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The base of an equilateral triangle with side 2a lies along the y axis such that the midpoint of the base is at the origin.find vertices of triangle? chapter 10 exercise10.1 2nd question in plus two state syllabus |
| Let ABC be the given equilateral triangle with side 2a. Accordingly, AB = BC = CA = 2a Assume that base BC lies along the y-axis such that the mid-point of BC is at the origin. i.e., BO = OC = a, where O is the origin. Now, it is clear that the coordinates of point C are (0, a), while the coordinates of point B are (0, -a). It is known that the line joining a vertex of an equilateral triangle with the mid-point of its opposite side is perpendicular. Hence, vertex A lies on the y-axis.  On applying Pythagoras theorem to triangleAOC, we obtain (AC)2 = (OA)2 + (OC)2 (2a)2 = (OA)2 + a2 4a2 - a2 = (OA)2 ie, (OA)2 = 3a2  |
