Compare the two quantities

First, since we are comparing areas, we can pick numbers and make our calculations easier. Suppose the radius of the circle is $2$ , thus the circumference is $4\pi$ which is equal to the perimeter of the square.

Thus, the side of the square is $\pi$ .

The area of the square is $\pi^2= 9.8696$ .

The area of the circle is $4 \times \pi = 12.5664$ .

We can conclude that the area of the circle is greater.

But we could also solve this more abstractly. Suppose the perimeter of the square and the circumference of the circle both equal to $4x$ . Then each side of the square would be $x$ and its area would be $x^2$ . The radius of the circle would be $\frac{4x}{2\pi}$ or $\frac{2x}{\pi}$ , so its area would be $\pi (\frac{2x}{\pi})^2$ or $\frac{4}{\pi}$ $x^2$ . Now, we can compare the area of the square, $x^2$ to the area of the circle $\frac{4}{\pi}$ $x^2$ .

Since $1< \frac{4}{\pi}$ , multiply both sides by $x^2$ to obtain $x^2 < \frac{4}{\pi} x^2$ .

Thus, the area of the circle is greater.

$4a=2\pi r$

$a=\dfrac{2\pi r}{4}=\dfrac{\pi r}{2}$

$Area_{\square}=a^2=\left(\dfrac{\pi r}{2}\right)^2=\dfrac{\pi^2r^2}{4}\approx2.48r^2<3.14r^2\approx\pi r^2=Area_{\bigcirc}$

Note: The picture is not drawn to scale.

One way of solving problems like this is to set a value for the radius of the circle or side length of the square.

Set $radius=1$ . Then the circumference is $2 \pi r=2 \pi (1) \approx 6.283185307$ and the area is $\pi r^2=\pi (1^2) \approx 3.141592654$

Since they have equal perimeters, compute for the side length of the square from the circumference of the circle.

$6.283185307=4x$ $\implies$ $x \approx 1.570796327$

So the area of the square is $1.570796327^2 \approx 2.4674011$

It can be seen that the area of the circle is larger than the area of the square.

$$ Let $r$ be the radius of the circle and $a$ be the side of the square.

We know that $P_c = 2 \pi r$ and $P_s = 4a$ are the same, and we want to know which area is larger: that of the circle or of the square.

We know that $A_c = \pi r^2$ and $A_s = a^2$ . Since we don't know the values of $r$ or $a$ , we rewrite them as:

$r = \frac{P_c}{2 \pi}$ and $a = \frac{P_s}{4}$ . Which allows to rewrite our area equations as:

$A_c = \pi (\frac{P_c}{2 \pi}) \ ^2$ and $A_s = (\frac{P_s}{4}) \ ^2$ .

Which leaves us with:

$A_c = \frac{P_c ^2}{4 \pi} \$ and $A_s = \frac{P_s ^2}{4^2} \$ .

Since $4 \pi < 4^2$ , we can conclude that when $P_c = P_s$ , then $A_c > A_s$ for any value of $P$ .

Let $r$ be the radius of the circle and $l$ the side of the square.

${ Perimeter }_{ circle }={ Perimeter }_{ square }\Leftrightarrow 2\pi r=4l\Leftrightarrow \pi r=2l\Leftrightarrow { \pi }^{ 2 }{ r }^{ 2 }=4{ l }^{ 2 }\Leftrightarrow \pi { A }_{ circle }=4{ A }_{ square }\Leftrightarrow { A }_{ circle }=\frac { 4 }{ \pi } { A }_{ square }>{ A }_{ square }\quad \blacksquare$

suppose,both of their perimeter is 1.

in case of square,4x=1

         0r,x=1/4=.25

in case of circle, 2πr=1

             or, r=1/2π=.16

now,the area of square is, x^2=(o.25)^2=1/16=0.06

and the area of circle is, πr^2=0.08

so,the area of circle is bigger.

Let $r$ be the radius of the circle and $x$ be the side of the square.

Let $P_{s}$ be perimeter of the square and $P_{c}$ be perimeter of the circle.

$P_{s} = P_{c} \implies 4x = 2\pi r \implies r = \dfrac{2x}{\pi} \implies Area_{\bigcirc} = \dfrac{4x^2}{\pi} > x^2 = Area_{\square}$

The isoperimetric inequality states that, given any simple closed curve in the plane (like both the circumpherence and the square are) with a fixed perimeter, the circumpherence is the curve that surrounds a bigger area.

https://en.wikipedia.org/wiki/Isoperimetric_inequality

I just remembered hearing years ago that a sphere maximizes volume based on surface area. Not sure why. So I just applied the same logic to conclude that the circle must maximize area based on perimeter. :-)

Suppose the perimeter of both shapes is 4 units.

Each side of the square is therefore 1 unit, so the square has an area of 1 $\times$ 1, i.e. 1.

The perimeter of the circle is $2\pi r$ . So $2\pi r$ = 4, and therefore $r$ = $\frac{4}{2\pi}$ = $\frac{2}{\pi}$ .

The area of the circle is $\pi r^{2}$ . Since $r$ = $\frac{2}{\pi}$ , the area is therefore $\pi \times \left(\frac{2}{\pi}\right)^{2}$ = $\pi \times \frac{4}{\pi^{2}}$ = $\frac{4}{\pi}$ = 1.27, which is larger than the area of the square.

$p = perimeter$

Area of square = $(p/4)^2$ = $p^2/16$

Perimeter of circle = $2*r*pi$

$p = 2*r*pi$

r = $p/2*pi$

Area of circle = $r^2*pi$ = $(p/2*pi)^2*pi$ = $p^2/4*pi$

$p^2/16 < p^2/4*pi$ -> Area of square < Area of circle

Let $r$ be the radius of the circle and $x$ be the side length of the square.

Now, $2\pi r = 4x \implies \pi r = 2x \implies \pi r^2 = \dfrac{\left(2x\right)^2}{\pi} = \dfrac4\pi x^2 > x^2$ , since $0 < \pi < 4$ .

Hence, the circle has larger area.

A circle's area is $A = 2\pi r$ , and its perimeter is $P=2\pi r$ . A rectangle's area is $A = s^2$ (where s = side length), and its perimeter is $P=4s$ .

$2\pi r=4s$ . Some math later... $r=(2s/\pi)$ .

Thus, the A of the circle is $4s^2/\pi$ , which is bigger than $s^2$ (since $4/\pi > 1$ ).

Let the side length of the square be $m$ and radius of the circle be $r$ . We have

$\begin{cases} 2\pi r=4m \\ \pi r^2 :m^2\end{cases}$

From the first eq. we have, by substitution:

$\Rightarrow m=\frac { \pi r }{ 2 } \Leftrightarrow m^2=\frac { \pi^2 r^2 }{ 4 }$

Plugging into the 2nd eq. ,as RHS represents the square's area and LHS represents the circle's area:

$\pi r^{ 2 }:\frac { \pi ^{ 2 }r^{ 2 } }{ 4 } \\ \Rightarrow 4\pi r^{ 2 }:\pi ^{ 2 }r^{ 2 }\\ \Leftrightarrow 4:\pi$

Now from this we conclude, easily, that the $LHS > RHS$ , or in other words:

$\boxed{A_{circle} > A_{square}}$

let r= radius of circle and a=side length of square. $2*pi*r=4*a$ (Given) implies that $(pi/2)=(a/r)$ squaring both sides, $(pi/2)^2 =(a/r)^2$ dividing the above equation by pi $(pi/4)=(a^2)/(pi*r^2)$ since $pi$ is less than $4$ , thus area of the circle is larger than the square.

How can I post my own questions ?

2pi r =4L
r/L = 4/(2pi) r =0.6366L

area of square L^2 area of circle pi r^2 =pi x (0.6366L)^2 = 1.273L^2

therefore a if a circle and a square have the same perimeter the circle will always have an area 27.3% larger than the square

Let the side of square be x and radius of the circle r. We are given 4x = 2πr, 2x = πr, and r = 2x/π. Then πr^2 = 2xr, or πr^2 = 4x^2/π. Since 4/π > 1, then πr^2 > x^2, so the area of the circle is larger.

Let's consider π = 3 for this problem to make the calculation easier.

It is known that perimeter of the circle is = 2πr = 6r, (r is radius). It is known that the perimeter of the square is = 4L, (L is each side of the square).

Since they have same perimeter, 4L = 6r and from this we get that L = $\frac{3r}{2}$ .

It is known that the area of the circle is πr², and the area of the square is L². Now let's suppose they have equal area and if we are wrong we will find what side of the equation is bigger. So 3r² = $\frac{9r²}{4}$ , and from this we get 3r² = 2.25r². -> 3r² > 2.25r²

So, for this problem, the area of the circle is larger ∀ r ∈ R.

For each shape, consider the perimeter P and area A.

For the square:

$P=4 * l$

$A=l^2$

Therefore A = $\frac{P^2}{4 * 4}$

For the circle:

$P=2 \pi r$

$A = \pi r^2$

Therefore A = $\frac{P^2}{4 \pi}$

$\pi < 4$ , so $A(circle)>A(square)$

El área del cuadrado esta definida por $\frac{P}{4}$ El área del circulo esta definida por $\frac{P}{π}$ Donde P, es el perímetro. Sabiendo que el perímetro es el mismo para ambos caso podemos crear la siguiente formula que nos permite ilustrar la relación de las áreas: A= $\frac{P}{ 4 or π }$ La relación del área es inversamente proporcional a la del denominador de la igualdad, por lo tanto al ser π (3.14...) menor a 4, entonces deducimos que la figura que mayor área tendrá sera la del circulo, debido a que el denominador relacionado en su ecuación es mas pequeño, que el del área del cuadrado, por lo tanto su área sera mayor.

Let P be the perimeter that is equal for both figures.

Square side length is $P/4$ , area is $P^2/16$

Circle diameter is $P/\pi$ , radius is $P/2\pi$ , area is $\pi (P/2\pi)^2 = P^2/ 4\pi$ ,

Since $\pi < 4$ it follows that $4(\pi) < 4(4) = 16$ and that the circle area is greater than the square area.

Given any closed 2 dimensional figure with a fixed value of perimeter the area of CIRCLE is the most .