Prove that the lengths of tangents drawn from an external point to a circle are equal. Use this theorem to solve:
Three circles of radii $1\text{ cm}, 2\text{ cm}$ and $3\text{ cm}$ touch each other externally. Find the perimeter of the triangle formed by joining their centres.
Step-by-Step Solution
Key Concept: Part 1: Proof of Theorem 10.2 ($TP = TQ$). Part 2: Let centres be $A, B, C$ with radii $r_1=1, r_2=2, r_3=3$. Distance between centres: $AB = 1+2 = 3\text{ cm}, BC = 2+3 = 5\text{ cm}, CA = 3+1 = 4\text{ cm}$. Perimeter $= 3 + 5 + 4 = 12\text{ cm}$.
Part 1: Complete Proof of Theorem 10.2 ($TP = TQ$). [2.5 Marks]
Part 2: Distance between centres $AB = 1 + 2 = 3\text{ cm}$, $BC = 2 + 3 = 5\text{ cm}$, $CA = 3 + 1 = 4\text{ cm}$. [1.5 Marks]
Perimeter of $\Delta ABC = 3 + 5 + 4 = 12\text{ cm}$. [1.0 Mark]
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🎯 Official CBSE Marking Scheme:
Proof of Theorem 10.2: 2.5 Marks
Calculating side lengths $AB=3, BC=5, CA=4$: 1.5 Marks
Evaluating total perimeter $= 12\text{ cm}$: 1.0 Mark
Correct Answer: