To find the centroid's distance from the center along the line of symmetry of a quarter circle, follow these steps:
Centroid of a Quarter Circle
Consider a quarter circle with radius R, centered at the origin (0,0) and lying in the first quadrant. The coordinates of its centroid $(x_c, y_c)$ are given by the standard formula:
$ x_c = \frac{4R}{3\pi} $
$ y_c = \frac{4R}{3\pi} $
Line of Symmetry and Distance
The line of symmetry for this quarter circle, passing through the center (origin) and the centroid, is the line where $x = y$.
The question asks for the distance of the centroid from the center *along this line of symmetry*. Since the centroid coordinates are equal ($x_c = y_c$), the centroid lies directly on the line $y=x$.
Therefore, the distance from the center (0,0) to the centroid $(\frac{4R}{3\pi}, \frac{4R}{3\pi})$ along the line $y=x$ is simply the value of the x-coordinate (or the y-coordinate).
Final Calculation
Distance = $ x_c = y_c = \frac{4R}{3\pi} $
Thus, the centroid on the line of symmetry from the center distance of a quarter circle is $ \frac{4R}{3\pi} $. This corresponds to Option D.
Important Questions from Centroid
In Δ ABC, the coordinates of B are (0, 0), AB = 2, ∠ABC = π/3 and the middle point of BC has the coordinates (2, 0). The centroid of triangle is:
The centroid of the triangle with vertices A(2, -3, 3), B(5, -3, -4) and C(2, -3, -2) is the point
What is the area of the triangle formed by these lines?
The centroid of the triangle is at which one of the following points?
If a vertex of a triangle is (1, 1) and the midpoints of two sides of the triangle through this vertex are (-1, 2) and (3, 2), then the centroid of the triangle is
