The sides of a triangle are `3x + 4y, 4x + 3y` and `5x+5y` units, where `x gt 0, y gt 0`. The triangle is

Options

A. right angled
B. acute angled
C. obtuse angled
D. isosceles

Answer

To determine the type of triangle formed by the sides \(3x + 4y\), \(4x + 3y\), and \(5x + 5y\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the sides of the triangle:**
Let:
- \( a = 3x + 4y \)
- \( b = 4x + 3y \)
- \( c = 5x + 5y \)

2. **Calculate the squares of the sides:**
- \( a^2 = (3x + 4y)^2 = 9x^2 + 24xy + 16y^2 \)
- \( b^2 = (4x + 3y)^2 = 16x^2 + 24xy + 9y^2 \)
- \( c^2 = (5x + 5y)^2 = 25x^2 + 50xy + 25y^2 \)

3. **Use the cosine rule to find \( \cos C \):**
According to the cosine rule:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]
Substitute the values of \( a^2 \), \( b^2 \), and \( c^2 \):
\[ a^2 + b^2 = (9x^2 + 24xy + 16y^2) + (16x^2 + 24xy + 9y^2) = 25x^2 + 48xy + 25y^2 \]
\[ c^2 = 25x^2 + 50xy + 25y^2 \]
Now, substitute these into the cosine formula:
\[ \cos C = \frac{(25x^2 + 48xy + 25y^2) - (25x^2 + 50xy + 25y^2)}{2ab} \]
Simplifying this gives:
\[ \cos C = \frac{(25x^2 + 48xy + 25y^2 - 25x^2 - 50xy - 25y^2)}{2ab} = \frac{-2xy}{2ab} = \frac{-xy}{ab} \]

4. **Determine the sign of \( \cos C \):**
Since \( x > 0 \) and \( y > 0 \), it follows that \( -xy < 0 \). Therefore, \( \cos C < 0 \).

5. **Conclusion about angle \( C \):**
If \( \cos C < 0 \), then angle \( C \) is greater than \( 90^\circ \). This indicates that triangle \( ABC \) is an obtuse triangle.

### Final Answer:
The triangle is an **obtuse triangle**.