O is the centre of the circle.
Chord AB Chord CD
Om = 3cm, MB = 4cm
ON = ------ cm (Reason:------------)
AB = CD = _______ cm
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B
9 cm
Given- AB and CD are two chords of a circle with centre O.
OM⊥AB and ON⊥CD
AM=AN=4.5 cm
OM=ON=2 cm
Solution-
AM=AN=4.5 cm
∴AB=AM+AN=(4.5+4.5) cm =9 cm
OM and ON are the perpendicular distances of AB and CD respectively.
Since OM⊥AB and ON⊥CD
Now OM=ON=2 cm
∴AB and CD are equidistant from O.
∴AB=CD
Since chords of a circle, which are equidistant from the centre of the same circle are equal in length.
∴AB=CD=9 cm
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