Show that the following points taken in order form the vertices of a rhombus.
(0, 0), (3, 4), (0, 8) and ( - 3, 4)
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Answered by
7
Let us consider the points as A(0,0) ; B(3,4) ; C(0,8) ; D(-3,4)
The distance b/w AB = √(3-0)²+(4-0)² = √25 = 5
The distance b/w BC = √(0-3)²+(8-4)² = √9+16 = 5
The distance b/w CD = √(-3-0)²+(4-8)² = √9+16 = 5
The distance b/w DA = √(-3-0)²+(4-0)² = √25 = 5
The distance b/w AC =√(0-0)²+(8-0)² = 8
The distance b/w BD = √(-3-3)²+(4-4)² = 6
As the distance of all sides are equal and diagonals are not equal.
Therefore the given points represents Rhombus
The distance b/w AB = √(3-0)²+(4-0)² = √25 = 5
The distance b/w BC = √(0-3)²+(8-4)² = √9+16 = 5
The distance b/w CD = √(-3-0)²+(4-8)² = √9+16 = 5
The distance b/w DA = √(-3-0)²+(4-0)² = √25 = 5
The distance b/w AC =√(0-0)²+(8-0)² = 8
The distance b/w BD = √(-3-3)²+(4-4)² = 6
As the distance of all sides are equal and diagonals are not equal.
Therefore the given points represents Rhombus
Answered by
3
Indirectly U r asking to prove that this points represent a rhombus or not
As we all know that rhombus has all its sides equal.So lets name the points as A(0,0),B(3,4),C(0,8),D(-3,4)
By taking out the difference from this points, we get,
AB=(0-3)+(0-4)=1
BC=(0-3)+(8-4)=1
CD=(-3-0)+(4-8)=1
AD=(-3-0)+(4-0)=1
we can see AB=CD=BC=AD This mean that the points are the edges of a rhombus.
As we all know that rhombus has all its sides equal.So lets name the points as A(0,0),B(3,4),C(0,8),D(-3,4)
By taking out the difference from this points, we get,
AB=(0-3)+(0-4)=1
BC=(0-3)+(8-4)=1
CD=(-3-0)+(4-8)=1
AD=(-3-0)+(4-0)=1
we can see AB=CD=BC=AD This mean that the points are the edges of a rhombus.
tithi:
good enough
you have shown only all sides are equal.we can't decide whether the given points form a rhombus.Becuase all sides equal fo both rhombus and square.If want to whether it is a square or rhombus you need find distance between AC and BD.If you get AC=BD it is a square. If you get AC ≠D it is a rhombus.
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