Question

The area of the region bounded by the lines \( \frac{x}{7\sqrt{3a}} + \frac{y}{b} = 4 \), \( x = 0 \), and \( y = 0 \) is:

• A. \( \frac{56}{\sqrt{3ab}} \)
• B. \( 56a \)
• C. \( \frac{ab}{2} \)
• D. \( 3ab \)

Answer 1

The Correct Option is A

The equation of the line is:

\[ \frac{x}{7\sqrt{3a}} + \frac{y}{b} = 4. \]

Rewriting:

\[ b \cdot x + 7\sqrt{3a} \cdot y = 28\sqrt{3ab}. \]

To find the intercepts:

1. When \( x = 0 \), \( y = 4b \).
2. When \( y = 0 \), \( x = 28\sqrt{3a} \).

The lines \( x = 0 \) and \( y = 0 \), together with \( \frac{x}{7\sqrt{3a}} + \frac{y}{b} = 4 \), form a triangle with vertices:

\[ (0, 0), \quad (28\sqrt{3a}, 0), \quad (0, 4b). \]

The area of the triangle is:

\[ \text{Area} = \frac{1}{2} \times \text{base} \times \text{height}, \]

where the base is \( 28\sqrt{3a} \) and the height is \( 4b \):

\[ \text{Area} = \frac{1}{2} \times (28\sqrt{3a}) \times (4b). \]

Simplify:

\[ \text{Area} = \frac{1}{2} \times 112\sqrt{3ab} = 56\sqrt{3ab}. \]

Thus, the area of the region is:

\[ 56\sqrt{3ab}. \]