Largest triangle in a semicircle (radius 6 m):\nFind the maximum possible area of a triangle inscribed in a semicircle of radius 6 m.

Introduction / Context:
Any triangle formed by the diameter as one side and a point on the semicircle is a right triangle (Thales’ theorem). Among right triangles with fixed hypotenuse, the area is maximized when the legs are equal (an isosceles right triangle), because the product of two numbers with fixed sum is maximized when they are equal.

Given Data / Assumptions:

• Radius r = 6 m ⇒ diameter (hypotenuse) = 12 m.
• Let legs be a and b; hypotenuse fixed = 12 m.

Concept / Approach:
Area = (1/2)*a*b. For fixed hypotenuse, a^2 + b^2 = 12^2. The product a*b is maximized when a = b. Thus a = b = 12/√2 = 6√2. Substitute into area.

Step-by-Step Solution:

Verification / Alternative check:
Pick any unequal legs satisfying a^2 + b^2 = 144; their product is less than when a = b, so the area is smaller. Hence 36 m² is indeed the maximum.

Why Other Options Are Wrong:
12 and 18 are sub-maximal; 72 mistakenly doubles the maximal area.

Common Pitfalls:
Using diameter*radius/2; the correct formula is (1/2)*a*b with a^2 + b^2 fixed.

Final Answer:
36 sq.m