Difficulty: Medium
Correct Answer: 42√3 cm
Explanation:
Introduction / Context:
As with similar problems, we connect the area of the inscribed circle to the triangle’s inradius r, then tie r to the triangle’s side a. From there we compute the perimeter 3a. The numbers are chosen so r comes out integral, making a and perimeter neat radicals.
Given Data / Assumptions:
Concept / Approach:
Find r from πr^2 = 154; then a = 6r/√3; finally P = 3a. Keep results exact until the end to avoid rounding.
Step-by-Step Solution:
r^2 = 154 * (7/22) = 49 ⇒ r = 7r = a√3/6 ⇒ a = 6r/√3 = 42/√3 = 14√3Perimeter P = 3a = 3 * 14√3 = 42√3 cm
Verification / Alternative check:
Back-calc area: r = 7 ⇒ πr^2 = (22/7)*49 = 154 sq. cm, confirming the setup.
Why Other Options Are Wrong:
21 cm and 42 cm ignore the √3 factor; 21√3 cm is one side, not the perimeter.
Common Pitfalls:
Confusing inradius with altitude or using area of triangle instead of the circle to start; dropping √3 during simplification.
Final Answer:
42√3 cm
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