Question

Let $l>0$ be a real number, $C$ denote a circle with circumference $l,$ and $T$ denotes a triangle with perimeter $l.$ Then.

• A. given any positive real number $\alpha ,$ we can choose $C$ and $T$ as above such that the ratio $\frac{\text{area}\left(C\right)}{\text{area}\left(T\right)}$ is greater than $\alpha$
• B. given any positive real number $\alpha ,$ we can choose $C$ and $T$ as above such that the ratio $\frac{\text{area}\left(C\right)}{\text{area}\left(T\right)}$ is less than $\alpha$
• C. given any $C$ and $T$ as above, the ratio $\frac{\text{area}\left(C\right)}{\text{area}\left(T\right)}$ is independent of $C$ and $T$
• D. There exist real numbers $a$ and $b$ such that for any circle $C$ and triangle $T$ as above, we must have $a<\frac{\text{area}\left(C\right)}{\text{area}\left(T\right)}<b$

Answer 1

The correct option is A given any positive real number $\alpha ,$ we can choose $C$ and $T$ as above such that the ratio $\frac{\text{area}\left(C\right)}{\text{area}\left(T\right)}$ is greater than $\alpha$

For any perimeter $l$,
Let the radius of circle be $r$
Then, $2\pi r=l\Rightarrow r=\frac{l}{2\pi }$
$\text{Area of circle}=\pi \times \frac{l^2}{4{\pi }^2}=\frac{l^2}{4\pi }=$ constant

$0<\text{Area}\left(T\right)<\infty \Rightarrow 0<\frac{1}{\text{Area}\left(T\right)}<\infty \Rightarrow 0<\frac{\text{Area}\left(C\right)}{\text{Area}\left(T\right)}<\infty (\text{As the area of circle is constant)}$

Now, For any perimeter $l$, the area of circle is greater than the area of triangle
$\Rightarrow \frac{\text{Area}\left(C\right)}{\text{Area}\left(T\right)}>1$

From both equations,
$\Rightarrow 1<\frac{\text{Area}\left(C\right)}{\text{Area}\left(T\right)}<\infty$

So, for any given $\alpha$, we can choose the triangle in such way that the ratio $\frac{\text{area}\left(C\right)}{\text{area}\left(T\right)}$ is greater than $\alpha$