Let incircle of radius $4$ units of a triangle $ABC$ touches the side $BC$ at $D$. If $BD = 6, DC = 8$ and $\Delta$ be the area of triangle, then $\sqrt[4]{\Delta - 3}$ = _______.
Step-by-Step Solution
Key Concept: The inradius formula $r = \frac{\Delta}{s}$ connects area, semi-perimeter, and inradius to find the unknown side.
Using the semi-perimeter formula with $s = x + 14$, the area is expressed as $\Delta = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)(8)(6)(x)}$. Setting $r = \frac{\Delta}{s} = 4$ gives $\sqrt{\frac{48x}{x+14}} = 4$, which simplifies to $16(x+14) = 48x$ or $x = 7$. Therefore $\Delta = \sqrt{48 \times 7 \times 21} = 84$ and $\sqrt{\Delta - 3} = 3$.
Correct Answer: 3