A triangle $ABC$ is drawn to circumscribe a circle of radius $4\text{ cm}$ such that the segments $BD$ and $DC$ into which $BC$ is divided by the point of contact $D$ are of lengths $8\text{ cm}$ and $6\text{ cm}$ respectively. Find the sides $AB$ and $AC$.
Step-by-Step Solution
Key Concept: Let $AF = AE = x$. Sides are $a = 14, b = x + 6, c = x + 8$. Semi-perimeter $s = x + 14$. Equate Heron's area to sum of areas of $\Delta OBC, \Delta OCA, \Delta OAB = r \cdot s$.
Tangents from vertices: $CD = CE = 6\text{ cm}, BD = BF = 8\text{ cm}, AF = AE = x$. [0.5 Mark]
Sides: $BC = 14, AC = x + 6, AB = x + 8$. Semi-perimeter $s = \dfrac{14 + x + 6 + x + 8}{2} = x + 14$. [0.5 Mark]
$\text{Area}(\Delta ABC) = r \cdot s = 4(x + 14)$. [0.5 Mark]
By Heron's formula: $\text{Area} = \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{(x+14)(x)(8)(6)} = \sqrt{48x(x+14)}$. [0.5 Mark]
Squaring: $48x(x+14) = 16(x+14)^2 \Rightarrow 3x = x + 14 \Rightarrow 2x = 14 \Rightarrow x = 7\text{ cm}$. [0.5 Mark]
Sides: $AB = 7 + 8 = 15\text{ cm}$, $AC = 7 + 6 = 13\text{ cm}$. [0.5 Mark]
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🎯 Official CBSE Marking Scheme:
Tangents setup and side expressions in $x$: 1.0 Mark
Heron's area and $r \cdot s$ area expressions: 1.0 Mark
Solving $3x = x + 14 \Rightarrow x = 7$: 0.5 Mark
Stating $AB = 15\text{ cm}, AC = 13\text{ cm}$: 0.5 Mark
Correct Answer: