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 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 $a = 14, b = x + 6, c = x + 8$. $s = 14 + x$. Area of $\Delta ABC = r \cdot s = 4(14 + x)$. Also Heron's formula: $\text{Area} = \sqrt{(14+x)(x)(8)(6)} = \sqrt{48x(14+x)}$. Equating: $16(14+x)^2 = 48x(14+x) \Rightarrow 14 + x = 3x \Rightarrow 2x = 14 \Rightarrow x = 7$. $AB = 7 + 8 = 15\text{ cm}, AC = 7 + 6 = 13\text{ cm}$.
Semi-perimeter $s = 14 + x$. $\text{Area}(\Delta ABC) = r \times s = 4(14 + x)$. [1.0 Mark]
By Heron's formula, $\text{Area} = \sqrt{48x(14 + x)}$. [1.0 Mark]
Equating areas: $16(14 + x)^2 = 48x(14 + x) \Rightarrow 14 + x = 3x \Rightarrow x = 7\text{ cm}$. $AB = 15\text{ cm}, AC = 13\text{ cm}$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Expressing area in terms of inradius $4(14+x)$: 1.0 Mark
Expressing area via Heron's formula $\sqrt{48x(14+x)}$: 1.0 Mark
Solving $x = 7 \Rightarrow AB = 15\text{ cm}, AC = 13\text{ cm}$: 1.0 Mark
Correct Answer: