Question

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

• A. \(\frac{1}{2}abc\)
• B. \(\frac{1}{2}(ab+bc+ca)\)
• C. \(\frac{1}{2}(ab^2+b^2+c^2)\)
• D. 0

Answer 1

The Correct Option is D

Step 1: Recall the formula for the area of a triangle given its vertices.

The area of a triangle with vertices \( (x_1, y_1) \), \( (x_2, y_2) \), and \( (x_3, y_3) \) is given by:

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

Step 2: Substitute the coordinates of the vertices into the formula.

The vertices are \( (a, b+c) \), \( (b, c+a) \), and \( (c, a+b) \). Substituting these into the 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 each term in the expression.

Expand the terms inside the absolute value:

\[ a((c+a) - (a+b)) = a(c + a - a - b) = a(c - b), \]

\[ b((a+b) - (b+c)) = b(a + b - b - c) = b(a - c), \]

\[ c((b+c) - (c+a)) = c(b + c - c - a) = c(b - a). \]

Substitute these back into the formula:

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

Step 4: Combine like terms.

Expand the terms:

\[ a(c - b) = ac - ab, \quad b(a - c) = ab - bc, \quad c(b - a) = bc - ca. \]

Combine all terms:

\[ ac - ab + ab - bc + bc - ca = 0. \]

Step 5: Conclude the result.

The area simplifies to:

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

Final Answer: The area of the triangle is \( \mathbf{0} \), which corresponds to option \( \mathbf{(4)} \).