The radii of two concentric circle are 13 cm and 8 cm. AB is a diameter of the bigger circle and BD is a tangent to the smaller circle touching it at D and the bigger circle at E. Point A is joined to D. Find the length of AD.

Explanation

We have `/_AEB=90^(@)` [angle in a semicircle]
Also, `OD bot BE `and `OD` bisects `BE`.
In right `DeltaOBD`, we have
`OB^(2)=OD^(2)+BD^(2)` [by Pythagoras' theorem]
`impliesBD=sqrt(OB^(2)-OD^(2))=sqrt(13^(2)-8^(2))cm` [`:' OB=13cm`, `OD=8cm`]
`:.BE=2BD=2sqrt(105)cm` [`:' D` is the midpoint of `BE`]
In right `DeltaAEB`, we have
`AB^(2)=AE^(2)+BE^(2)` [by Pythagoras' theorem]
`impliesAE=sqrt(AB^(2)-BE^(2))=sqrt(26^(2)-(2sqrt(105)^(2)))cm=sqrt(256)cm=16cm`
[`:' AB=` diameter `=2xxOB=2xx13cm=26cm`]
In right `DeltaAED`, we have
`AD^(2)=AE^(2)+DE^(2)` [by Pythagoras' theorem]
`impliesAD=sqrt(AE^(2)+DE^(2))=sqrt(16^(2)+(sqrt(105))^(2))cm`
`=19cm` `[:' DE=BD=sqrt(105)cm]`.
[Note We can also find `AE` by using midpoint theorem, since in `DeltaABE`, `O` is the midpoint of `AB` and `D` is the midpoint of `BE` and so `OD||AE` and `AE=2xxOD=16cm`]