Draw perpendicular bisector of line AE which divides it into two equal parts i.e. AF and FE
⟹AF=252=FE ........... (i)
We know that OD=5 cm and OE=13 cm ....... Radius of two circlesNow, △OFE=90o
∴ By Pythagoras theorem,
OE2=OF2+EF2
⟹OF2=169−(252)2=169−6254
=514
∴OF=√512 ........... (ii)
Now, △OFD=90o
∴ By Pythagoras theorem,
OD2=OF2+FD2
⟹FD2=OD2−OF2
=25−514=100−514=794
∴FD=72=FC ...........(iii) [OF bisects chords AE and CD]
Now, CD=DF+FC=72+72=7 cm
∴CD=7 cm ......... (iv)
By using the theorem,
(Two concentric circles having centre O. A line l intersect outer circle at A and B and inner circle at C and D. If AB=x and CD=y then AC=BD=12(x−y))
We get, AC=DE=12(AE−CD) .............. (v)
And AE=AD+DE
⟹AE=AD+12(AE−CD) ......... From (v)
⟹AD=AE−12(AE−CD) ......... From (v)
=25−12(25−7)
=25−9=16
Hence, AD=16 cm