In a circle of radius `5\ c m ,\ A B\ a n d\ A C` are two chords such that `A B=A C=6 cm`.
Find the length of the chord `B C`

Explanation

Let `AD` be the bisector of `/_ BAC.`
Then , `AD` is the perpendicular bisector of `BC` and passes through the centre `O`. Join `CO`
Let `AD` meet `BC` in `M.`
Then , `BM` `=` `CM.`
From the right `Delta AMC,` we have
`MC^(2)=AC^(2)-AM^(2)=36cm^(2)-AM^(2)` ....(i)
From the right `Delta OMC `, we have
`MC^(2)=OC^(2)-OM^(2)=25cm^(2)- ( OA - AM)^(2)`
`= 25cm^(2)- ( 5 cm-AM)^(2) ` ....(ii)
`:. 36cm^(2)-AM^(2) =25cm^(2)=- ( 5 cm - AM)^(2) `[from (i) and (ii) ]
`implies 36 cm^(2) - AM^(2) = 25 cm^(2)- (25 cm^(2)-AM^(2)+10 xxAM)`
`implies 36cm^(2)-AM^(2)=25cm^(2)-25cm^(2)-AM^(2)+10cm xx AM`
`implies 36cm^(2)=10xx AM`
`implies AM = 3.6 CM`.
`:. MC = sqrt(36-(3.6)^(2))cm= sqrt(23.04)cm= 4.8 cm`
`:. BC = 2 xx MC = 9.6 cm`