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Show that the quadrilateral 'HOPE' with vertices $H(-2, 1)$, $O(-1, 2)$, $P(0, 1)$ and $E(-1, 0)$ is a rhombus. Is it a square ? Justify.
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Solution: Getting, $HO = \sqrt{2}$, $OP = \sqrt{2}$, $PE = \sqrt{2}$, $EH = \sqrt{2}$ (4
times 1/2 Mark)
Since, $HO = OP = PE = EH$, so HOPE is a rhombus. (1 Mark)
Length of diagonals
$HP = \sqrt{(2)^2 + (0)^2} = 2$, $OE = \sqrt{(0)^2 + (2)^2} = 2$ (1/2 Mark)
As $HP = OE$ (Diagonals equal)
Yes, HOPE is a square (1/2 Mark)
times 1/2 Mark)
Since, $HO = OP = PE = EH$, so HOPE is a rhombus. (1 Mark)
Length of diagonals
$HP = \sqrt{(2)^2 + (0)^2} = 2$, $OE = \sqrt{(0)^2 + (2)^2} = 2$ (1/2 Mark)
As $HP = OE$ (Diagonals equal)
Yes, HOPE is a square (1/2 Mark)