AB and AC are two equal chords of a circle having the radius of 5 cm . The centre of the circle is situated at the outside of the triangle ABC.If AB= AC = 6 cm, then calculate the length of the chords BC.

Explanation

Let O is the centre . Let us join O,A .
Also , let OA intersects BC at D.
Now , in `DeltaBOD` and `DeltaCOD , OB = OC , [ because " radii of same circle"]`
`angle BOD = angle COD [ because ` equal chords subtends equals angles at the centre] and OD is common to both .
`therefore Delta ~= Delta COD therefore BD = CD , therefore D`, is the mid - point of BC `therefore OD bot BC`
` therefore BOD` is a right - angle triangle.
`therefore OB^(2) = BD^(2) + OD^(2)`
or ` (5)^(2) = = BD^(2) + OD^(2) or , 25 = BD^(2) + OD^(2) .........(1)`
Again , in the angled triangle ABD,
`AB^(2) = BD^(2) + AD^(2) `
or ` (6)^(2) = BD^(2) + AD^(2)`
or ` 36 = BD^(2) + AD^(2) ...........(2)`
`therefore ` Substracting (1) from (2) we get `AD^(2) - OD^(2) = 11`
` or (AD + OD) (AD - OD) = 11 or , OA (AD- OD) = 11`
` or 5 (AD - OD) = 11 or OA (AD - OD) = 11`
or ` 5(AD - OD) = 11 or , AD - OD = (11)/(5)+5`
` or 2AD = (36)/(5) or AD =(18)/(5) ` d
`therefore ` form the right- angled triangle ABD we get `AB^(2) = BD^(2) + AD^(2)`
or `(6)^(2) = BD^(2) = 36 - (324)/(25) or BD^(2) = (900-324)/(25) = (576)/(25)`
`therefore BD = sqrt((576)/(25)) = (24)/(5)`
`therefore BC = 2 xx BD [ because ` D is the mid-point of BC ]
` = 2 xx (24)/(g) cm = (48)/(5) cm = 9.6`
`therefore` length of the chord BC = 9.6 cm .