Difficulty: Medium
Correct Answer: 126 cms
Explanation:
Introduction / Context:
An equilateral triangle has a well-known relation between its side length and the inradius (radius of the inscribed circle). If the inradius is known via the circle’s area, we can back-compute the side length and hence the perimeter of the triangle. The algebra is clean when π = 22/7 and the numbers are crafted to be exact.
Given Data / Assumptions:
Concept / Approach:
From A_circ = πr^2, compute r. Then use r = (a√3)/6 to find a. Finally compute P = 3a. The numbers are chosen so that r comes out as a neat radical, and a, P are integers.
Step-by-Step Solution:
r^2 = A_circ / π = 462 * (7/22) = 147r = √147 = 7√3r = (a√3)/6 ⇒ a = 6r/√3 = 6 * (7√3) / √3 = 42Perimeter P = 3a = 3 * 42 = 126 cm
Verification / Alternative check:
Compute area from a = 42: inradius r = a√3/6 = 7√3; circle area = πr^2 = (22/7) * 147 = 462 sq. cm, consistent.
Why Other Options Are Wrong:
42√3 cms (option A) is the triangle’s side expressed differently (not the perimeter). 72.6 and 168 cms do not match the consistent inradius–side relationships.
Common Pitfalls:
Mixing up inradius with circumradius, or using the area formula of the triangle instead of the inscribed circle when working backward; also forgetting that perimeter is 3a.
Final Answer:
126 cms
Discussion & Comments