Let l > 0 be a real number, C denote a circle with circumference l an T denote a triangle with perimeter l, Then

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Given any positive real number $α$ , we can choose C and T as above such that ratio $\frac{a r e a (C)}{a r e a (T)}$ is less than ‘ $α$ ’

Given any positive real number $α$ , we can choose C and T as above such that ratio $\frac{a r e a (C)}{a r e a (T)}$ is greater than ‘ $α$ ’

There exists real numbers ‘a’ and ‘b’ such that for any circle C and Triangle T as above must have a < $\frac{a r e a (C)}{a r e a (T)}$ < b.

Given any C and T as above the ratio $\frac{a r e a (C)}{a r e a (T)}$ is independent of C and T

answer is A.

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Detailed Solution

It is given that circumference of circle C is l and the perimeter of triangle T is l. Now, let the radius of circle C is r, so $2 π r = l \Rightarrow r = \frac{l}{2 π}$
area of circle C is $A_{1} = π r^{2} = \frac{l^{2}}{4 π}$
Now, as we know that area of triangle will be maximum for given perimeter if it is an equilateral triangle, let the length of side of equilateral triangle is ‘a’, then $3 a = l \Rightarrow a = \frac{l}{3}$ and area of equilateral triangle is $\begin{array}{l} A_{2} = \frac{\sqrt{3}}{4} a^{2} \\ S o, A_{2} = \frac{\sqrt{3}}{4} (\frac{l^{2}}{9}) \frac{l^{2}}{12 \sqrt{3}} \\ ∵ \frac{A_{1}}{A_{2}} = \frac{\frac{l^{2}}{4 π}}{\frac{l^{2}}{12 \sqrt{3}}} = \frac{3 \sqrt{3}}{π} > 1 \end{array}$
Since, as we took an equilateral triangle, which has maximum area. But we can take a triangle T such that the ratio $\frac{a r e a (C)}{a r e a (T)}$ is greater than any positive real number $α$