Quantitative Aptitude Ques 867
Question: A copper wire is bent in the form of an equilateral triangle and has an area $ 121\sqrt{3}cm^{2}. $ If the same wire is bent into the form of a circle, then the area $ (incm^{2}) $ enclosed by the wire is
Options:
A) 364.5
B) 346.5
C) 693.5
D) 639.5
Show Answer
Answer:
Correct Answer: B
Solution:
- Let a be the side of equilateral triangle.
$ \therefore $ $ \frac{\sqrt{3}}{4}a^{2}=121\sqrt{3} $
$ \Rightarrow $ $ a^{2}=121\times 4 $
$ \Rightarrow $ $ a=11\times 2=22cm $
$ \therefore $ Perimeter of triangle $ =3\times 22=66cm $
$ \Rightarrow $ Circumference of circle = Perimeter of equilateral triangle
$ \Rightarrow $ $ 2\pi r=66 $
$ \Rightarrow $ $ r=\frac{66\times 7}{2\times 22}=\frac{21}{2}=10.5 $ Hence, area of circle $ =\pi (r^{2}) $ $ =\frac{22}{7}\times {{(10.5)}^{2}}=346.5cm^{2} $
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