# 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. - Mathematics

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 cm2, then find the lengths of sides AB and AC.

#### 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.
Is there an error in this question or solution?
2014-2015 (March) All India Set 3

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