In Fig. 3, a triangle ABC is drawn to circumscribe a circle of radius 4 cm such that the segments BD and DC into which BC is divided by the point of contact D are of lengths 6 cm and 8 cm respectively. If the area of $\triangle ABC$ is 84 cm$^2$, find the lengths of sides AB and AC.
Generated by claude-sonnet-4-6 · 2026-06-15 10:28 · grounding rag
Model Answer
Given: BD = 6 cm, DC = 8 cm, radius = 4 cm, Area of △ABC = 84 cm²
Using the property that tangents from an external point are equal:
Let the circle touch AB at E and AC at F.
- From B: BD = BE = 6 cm
- From C: CD = CF = 8 cm
- From A: AE = AF = x (say)
So, AB = x + 6, AC = x + 8, BC = 6 + 8 = 14 cm
Using Area:
Area of △ABC = Area of △OAB + Area of △OBC + Area of △OCA
$$84 = \frac{1}{2}(AB)(4) + \frac{1}{2}(BC)(4) + \frac{1}{2}(AC)(4)$$
$$84 = 2(AB + BC + AC) = 2(x+6 + 14 + x+8)$$
$$84 = 2(2x + 28)$$
$$84 = 4x + 56$$
$$4x = 28 \Rightarrow x = 7 \text{ cm}$$
Therefore:
$$AB = 7 + 6 = \textbf{13 cm}, \quad AC = 7 + 8 = \textbf{15 cm}$$
Source: Chapter 10, Section 10.3 (Theorem 10.2)
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Explanation
- The key property used is Theorem 10.2: tangents from an external point are equal. This lets you express all sides in terms of one unknown x.
- Then, the area of the triangle = sum of areas of three smaller triangles formed by joining the incentre O to each vertex, each with height = radius = 4 cm.
- Examiners expect you to clearly state the tangent-length property, set up the equation using areas, and show working step by step. Don't skip the step where you split the area.