In the circle shown below AB is the chord having length 8 cm and being bisected by the line OD from centre. If the radius of circle is 5 cm, then find the distance of OD.

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Solution

Here OD is bisecting AB, so AD = DB = 4 cm.

$∠ O D B = 90^{\circ}$ [Line through center that bisect the chord is perpendicular to chord]

So, $O B^{2} = O D^{2} + D B^{2}$ [ $∵$ ODB is right angled triangle]
$O D^{2} = O B^{2} - D B^{2}$
$O D^{2} = (5)^{2} - (4)^{2}$ [OB is radius]
$O D^{2} = 25 - 16$
$O D^{2} = 9$
OD = 3 cm

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In the circle shown below AB is the chord having length 8 cm and being bisected by the line OD from centre. If the radius of circle is 5 cm, then find the distance of OD.

Here OD is bisecting AB, so AD = DB = 4 cm. ∠ODB=90∘ [Line through center that bisect the chord is perpendicular to chord] So, OB2=OD2+DB2 [∵ ODB is right angled triangle] OD2=OB2−DB2 OD2=(5)2−(4)2 [OB is radius] OD2=25−16 OD2=9 OD = 3 cm