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Question
The base PQ of two equilateral triangles PQR and PQR' with side 2a lies along y-axis such that the mid-point of PQ is at the origin. Find the coordinates of the vertices R and R' of the triangles.
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Solution
In an equilateral triangle, the height ‘h’ is given by
`h =(sqrt3`("Side of the equilateral triangle"))/2`
Here it is given that 'PQ' forms the base of two equilateral triangles whose side measures '2a' units.
The height of these two equilateral triangles has got to be
`h = (sqrt3("Side of the equilateral triangle"))/2`
`= (sqrt3(2a))/2`
`h = asqrt3`
In an equilateral triangle, the height drawn from one vertex meets the midpoint of the side opposite this vertex.
So here we have ‘PQ’ being the base lying along the y-axis with its midpoint at the origin, that is at (0, 0)
So the vertices ‘R’ and ‘R’’ will lie perpendicularly to the y-axis on either side of the origin at a distance of `asqrt3` units
Hence the co-ordinates of ‘R’ and ‘R’’ are
`R(asqrt3,0)`
`R'(-a sqrt3, 0)`
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