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Question

Two circles of radius 3 cm and 5cm have a common centre, O. AB is a chord to both the circles and length of CD is 25 cm. Find the distance of the chord from the centre and the length AC.



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Solution


Given, CD = 25 cm
Since OECD, CE=DE=5 cm.
(Perpendicular from the centre to a chord bisects the chord.)
Apply Pythagoras' theorem in ΔOEC. We get
OE2+CE2=OC2
OE2+(5)2=9
OE2=95
OE2=4
OE=2 cm
In ΔOEA,
OE2+AE2=OA2
22+AE2=52
AE=21 cm
Now,
AC=AECE =215
=4.582 cm2.236 cm
=2.346 cm


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