The area of a triangle with vertices (a,b+c), (b,c+a) and (c,a+b) is

Options

A. `(a+b+c)^(2)`
B. 0
C. (a + b + c)
D. abc

Explanation

To find the area of the triangle with vertices at the points \( (a, b+c) \), \( (b, c+a) \), and \( (c, a+b) \), we can use the formula for the area of a triangle given its vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

\[ \text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right| \]

### Step 1: Assign the vertices
Let:
- \( A = (a, b+c) \) → \( (x_1, y_1) = (a, b+c) \)
- \( B = (b, c+a) \) → \( (x_2, y_2) = (b, c+a) \)
- \( C = (c, a+b) \) → \( (x_3, y_3) = (c, a+b) \)

### Step 2: Substitute the coordinates into the area formula
Substituting the coordinates into the area formula:

\[ \text{Area} = \frac{1}{2} \left| a((c+a) - (a+b)) + b((a+b) - (b+c)) + c((b+c) - (c+a)) \right| \]

### Step 3: Simplify the expressions
Now, simplify each term inside the absolute value:

1. For the first term:
\[ a((c+a) - (a+b)) = a(c - b) \]

2. For the second term:
\[ b((a+b) - (b+c)) = b(a - c) \]

3. For the third term:
\[ c((b+c) - (c+a)) = c(b - a) \]

Now, substituting these back into the area formula gives:

\[ \text{Area} = \frac{1}{2} \left| a(c-b) + b(a-c) + c(b-a) \right| \]

### Step 4: Combine the terms
Combining the terms:

\[ = \frac{1}{2} \left| ac - ab + ab - bc + cb - ca \right| \]

Notice that \( ac \) and \( -ca \) cancel out, and \( -ab \) and \( ab \) cancel out, leaving us with:

\[ = \frac{1}{2} \left| 0 \right| = 0 \]

### Conclusion
Thus, the area of the triangle with vertices \( (a, b+c) \), \( (b, c+a) \), and \( (c, a+b) \) is:

\[ \text{Area} = 0 \]

This indicates that the three points are collinear.