Question

If an area enclosed by a circle or a square or an equilateral triangle is the same, Then the maximum perimeter is possessed by:
(a) The circle
(b) The square
(c) The equilateral triangle
(d) The triangle and square have equal perimeters greater than that of the circle.

Answer 1

Hint: Start by letting the area of the given figures be A. Start by finding the side of the square using the fact that area of square is equal to square of its sides, then the radius of the circle using the formula $A=\pi {{r}^{2}}$ and the side of the equilateral triangle using the formula that area of equilateral triangle is equal to $\dfrac{\sqrt{3}}{4}{{l}^{2}}$ . Now use the formula of perimeters of each figure and compare the result to get the answer.

Complete step by step solution:
Let us start by letting the area of each of the figures be A. We also let the side of the square be a, side length of the triangle be l and the radius of the circle be r.
First we will try to find the Perimeter of the square in terms of A.

 We know the area of the square is equal to the square of its sides.
$\begin{align}
  & A={{a}^{2}} \\
 & \Rightarrow a=\sqrt{A} \\
\end{align}$
We also know that the perimeter of the square is 4 times its side length. And if we use the result $a=\sqrt{A}$ , we get
$\text{Perimeter of square}=4a=4\sqrt{A}............(i)$
Now, we will try to find the Perimeter of the equilateral triangle in terms of A.

 We know the area of the equilateral triangle with side l is equal to $\dfrac{\sqrt{3}}{4}{{l}^{2}}$ .
$\begin{align}
  & A=\dfrac{\sqrt{3}}{4}{{l}^{2}} \\
 & \Rightarrow {{l}^{2}}=\dfrac{4A}{\sqrt{3}} \\
 & \Rightarrow l=\sqrt{\dfrac{4A}{\sqrt{3}}} \\
\end{align}$
We also know that the perimeter of the equilateral triangle is 3 times its side length. And if we use the result $l=\sqrt{\dfrac{4A}{\sqrt{3}}}$ , we get
$\text{Perimeter of triangle}=3l=3\sqrt{\dfrac{4A}{\sqrt{3}}}=\sqrt{12\sqrt{3}A}\simeq 4.51\sqrt{A}.........(ii)$
Finally, we will try to find the Perimeter of the circle in terms of A.

 We know the area of the circle with radius r is equal to $\pi {{r}^{2}}$ .
$\begin{align}
  & A=\pi {{r}^{2}} \\
 & \Rightarrow \dfrac{A}{\pi }={{r}^{2}} \\
 & \Rightarrow \sqrt{\dfrac{A}{\pi }}=r \\
\end{align}$
We also know that the perimeter of the circle is given by $2\pi r$ . And if we use the result $\sqrt{\dfrac{A}{\pi }}=r$ , we get
$\text{Perimeter of circle}=2\pi r=2\pi \sqrt{\dfrac{A}{\pi }}=2\sqrt{\pi A}=2\times 1.77\sqrt{A}=3.54\sqrt{A}.....(iii)$
Looking at the three equations (i), (ii) and (iii), we can easily conclude that the equilateral triangle has the largest perimeter among the three. Also, if we see the option (d), the perimeter of the square and triangle are greater than that of the circle, but the perimeter of the square and the triangle is not equal, so option (d) is not the correct option.
Hence, the answer to the above question is option (c).

Note: You need to remember the approximate values of square roots of single digit numbers as they are used very often. Also, you should be familiar with the value of $\pi $ , i.e., $\dfrac{22}{7}\text{ or 3}\text{.14}$ . Remembering the formulas of areas and perimeter of different geometric figures is also important. Also, if you have noticed, I have represented the perimeter of each of the shapes in terms of A, which makes it easier to compare, as we just have to compare the real term that is getting multiplied with the variable.