AB and CD are two parallel chords of a circle lying on the opposite side of the centre O such that AB= 24 cm and CD = 10 cm if the chord AB is at a distance of 5 cm from the centre ,find the distance of the chord CDfrom the centre of the circle
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Answered by
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AM=1/2 of AB
AM=12cm
In∆AMO,
by Pythagoras theorem,
OA²=OM²+AM²
=5²+12²
=169
OA=13
OA=OC=radius=13
CN=1/2of CD
CN=5cm.
In∆ONC,
by Pythagoras theorem,
OC²=ON²+CN²
13²=ON²+5²
ON²=144
ON=12
distance of chord CD from centre is 12cm
AM=12cm
In∆AMO,
by Pythagoras theorem,
OA²=OM²+AM²
=5²+12²
=169
OA=13
OA=OC=radius=13
CN=1/2of CD
CN=5cm.
In∆ONC,
by Pythagoras theorem,
OC²=ON²+CN²
13²=ON²+5²
ON²=144
ON=12
distance of chord CD from centre is 12cm
Lightbearer:
wrong answer... equal chords are equidistant from the centre even if it's not parallel
i corrected it
still there is correction... u got radius 13 n still u took radius as 12 for the second triangle
i corrected it
Answered by
1
The distance of the chord CD from the centre of the circle is 12 cm.
Given,
AB and CD are two parallel chords of a circle lying on the opposite side of the centre O such that AB= 24 cm and CD = 10 cm if the chord AB is at a distance of 5 cm from the centre.
To find,
Find the distance of the chord CD from the centre of the circle.
Solution,
AM=1/2 of AB
AM=12cm
In∆AMO,
by Pythagoras theorem,
OA²=OM²+AM²
=5²+12²
=169
OA=13
OA=OC=radius=13
CN=1/2of CD
CN=5cm.
In∆ONC,
by Pythagoras theorem,
OC²=ON²+CN²
13²=ON²+5²
ON²=144
ON = 12 cm
#SPJ3
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