Question

The plane \(\frac{x}{2}+\frac{y}{3}+\frac{z}{4} = 1\) cuts the co-ordinate axes at the points A, B, C respectively. Then the area of triangle ABC is

• A. \(\sqrt{61}\) sq. units
• B. \(\frac{\sqrt{61}}{2}\) sq. units
• C. \(\frac{\sqrt{61}}{4}\) sq. units
• D. \(\frac{\sqrt{71}}{2}\) sq. units

Hint:

Find the intercept points A, B and C, then compute half the magnitude of the cross product of two sides.

Answer 1

The Correct Option is A

Step 1: Use projections.
The areas of the projections of triangle ABC on the coordinate planes are $\tfrac{1}{2}\cdot 2\cdot 3 = 3$ (xy), $\tfrac{1}{2}\cdot 3\cdot 4 = 6$ (yz) and $\tfrac{1}{2}\cdot 2\cdot 4 = 4$ (zx).

Step 2: Combine.
The true area is the root of the sum of squares of the projected areas.
\[ \text{Area} = \sqrt{3^2 + 6^2 + 4^2} = \sqrt{9 + 36 + 16} = \sqrt{61} \]

Final Answer:
Option (A). \[ \boxed{\sqrt{61}\text{ sq. units}} \]