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Sum

the below given figure, a triangle ABC is drawn to circumscribe a circle of radius 3 cm, such that the segments BD and DC are respectively of lengths 6 cm and 9 cm. If the area of ΔABC is 54 cm^{2}, then find the lengths of sides AB and AC.

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#### Solution

**Given : **OD = 3cm**Construction : **Join OA, OB and X

**Proof :**Area of the ΔABC = area of ΔOBC + area of ΔOAC + arc of ∠ OAB.

BD = 6 cm : BE = 6 cm ( equal tangents )

DC = 9 cm : CF = 9 cm ( equal tangents )

area of Δ OBC = `1/2 xx b xx h`

= `1/2 xx 15 xx 3`

= `45/2 cm^2`

area of Δ OAC = `1/2 xx ( x + 9 ) xx 3 = [ 3( x + 9 )]/2 cm^2`

area of Δ OAB = `1/2 xx ( x + 6 ) xx 3 = [ 3( x + 6 )]/2 cm^2`

area of the Δ ABC = `sqrt[ s ( s - a )( s - b )( s - c )]cm^2`

15cm,12cm,9cm">`s = [( x + 9 ) + ( x + 6 ) + 15 ]/2 = [ 2x + 30 ]/2 = x + 15 `

∴ Δ = `sqrt[( x + 15 )( x + 15 - 15 )( x - 15 - x - 4)( x + 15 - x - 6)]`

= `sqrt[( x + 15 )(x)(6)(9)]`

= `sqrt[ x( x + 15 )(54)] = [3( x + 9 )]/2 + [3( x + 16 )]/2 + 45/2`

= `sqrt[ x( x + 15)(54)] = 3/2[( x + 9) + ( x + 6 ) + 15]`

= `sqrt[ x( x + 15)(54)] = 3/2( 2x + 30 )`

= `sqrt[ x( x + 15)(54)] = ( 3x + 15 )`

Squaring on both sides we get

x( x + 15)(54) = 9( x + 15 )2

6x = x +15

5x = 15

x = 3 cm

Hence the sides are 15cm, 12cm, 9cm

AC = 3 + 9 = 12

AB = AF + FB = 6 + x = 6 + 3 = 9.

AB = AF + FB = 6 + x = 6 + 3 = 9.

Concept: Triangles Examples and Solutions

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