Let C1 and C2 are two concentric circles with center O and radius of outer circle is 5 cm.
Given : AC is a chord with length 8 cm that is tangent to inner circle .
To find : Radius of inner circle i.e. OD
AC is a tangent for C1 at D so,
OD ⏊ AC [Tangent at a point on the circle is perpendicular to the radius through point of contact ]
So, OAD is a right-angled triangle at D .
Also it implies that OD is perpendicular to the chord AC in C2
So we have,
AD = DC [perpendicular from the center to the chord bisects the chord]
AC = AD + DC
8 = AD + AD
AD = 4 cm
In △OAD By Pythagoras Theorem
(OA)2 = (OD)2 + (AD)2
(5)2 = (OD)2+ (4)2
25 = (OD)2 + 16
OD = 3 cm
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