Question

If the plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1$ cuts the axes of coordinates at points $A,B$ and $C$, then find the area of the triangle $ABC$.

• A. $18$ sq unit.
• B. $36$ sq unit.
• C. $3\sqrt{14}$ sq unit.
• D. $2\sqrt{14}$ sq unit

Answer 1

Answer: (C)

$3\sqrt{14}$ sq unit.

Coordinates of $A,B$ and $C$ are $(2,0,0),(0,3,0)$ and $(0,0,6)$
Using formula, Area of triangle formed by intersection of axes and a plane $=\frac{1}{2}\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}$
So, area of triangle $ABC$ is $=\frac{1}{2}\sqrt{2^2{.3}^2+3^2{.6}^2+6^2{.2}^2}=3\sqrt{14}$sq unit.