AB and CD are equal chords of a circle of radius 20 cm. If O is the centre of circle and OE = 16 cm, find EB.

10 cm

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12 cm

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16 cm

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20 cm

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Solution

The correct option is B
12 cm
Given: AB = CD, OF = 16 cm

OE = 16 cm ...... (1)

In $Δ O E B$
Using Pythagoras theorem,
$O B^{2} = O E^{2} + E B^{2}$

$E B^{2} = O B^{2} - O E^{2}$

$E B^{2} = (20)^{2} - (16)^{2} = 144$

EB = 12 cm

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AB and CD are equal chords of a circle of radius 20 cm. If O is the centre of circle and OE = 16 cm, find EB.

The correct option is B 12 cmGiven: AB = CD, OF = 16 cm OE = 16 cm ...... (1) In ΔOEB Using Pythagoras theorem, OB2=OE2+EB2 EB2=OB2−OE2 EB2=(20)2−(16)2=144 EB = 12 cm

The correct option is B
12 cm
Given: AB = CD, OF = 16 cm

OE = 16 cm ...... (1)

In $Δ O E B$
Using Pythagoras theorem,
$O B^{2} = O E^{2} + E B^{2}$

$E B^{2} = O B^{2} - O E^{2}$

$E B^{2} = (20)^{2} - (16)^{2} = 144$

EB = 12 cm