AB & CD are two equal chords of a circle with Centre O, which intersect each other at a 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.

Also, OM=PN & ON= MP (geometry).

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 $90^{\circ}$ ..

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Similar questions

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

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AB & CD are two equal chords of a circle with Centre O, which intersect each other at a 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_______.

The correct option is A square. Two equal chords in a circle are equidistant from the centre. Therefore OM=ON. Also, OM=PN & ON= MP (geometry). ON = PN = MP = MO Moreover all angles are 900 . Therefore the figure OMNP forms a square as all sides are equal and angles are 90∘..

The correct option is A
square.

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

Also, OM=PN & ON= MP (geometry).

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 $90^{\circ}$ ..