A) circle
B) square
C) equilateral triangle
D) Both triangle and square
Correct Answer: C
Solution :
| [c] \[\pi {{R}^{2}}={{s}^{2}}=\frac{\sqrt{3}}{4}{{a}^{2}}=A\] \[\Rightarrow \] \[R=\sqrt{\frac{A}{\pi }},\]\[s=\sqrt{A}\] and \[a=\sqrt{\frac{4A}{\sqrt{3}}}\] Perimeter of circle\[=2\pi R=2\pi .\sqrt{\frac{A}{\pi }}\] \[=2\sqrt{\pi }.\sqrt{A}=2\times \sqrt{3.14}\times \sqrt{A}\] \[=(2\times 1.77\times \sqrt{A})=(3.54\times \sqrt{A})\] Perimeter of square \[=4\sqrt{A}\] Perimeter of triangle \[=3a\] \[=3\sqrt{\frac{4A}{\sqrt{3}}}\] \[=\frac{6}{{{3}^{1/4}}}\sqrt{A}=\frac{{{3}^{3/4}}}{4}\cdot 2\sqrt{A}\] \[={{(27)}^{1/4}}\cdot 2\sqrt{A}>4\sqrt{4}\] \[\therefore \]Perimeter of triangle is maximum. |
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