
= AB2
-
BO2
= 31/2
a
∴ The point A lying on x-axis is (31/2 a, 0).
The point B is (0, a) and the point C is (0, -a). When A lies on the left of y-axis, the vertices of the triangle are (-31/2 a, 0), (0,a) and (0, -a).
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