Centroid of a triangle — intersection of medians: For a uniform triangular lamina or area, the centre of gravity (centroid) lies at the point where all three medians intersect (the centroid). Is this statement true?

Introduction / Context:Finding the centroid of basic shapes underpins calculations of bending stresses, deflections, and resultant forces. The triangle’s centroid has a celebrated geometric property that simplifies many engineering analyses.

Given Data / Assumptions:

• Homogeneous triangular area (uniform thickness and density if treated as a lamina).
• Medians are the line segments from each vertex to the midpoint of the opposite side.

Concept / Approach:The centroid of a triangle lies at the intersection of its three medians and divides each median in a 2:1 ratio (measured from the vertex). This location is independent of coordinate orientation and works for any triangle (acute, obtuse, right).

Step-by-Step Solution:Draw all three medians; they are concurrent at a single point.Use coordinate geometry: with vertices (x1,y1), (x2,y2), (x3,y3), the centroid is at ( (x1 + x2 + x3)/3 , (y1 + y2 + y3)/3 ).This is precisely the intersection point of the medians.Therefore, the statement is true.

Verification / Alternative check:Physical reasoning: uniform density implies balance occurs where equal triangular masses balance along the medians; experiments with a cardboard triangle and a plumb line confirm the same point.

Why Other Options Are Wrong:

• False: contradicts well-established geometry and engineering handbooks.

Common Pitfalls:

• Confusing centroid (area) with circumcenter, incenter, or orthocenter; only the centroid is the intersection of medians and always lies inside the triangle.

Final Answer:True