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
If the centroid of a triangle formed by (7, x), (y, -6) and (9, 10) is (6, 3), then the values of x and y are respectively
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 \(\vec a, \vec b, \vec c\) , are the position vectors of the vertices A, B, C respectively of a triangle ABC and G is the centroid of the triangle, then what is \(\overrightarrow{AG}\) equal to ? .
The centroid of the triangle with vertices A(2, -3, 3), B(5, -3, -4) and C(2, -3, -2) is the point
