A) \[\sqrt{29}\] sq unit
B) \[\sqrt{41}\,\text{sq}\,\text{unit}\]
C) \[\sqrt{61}\,\text{sq}\,\text{unit}\]
D) none of these
Correct Answer: C
Solution :
Key Idea: 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 \[\Delta ABC=\frac{1}{2}\sqrt{{{a}^{2}}{{b}^{2}}+{{b}^{2}}{{c}^{2}}+{{c}^{2}}{{a}^{2}}}\] \[=\frac{1}{2}\sqrt{{{2}^{2}}{{.3}^{2}}+{{3}^{2}}{{.4}^{2}}+{{4}^{2}}{{.2}^{2}}}\] \[=\frac{1}{2}\sqrt{36+144+64}\] \[=\frac{1}{2}\sqrt{244}=\frac{1}{2}.2\sqrt{61}\] \[=\sqrt{61}\]
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