Question

The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the coordinate axes in $A,B,C$, then the area of the $△ABC$ is :

• A. $\sqrt{29}$ sq unit
• B. $\sqrt{41}$ sq unit
• C. $\sqrt{61}$ sq unit
• D. none of these

Answer 1

Answer: (C)

If the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
cuts the coordinate axes in $A,B$ and $C$, then area of triangle $ABC$
$=\frac{1}{2}\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}$
Given equation of plane is
$\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$
On comparing with $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
$\Rightarrow a=2,b=3,c=4$
$\therefore$ Area of $\Delta ABC=\frac{1}{2}\sqrt{a^2{b}^2+b^2{c}^2+c^2{a}^2}$
$=\frac{1}{2}\sqrt{2^2\cdot 3^2+3^2\cdot 4^2+4^2\cdot 2^2}$
$=\frac{1}{2}\sqrt{36+144+64}$
$=\frac{1}{2}\sqrt{244}=\frac{1}{2}\cdot 2\sqrt{61}$
$=\sqrt{61}$