In `Delta ABC, ` a=5, b=4, c=3. G is the centroid of triangle. If `R_(1)` be the circum radius of triangle GAB then the value of `(a)/(65) R_(1)^(2)` must be

Answer

To solve the problem, we will follow a systematic approach to find the circumradius \( R_1 \) of triangle \( GAB \) and then compute the value of \( \frac{a}{65} R_1^2 \).

### Step-by-Step Solution:

1. **Identify the Coordinates of Points A, B, and C**:
- Given \( a = 5 \), \( b = 4 \), \( c = 3 \).
- We can assign coordinates based on the lengths:
- Let \( A(0, 0) \)
- Let \( B(3, 0) \)
- Let \( C(0, 4) \)

2. **Calculate the Centroid G of Triangle ABC**:
- The centroid \( G \) 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:
\[ G\left(\frac{0 + 3 + 0}{3}, \frac{0 + 0 + 4}{3}\right) = G\left(1, \frac{4}{3}\right) \]

3. **Calculate the Lengths of Sides GA, GB, and AB**:
- Length \( GA \):
\[ GA = \sqrt{(1 - 0)^2 + \left(\frac{4}{3} - 0\right)^2} = \sqrt{1 + \frac{16}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3} \]
- Length \( GB \):
\[ GB = \sqrt{(1 - 3)^2 + \left(\frac{4}{3} - 0\right)^2} = \sqrt{(-2)^2 + \left(\frac{4}{3}\right)^2} = \sqrt{4 + \frac{16}{9}} = \sqrt{\frac{52}{9}} = \frac{2\sqrt{13}}{3} \]
- Length \( AB \) (already given):
\[ AB = 3 \]

4. **Calculate the Area of Triangle GAB**:
- Using the formula for the area of a triangle with vertices at coordinates:
\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]
- Substituting \( G(1, \frac{4}{3}), A(0, 0), B(3, 0) \):
\[ \text{Area} = \frac{1}{2} \left| 1(0 - 0) + 0(0 - \frac{4}{3}) + 3(\frac{4}{3} - 0) \right| = \frac{1}{2} \left| 0 + 0 + 4 \right| = \frac{4}{2} = 2 \]

5. **Calculate the Circumradius \( R_1 \)**:
- The formula for the circumradius \( R \) of triangle \( ABC \) is:
\[ R = \frac{abc}{4 \Delta} \]
- Here, \( a = GA = \frac{5}{3}, b = GB = \frac{2\sqrt{13}}{3}, c = AB = 3, \Delta = 2 \):
\[ R_1 = \frac{\left(\frac{5}{3}\right) \left(\frac{2\sqrt{13}}{3}\right)(3)}{4 \cdot 2} = \frac{5 \cdot 2\sqrt{13}}{12} = \frac{5\sqrt{13}}{12} \]

6. **Calculate \( \frac{a}{65} R_1^2 \)**:
- First, calculate \( R_1^2 \):
\[ R_1^2 = \left(\frac{5\sqrt{13}}{12}\right)^2 = \frac{25 \cdot 13}{144} = \frac{325}{144} \]
- Now substitute \( a = 5 \):
\[ \frac{a}{65} R_1^2 = \frac{5}{65} \cdot \frac{325}{144} = \frac{1}{13} \cdot \frac{325}{144} = \frac{325}{1872} \]

### Final Answer:
\[ \frac{a}{65} R_1^2 = \frac{325}{1872} \]