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If an area enclosed by a circle or a square or an equilateral triangle is the same, Then the maximum perimeter is possessed by:

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The circle

The square

The equilateral triangle

The triangle and square have equal perimeters greater than that of the circle.

answer is D.

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Detailed Solution

First let the area of each figure be A. We also allow the side of the place to be a, the length of the side of the triangle to be

l

and the radius of the circle to be r.
Let's first try to find the perimeter of the square in terms of A.
Question Image

We know that the area of the square is equal to the square of its sides.

A = a^{2}

\Rightarrow a = \sqrt{A}

We also know that the perimeter of the square is four times its side length.
And if we use the outcome,

a = \sqrt{A}

, we get
Perimeter of square =

4 a = 4 \sqrt{A}

...........(i)
Now, let's attempt to find the perimeter of the equilateral triangle in terms of A.
Question Image

We know that the equilateral triangle with side

l

has an area equal to

\frac{\sqrt{3}}{4} l^{2}

A = \frac{\sqrt{3}}{4} l^{2}

⇒

l^{2}

\frac{4 A}{\sqrt{3}}

⇒

l^{}

\sqrt{\frac{4 A}{\sqrt{3}}}

The perimeter of the equilateral triangle is also known to be 3 times its lateral length.
And if we use the outcome,

l^{}

\sqrt{\frac{4 A}{\sqrt{3}}}

, we get
Perimeter of triangle = 3

l^{}

= 3

\sqrt{\frac{4 A}{\sqrt{3}}}

\sqrt{12 \sqrt{3} A} ≃

4.51

\sqrt{A}

…….. (ii)
Finally, let's try and find the perimeter of the circle in terms of A.
Question Image

The area of the circle with radius r is known to be πr2. .

A = π r^{2}

⇒

\frac{A}{π}

= r2

\sqrt{\frac{A}{π}} = r

We also know that the circumference of the circle is given by 2 πr.
And if we use the outcome,

\sqrt{\frac{A}{π}} = r

, we get
Perimeter of circle =

2 πr

= 2π

\sqrt{\frac{A}{π}}

= 2π

\sqrt{πA}

=2×1.77

\sqrt{A}

=3.54

\sqrt{A}

……. (iii)
Looking at the three equations (i), (ii) and (iii), it is easy to conclude that the equilateral triangle has the largest perimeter of the three.
This means that the answer to the above question is option 3.

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