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 will analyze the sides and apply the triangle inequality and properties of angles.

### Step 1: Assign the sides
Let:
- \(a = 3x + 4y\)
- \(b = 4x + 3y\)
- \(c = 5x + 5y\)

### Step 2: Calculate the squares of the sides
We will calculate \(a^2\), \(b^2\), and \(c^2\):
- \(a^2 = (3x + 4y)^2 = 9x^2 + 24xy + 16y^2\)
- \(b^2 = (4x + 3y)^2 = 16x^2 + 24xy + 9y^2\)
- \(c^2 = (5x + 5y)^2 = 25x^2 + 50xy + 25y^2\)

### Step 3: Check the relationship between the sides
To determine the type of triangle, we will check if the triangle is acute, right, or obtuse by using the cosine rule:
\[ \cos C = \frac{a^2 + b^2 - c^2}{2ab} \]

### Step 4: Calculate \(a^2 + b^2 - c^2\)
Now, we calculate \(a^2 + b^2 - c^2\):
\[ a^2 + b^2 = (9x^2 + 24xy + 16y^2) + (16x^2 + 24xy + 9y^2) = 25x^2 + 48xy + 25y^2 \]
\[ c^2 = 25x^2 + 50xy + 25y^2 \]
Now, subtract \(c^2\) from \(a^2 + b^2\):
\[ a^2 + b^2 - c^2 = (25x^2 + 48xy + 25y^2) - (25x^2 + 50xy + 25y^2) = -2xy \]

### Step 5: Substitute into the cosine formula
Now substitute into the cosine formula:
\[ \cos C = \frac{-2xy}{2(3x + 4y)(4x + 3y)} \]
Since \(x > 0\) and \(y > 0\), \(xy > 0\). Thus, \(-2xy < 0\).

### Step 6: Determine the sign of \(\cos C\)
The denominator \(2(3x + 4y)(4x + 3y)\) is positive because both \(3x + 4y > 0\) and \(4x + 3y > 0\).

Thus, we have:
\[ \cos C < 0 \]
This implies that angle \(C\) is obtuse (greater than \(90^\circ\)).

### Conclusion
Since angle \(C\) is obtuse, the triangle formed by the sides \(3x + 4y\), \(4x + 3y\), and \(5x + 5y\) is an obtuse triangle.

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