A chord 6 cm long is drawn in a circle with a diameter equal to 10 cm. Then the perpendicular distance from the centre is​

A chord 6 cm long is drawn in a circle with a diameter equal to 10 cm. Then the perpendicular distance from the centre is​

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  1. O is the centre of circle, radius of the circle=OA=r=10 cm, AB = ?, OC = 6 cm, C is point on chord AB & OC perpendicular to AB.

    [tex]\text{Now, In right triangle OAC,}[/tex]

    [tex]\boxed{\bf{By\: Pythagoras\: Theorem}}[/tex]

    [tex]\sf OA ² = AC² + CO²[/tex]

    [tex]\sf Or, AC² = OA² – CO²[/tex]

    [tex]\sf Or, AC² = 10² – 6² = 100 – 36 = 64[/tex]

    [tex]\sf Or, AC² = 100 – 36[/tex]

    [tex]\sf Or, AC² = 64[/tex]

    [tex]\sf Or, AC² = 8²[/tex]

    [tex]\sf Or, AC = 8[/tex]

    [tex]\sf Or AB = AC + CB[/tex]

    [tex]\sf Or AB = 8 + 8 ———– (AC = CB)[/tex]

    Or AB = 16 cm

    Therefore, length of chord = AB = 16 cm

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