To find the centroid's distance from the center along the line of symmetry of a quarter circle, follow these steps:
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} $
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).
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.
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?
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 ? .
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:
Consider the following statements regarding the centre of gravity and centroid:
1. Centroid of an area does not lie on the axis of symmetry if it exits.
2. Centre of gravity of a body is a point through which the resultant gravitational force acts for any orientation of the body.
3. Centroid is a point in a line plane area volume such that the moment of area about any axis through that point is zero.
Which of the above statements are correct?