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प्रश्न
Prove that the points (3, 0), (4, 5), (-1, 4) and (-2, -1), taken in order, form a rhombus.
Also, find its area.
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उत्तर
The distance d between two points `(x_1, y_1)` and `x_2,y_2)` is given by the formula
`d = sqrt((x_1 - x_2)^2 + (y_1 - y_2)^2)`
In a rhombus, all the sides are equal in length. And the area ‘A’ of a rhombus is given as
`A = 1/2("Product of both diagonals")`
Here the four points are A(3,0), B(4,5), C(−1,4) and D(−2,−1).
First, let us check if all the four sides are equal.
`AB = sqrt((3 -4)^2 + (0 - 5)^2)`
`= sqrt((-1)^2 + (-5)^2)`
`= sqrt(1 + 25)`
`= sqrt(25 + 1)`
`BC=sqrt26`
`CD = sqrt((-1+2)^2 + (4 + 1)^2)`
`= sqrt((1)^2 +(5)^2)`
`= sqrt(26)`
`AD = sqrt((3 + 2)^2 + (0 + 1)^2)`
`= sqrt((5)^2 + (1)^2)`
`= sqrt(25 + 1)`
`AD = sqrt26`
Here, we see that all the sides are equal, so it has to be a rhombus.
Hence we have proved that the quadrilateral formed by the given four vertices is a rhombus.
Now let us find out the lengths of the diagonals of the rhombus.
`AC = sqrt((3 + 1)^2 + (0 - 4)^2)`
`= sqrt((4)^2 + (-4)^2)`
`= sqrt((6)^2 + (6)^2)`
`= sqrt(36 + 36)`
`BD = 6sqrt2`
Now using these values in the formula for the area of a rhombus we have,
`A = ((6sqrt2)(4sqrt2))/2`
`= ((6)(4)(2))/2`
A = 24
Thus the area of the given rhombus is 24 square units.
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