Question

triangle ABC is drawn to circumscribe a circle of radius 4 centimetre such that the segments BD and DC into which BC is divided by the point of contact D are of length 8 cm and 6 CM respectively find the sides ab and ac

Answer 1

Let there is a circle having centre O touches the sides AB and AC of the triangle at point E and F respectively.
Let the length of the line segment AE is x.
Now in ΔABC,
CF = CD = 6 (tangents on the circle from point C)
BE = BD = 6 (tangents on the circle from point B)
AE = AF = x (tangents on the circle from point A)
Now AB = AE + EB
=> AB = x + 8
BC = BD + DC
=> BC = 8+6 = 14
CA = CF + FA
=> CA = 6 + x
Now
s = (AB + BC + CA )/2
=> s = (x + 8 + 14 + 6 +x)/2
=> s = (2x + 28)/2
=> s = x + 14
Area of the ΔABC = √{s*(s-a)*(s-b)*(s-c)}
= √[(14+x)*{(14+x)-14}*{(14+x)-(6+x)}*{(14+x)-(8+x)}]
= √[(14+x)*x*8*6]
= √[(14+x)*x*2*4*2*3]
=> Area of the ΔABC = 4√[3x(14+x)] .............................1
Now area of ΔOBC = (1/2)*OD*BC
= (1/2)*4*14
= 56/2
= 28
Area of ΔOBC = (1/2)*OF*AC
= (1/2)*4*(6+x)
= 2(6+x)
= 12 + 2x
Area of ΔOAB = (1/2)*OE*AB
= (1/2)*4*(8+x)
= 2(8+x)
= 16 + 2x
Now Area of the ΔABC = Area of ΔOBC + Area of ΔOBC + Area of ΔOAB
=> 4√[3x(14+x)] = 28 + 12 + 2x + 16 + 2x
=> 4√[3x(14+x)] = 56 + 4x
=> 4√[3x(14+x)] = 4(14 + x)
=> √[3x(14+x)] = 14 + x
On squaring bothe side, we get
3x(14 + x) = (14 + x)2
=> 3x = 14 + x (14 + x = 0 => x = -14 is not possible)
=> 3x - x = 14
=> 2x = 14
=> x = 14/2
=> x = 7
Hense
AB = x + 8
=> AB = 7+8
=> AB = 15
AC = 6 + x
=> AC = 6 + 7
=> AC = 13
So value of AB is 15 cm and value of AC is 13 cm