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Show that the quadrilateral ABCD with vertices A(0, 3), B(-2, 0), C(0, -5) and D(2, 0) is a kite. Also, find the length of each diagonal of the kite ABCD.
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Solution: $AB = \sqrt{(0+2)^2 + (3 – 0)^2} = \sqrt{13}$ (1/2 Mark)
$BC = \sqrt{(-2-0)^2 + (0 + 5)^2} = \sqrt{29}$ (1/2 Mark)
$CD = \sqrt{(0 – 2)^2 + (-5 – 0)^2} = \sqrt{29}$ (1/2 Mark)
$AD = \sqrt{(2 – 0)^2 + (0 – 3)^2} = \sqrt{13}$ (1/2 Mark)
Since, $AB = AD$ and $BC = CD$, so ABCD is a kite.
Length of diagonals
$AC = \sqrt{(0-0)^2 + (3 + 5)^2} = \sqrt{64} = 8$ (1/2 Mark)
$BD = \sqrt{(-2-2)^2 + (0 – 0)^2} = \sqrt{16} = 4$ (1/2 Mark)
$BC = \sqrt{(-2-0)^2 + (0 + 5)^2} = \sqrt{29}$ (1/2 Mark)
$CD = \sqrt{(0 – 2)^2 + (-5 – 0)^2} = \sqrt{29}$ (1/2 Mark)
$AD = \sqrt{(2 – 0)^2 + (0 – 3)^2} = \sqrt{13}$ (1/2 Mark)
Since, $AB = AD$ and $BC = CD$, so ABCD is a kite.
Length of diagonals
$AC = \sqrt{(0-0)^2 + (3 + 5)^2} = \sqrt{64} = 8$ (1/2 Mark)
$BD = \sqrt{(-2-2)^2 + (0 – 0)^2} = \sqrt{16} = 4$ (1/2 Mark)