Answer:
The given figure:
![](https://www.learnatnoon.com/s//wp-content/uploads/2021/09/n-1.png)
Considering the △ABC,
We know that any two tangents drawn from the same point to the circle have the same length.
As a result,
(i) BE = BD = 8 cm
(ii) CF = CD = 6 cm
(iii) AE = AF = x
It is observed that,
(i) CA = CF+FA = 6+x
(ii) BC = DC+BD = 6+8 = 14
(iii) AB = EB+AE = 8+x
Now compute the semi perimeter “s” as follows:
2s = AB+CA+BC
Putting the values respectively we get,
2s = 28+2x
s = 14+x
Area of triangle ABC
On solving the above equation we get,
= √(14+x)48x ……… (i)
Again, the area of triangle ABC = 2 × area of (△AOF + △COD + △DOB)
= 2×[(½×OF×AF)+(½×CD×OD)+(½×DB×OD)]
= 2×½(4x+24+32) = 56+4x …………..(ii)
Now from eq.(i) and eq.(ii) we get,
√(14+x)48x = 56+4x
On squaring both the sides,
48x(14+x) = (56+4x)2
48x = [4(14+x)]2/(14+x)
48x = 16(14+x)
48x = 224+16x
32x = 224
x = 7 cm
As a result, AB = 8+x Therefore, AB = 15 cm and CA = x+6 =13 cm.