To find the center of gravity (centroid) of a triangular lamina with each side measuring [tex]\(9 \sqrt{3}\)[/tex] cm, we can follow these steps:
1. Identify the Type of Triangle and Initial Setup:
Since the triangle is equilateral (all sides are equal), the centroid will be at an equal distance from all three vertices.
Let's identify the vertices of the triangle in a coordinate system for easier calculation:
- Vertex [tex]\(A\)[/tex] at [tex]\((0, 0)\)[/tex]
- Vertex [tex]\(B\)[/tex] at [tex]\((9 \sqrt{3}, 0)\)[/tex]
- Vertex [tex]\(C\)[/tex] can be calculated using the properties of an equilateral triangle: the x-coordinate is half the side length and the y-coordinate is the height of the triangle.
2. Calculate the Coordinates for Vertex [tex]\(C\)[/tex]:
- The x-coordinate of [tex]\(C\)[/tex] is [tex]\( \frac{9 \sqrt{3}}{2} \)[/tex]
- The height (h) of an equilateral triangle with side length [tex]\(a\)[/tex] can be found using the formula:
[tex]\[
h = \frac{\sqrt{3}}{2} a = \frac{\sqrt{3}}{2} \times 9 \sqrt{3}
\][/tex]
Simplifying this,
[tex]\[
h = \frac{\sqrt{3}}{2} \times 9 \sqrt{3} = \frac{3 \times 9}{2} = \frac{27}{2} = 13.5 \text{ cm}
\][/tex]
Thus, vertex [tex]\(C\)[/tex] is at [tex]\(\left(\frac{9 \sqrt{3}}{2}, 13.5\right)\)[/tex].
3. Find the Centroid Coordinates:
For any triangle, the centroid [tex]\((x_g, y_g)\)[/tex] is found by averaging the coordinates of its vertices:
[tex]\[
x_g = \frac{x_1 + x_2 + x_3}{3}
\][/tex]
[tex]\[
y_g = \frac{y_1 + y_2 + y_3}{3}
\][/tex]
Substituting the coordinates of vertices [tex]\(A\)[/tex], [tex]\(B\)[/tex], and [tex]\(C\)[/tex]:
- [tex]\(A(0, 0)\)[/tex]
- [tex]\(B(9 \sqrt{3}, 0)\)[/tex]
- [tex]\(C(\frac{9 \sqrt{3}}{2}, 13.5)\)[/tex]
The x-coordinate of the centroid:
[tex]\[
x_g = \frac{0 + 9\sqrt{3} + \frac{9\sqrt{3}}{2}}{3} = \frac{9\sqrt{3}(1 + \frac{1}{2})}{3} = \frac{9\sqrt{3} \times \frac{3}{2}}{3} = \frac{27\sqrt{3}}{6} = 7.794228634059947 \text{ cm}
\][/tex]
The y-coordinate of the centroid:
[tex]\[
y_g = \frac{0 + 0 + 13.5}{3} = \frac{13.5}{3} = 4.5 \text{ cm}
\][/tex]
4. Conclusion:
The center of gravity (centroid) of the given triangular lamina, each side of which is [tex]\(9 \sqrt{3}\)[/tex] cm, has coordinates:
[tex]\[
(x_g, y_g) = \left(7.794228634059947, 4.5 \text{ cm}\right)
\][/tex]
So, the final answer is that the centroid of the triangle is at approximately [tex]\( \left( 7.79 \, \text{cm}, 4.50 \, \text{cm} \right) \)[/tex], with side length [tex]\(15.59 \, \text{cm}\)[/tex].
