We are tasked with finding $BG$, the distance between point $B$ and the centroid $G$ of $\triangle ABC$.

Step 1: Coordinates of Points

The given coordinates of the vertices of the triangle are:

• $A(5, 0)$
• $B(0, 0)$
• $C(0, 3)$

Step 2: Centroid Formula

The centroid $G$ of a triangle with vertices $A(x_1, y_1)$, $B(x_2, y_2)$, and $C(x_3, y_3)$ is given by the formula: $G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3} \right)$ Substituting the coordinates of $A$, $B$, and $C$: $G = \left( \frac{5 + 0 + 0}{3}, \frac{0 + 0 + 3}{3} \right) = \left( \frac{5}{3}, 1 \right)$

Step 3: Distance Formula

The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$ Using the coordinates $B(0, 0)$ and $G\left(\frac{5}{3}, 1\right)$, we can calculate $BG$: $BG = \sqrt{\left( \frac{5}{3} - 0 \right)^2 + \left( 1 - 0 \right)^2} = \sqrt{\left( \frac{5}{3} \right)^2 + 1^2}$ $BG = \sqrt{\frac{25}{9} + 1} = \sqrt{\frac{25}{9} + \frac{9}{9}} = \sqrt{\frac{34}{9}} = \frac{\sqrt{34}}{3}$

Thus, the distance $BG$ is $\frac{\sqrt{34}}{3}$.

Would you like further details or have any questions?

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Here are 5 related questions:

1. What is the formula for finding the centroid of a triangle?
2. How would you calculate the centroid if the triangle had different coordinates?
3. How does the centroid divide each median in a triangle?
4. Can you explain why the centroid lies inside the triangle for all triangles?
5. What is the significance of the centroid in applications like physics (center of mass)?

Tip: The centroid divides each median of a triangle in the ratio 2:1, with the longer segment being closer to the vertex.