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If the point of tangent to the inner circle is J and center is O and if you draw a diagram, you would see that OJK and OJH are identical right angles with OJ and OK are being hypotenuses. Also HJ is equal to JK.
Consider the triangle OJK. We can apply Pythagorean theorem to it.
`(OK)^2 = (OJ)^2 + (JK)^2`
`25.5 ^2 = 12^2 + (JK)^2`
`(JK)^2 = 506.25`
`JK = 22.5`
Therefore `JK = 22.5 cm` and `HK = 2 xx JK`
`HK = 2 xx 22.5 cm = 45 cm.`
Therefore the answer is 45 cm.
Let the tangent to the inner circle is drawn at point P of the smaller circle which meets the larger circle at points H and K and O be the common center of the two circles. The line OP will be the radious of the smaller circle and OK will be the radious of the larger circle i.e. OP=12 and OK=25.3. We know that the radious at the point of contact is perpendicular to the tangent hence, OP is perpendicular HK . OP also bisects Hk ( this we can prove by proving the two triangles OPK and OPH congurent) i.e PK=PH or HK= 2*PK. We take the right angled triangle OPK in which, OP=12 and OK=25.5.
OK^2= OP^2 + PK^2 [ Pythagoras theroem]
Or, PK^2= OK^2 - OP^2
Or, PK^2= (25.5)^2 - (12)^2
Or, PK^2= (25.5+12)(25.5-12) [Using the formula (a-b)^2=a^2-b^2]
Or, PK^2=37.5 * 13.5
Or, Pk= 22.5
And Hk= 2*PK= 2*22.5= 45
HK= 45 Answer
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