A triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is…

Area ($\triangle$ ABC) = Area ($\triangle$ BOC) + Area ($\triangle$ AOC) + Area ($\triangle$ AOB)
$= \frac{1}{2} \times 4 \times 14 + \frac{1}{2} \times 4 \times (6 + x) + \frac{1}{2} \times 4 \times (8 + x)$
$= (56 + 4x)$ or $4(14 + x)$ --- (1)
Semi perimeter of $\triangle$ ABC = $\frac{14+(6+x)+(8+x)}{2} = (14 + x)$
Also, area ($\triangle$ ABC) = $\sqrt{(14 + x)(14 + x – 14)[(14 + x) – (6 + x)][(14 + x) – (8 + x)]}$
$= \sqrt{48x(14 + x)}$ --- (2)
From (1) and (2), we get
$x=7$
$\therefore AB = 15$ cm and $AC = 13$ cm