AB & CD are two equal chords of a circle with Centre O which intersect each other at the right angle at point P. If the perpendiculars from the centre to AB and CD meet them at M and N respectively, then MONP is a_______.

square.

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rectangle.

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kite.

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rhombus.

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Solution

The correct option is A
square.

Two equal chords in a circle are equidistant from the centre. Therefore OM = ON.

Join OP. Now $△ O P M$ and $△ O P N$ are congruent (RHS congruency)

$\Rightarrow$ OM=PN & ON= MP (cpct)

Hence, ON = PN = MP = MO

Moreover all angles are $90^{0}$ . Therefore the figure OMNP forms a square as all sides are equal and angles are $90^{0}$ .

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