The sides of a triangle are in geometric progression with common ratio `rgt1`. If the triangle is a right angled triangle, then `r^(2)` is given by ?

Answer

To solve the problem, we need to find the value of \( r^2 \) given that the sides of a triangle are in geometric progression and the triangle is a right-angled triangle.

### Step-by-Step Solution:

1. **Define the sides of the triangle**:
Let the sides of the triangle be \( a \), \( ar \), and \( ar^2 \), where \( a \) is the first term and \( r \) is the common ratio (with \( r > 1 \)).

2. **Identify the hypotenuse**:
Since \( r > 1 \), the largest side \( ar^2 \) will be the hypotenuse of the right-angled triangle.

3. **Apply the Pythagorean theorem**:
According to the Pythagorean theorem, for a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides:
\[ (ar^2)^2 = a^2 + (ar)^2 \]

4. **Expand the equation**:
Expanding both sides gives:
\[ a^2 r^4 = a^2 + a^2 r^2 \]

5. **Factor out \( a^2 \)**:
We can factor \( a^2 \) from the right side:
\[ a^2 r^4 = a^2 (1 + r^2) \]

6. **Cancel \( a^2 \)** (assuming \( a \neq 0 \)):
Dividing both sides by \( a^2 \) gives:
\[ r^4 = 1 + r^2 \]

7. **Rearrange the equation**:
Rearranging the equation results in:
\[ r^4 - r^2 - 1 = 0 \]

8. **Let \( x = r^2 \)**:
Substitute \( x \) for \( r^2 \):
\[ x^2 - x - 1 = 0 \]

9. **Use the quadratic formula**:
The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be applied here, where \( a = 1 \), \( b = -1 \), and \( c = -1 \):
\[ x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} \]
\[ x = \frac{1 \pm \sqrt{1 + 4}}{2} \]
\[ x = \frac{1 \pm \sqrt{5}}{2} \]

10. **Select the positive root**:
Since \( r > 1 \), we only consider the positive root:
\[ r^2 = \frac{1 + \sqrt{5}}{2} \]

### Final Answer:
Thus, the value of \( r^2 \) is:
\[ \boxed{\frac{1 + \sqrt{5}}{2}} \]