Question

If \( A(a,0) \), \( B(0,0) \), and \( C(0,b) \) are the vertices of \( \triangle ABC \), then the area of \( \triangle ABC \) is:

• A. \( ab \)
• B. \( \frac{1}{2} ab \)
• C. \( \frac{1}{2} a^2 b^2 \)
• D. \( \frac{1}{2} b^2 \)

Hint:

For right-angled triangles: - Base is the horizontal side. - Height is the vertical side.

Answer 1

The Correct Option is B

Step 1: Using the area formula for a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height}. \] Step 2: Identifying base and height:
\[ \text{Base} = a, \quad \text{Height} = b. \] Step 3: Computing area:
\[ \text{Area} = \frac{1}{2} \times a \times b = \frac{1}{2} ab. \]