A triangle $ABC$ is drawn to circumscribe a circle of radius 4 cm such that the segments $BD$ and $DC$ are of lengths 10 cm and 8 cm respectively. Find the lengths of the sides $AB$ and $AC$, if it is given that the area of $\triangle ABC = 90$ cm$^2$.
Generated by claude-sonnet-4-6 · 2026-06-15 10:24 · grounding rag
Model Answer
Let the circle touch AB at E and AC at F.
By the property that tangents from an external point are equal:
- From B: BD = BE = 10 cm
- From C: CD = CF = 8 cm
- From A: AE = AF = x (unknown)
So the sides are:
$$BC = BD + DC = 10 + 8 = 18 \text{ cm}$$
$$AB = BE + AE = 10 + x$$
$$AC = CF + AF = 8 + x$$
Using Area of Triangle:
$$\text{Area of } \triangle ABC = \frac{1}{2} \times r \times (AB + BC + CA)$$
$$90 = \frac{1}{2} \times 4 \times (10 + x + 18 + 8 + x)$$
$$90 = 2(36 + 2x)$$
$$45 = 36 + 2x$$
$$2x = 9 \Rightarrow x = 4.5 \text{ cm}$$
Therefore:
$$AB = 10 + 4.5 = \textbf{14.5 cm}$$
$$AC = 8 + 4.5 = \textbf{12.5 cm}$$
Source: Chapter 10, Section 10.3 (Theorem 10.2)
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Explanation
Key concepts examiners look for:
- Tangent length property (Theorem 10.2): Tangents from an external point are equal — this lets you express all three sides in terms of one unknown $x$.
- Area formula for a triangle circumscribing a circle: Area = $\frac{1}{2} \times r \times \text{perimeter}$. This is derived because the incircle touches all three sides, and the triangle splits into three smaller triangles (each with base = one side and height = radius $r$).
- Note: The question states BD = 10, DC = 8 (unlike Exercise 10.2 Q.12 in the textbook which uses 8 and 6). Set up the equation carefully with the given values.
- Always state the final answer clearly with units.