I was given a question which was like:

Suppose that a circle, square, rectangle and triangle have same perimeter. How area their areas related??

My work:

I broke the question in parts and tried to prove it seperately:

STEP $1$. Suppose the given perimeter is $P$.So the radius of circle will be ${r=P/2\pi}$.

And hence area of the circle will be: ${P^2/4\pi}$.

Side of square will be $P/4$ and hence area of square will be $P^2/16$.

It is obvious that ${P^2/4\pi}$ > $P^2/16$.

So Area of circle is more than area of square.

STEP $2$: Whether by using $AM>GM$ inequality or by using a bit of differentiation, we can get that product off two quantities is maximum when they are equal.

So, Area of square is more than Area of rectangle.

Here, the hardest part come (at least for me). I can't relate the area of triangle to any of Circle, Square or Rectangle. It would have been easy if the triangle was Equilateral, Right angle but the question is talking about a general triangle.

However, when I considered the situation with different examples (Assuming the perimeter and calculating the area, I got the relation:

Circle > Square > Rectangle > Triangle.

I shall be highly thankful if you guys can establish the relation in purely mathematical way.

EDIT: This question does not ask that whether circle have the largest area among all 2-D figures or not. The OP just want to relate the areas of Circle, Square, Rectangle, Triangle in a pure mathematical way.

  • 1
    $\begingroup$ The circle contains the greatest area of any closed curve of fixed perimeter, is that what you are asking? See the Isoperimetric Inequality. $\endgroup$ – lulu Nov 25 '16 at 13:27
  • $\begingroup$ @lulu No. I know this already. I just want to relate the areas of four figures that I have mentioned, in a completely mathematical way. I have also read Isoperimetric Inequality before posting the question. $\endgroup$ – I am Back Nov 25 '16 at 13:29
  • $\begingroup$ Well, then I don't understand the question. Specifying the perimeter of a rectangle (or triangle) does not specify the area. $\endgroup$ – lulu Nov 25 '16 at 13:30
  • $\begingroup$ But we still know that rectangle have less area than square (with a complete mathematical way). $\endgroup$ – I am Back Nov 25 '16 at 13:32
  • $\begingroup$ Yes, of course. And, similarly, equilateral triangles have the greatest area amongst triangles of fixed perimeter, see, e.g. this. $\endgroup$ – lulu Nov 25 '16 at 13:34

I think you are operating under the mistaken and oversimplified belief that the perimeter of a shape provides more information than it does in determining the area of the shape.

In theory, you need to know every thing about the curve of the shape-- not just it's length/perimeter but every twist turn and angle.

Of course, if the shape has a general pattern or formula than one can probably determine a lot of this information for relatively little data. For example if you know the shape is a polygon, it's enough to know the lengths of all of the sides and the measurement of all the angles. And if you know all but one of the angles you can determine that final angle and if you know all the angles and all but one of the sides you can determine the final side.

But, in general you need to know a lot more than just the total perimeter and the number of sides.

But then again if you know more information about the shape you can often determine more information. If for example you know it is an equilateral $n$-gon and you know the perimeter is $P$ you know the sides are all $P/n$. (However if $n > 3$ you don't know the angles so you don't know the area.) If you know it is an equiangular $n$-gon you know the the total angle and there for every angle. If you know it is a rectangle then you know all the angles are right angles and it is a parallelagram so all you need for the area is the two side (width and height). If you know the perimeter you only need one side and you can determine the other side.

So, your question.

Perimeter = $P$.

Triangle: With sides $a,b,c: a+b+c = P$ and angles $A,B,C: A+B+C = 180$. Angle $A$ is opposite $a$, $B$ opposite $b$ and $C$ opposite $c$.

We don't have enough information. If we had two sides, we can get the third from the perimeter and from that we can use trig identities to get the angles. Or if we had two angles, we could get the third and then we could get the trig propotional lengths and from the perimeter we can get the sides. Or if we had one side and one angle using trig and the restriction of the perimeter we can the rest.

Basically: If we have $2$ out of the six pieces of information and the perimeter, we can get the remaining four pieces of information, and from the six pieces of information we can get the area.

The six pieces of information are related via:

$\frac {\sin A}a = \frac {\sin B}b = \frac{\sin C}c; A+B+C = 180; a+b+c = P$ and also $a^2 + b^2 + 2ab\cos C = c^2$, $b^2 + c^2 + 2bc\cos A = a^2$, or $a^2 + c^2 + 2ac\cos B$ . (Any two of the variables will get you the remaining four.)

The area of the triangle = $\frac 12 *base*height = \frac 12 *a* ( {\sin C}*b)$ or any of the other combinations of two sides and the angle between them.

Equilateral triangle:

In this case $s= a = b = c = P/3$ and $A = B = C = 60$ so the area is $\frac 12(\frac P3)^2\sin 60 = (\frac P3)^2\frac{\sqrt{3}}4= s^2 \frac{\sqrt{3}}4$.


If you don't know whether it is a rectangle or a parallegram there is no formula in general. But you can know that if the sides are $a,b,c,d$ and the angles are $A,B,C,D$ (capitals opposite lower case) you know: $a + b + c + d = P$ and $A+B+C+D= 360$. And you can calculate the area by "cutting" it into two triangles and figuring the area of those and adding them.


You know the angles are right angles and that it is a parallelagram so $a =c$ and $b = d$ so $2a + 2b = P$ and $b = \frac {P- 2a}2$ and the area is therefore $a*b = a*\frac{P - 2a}2 = \frac P{2a} - a^2$.


$s = P/4$ and area is $s^2 = \frac {P^2}{16}$

It's worth noting the area of a regular $n$-gon.

An regualar $n$-gon has $n$ sides all equal so $s = \frac Pn$.

And regular $n$-gon has $n$ angles. These angles add up to $({n-2})*180$ because we can can cut the $n$-gon into $n-2$ triangles whose angles are each $180$. So each angle of the $n$-gon is $\theta = \frac{n-2}n * 180$.

We can slice the $n$-gon into isoceles triangle wedges. Each isoceles tringle slice has a base of length $s = \frac Pn$. The base angle is $\frac {\theta}2 = \frac{n-2}n*90$. So the height of the triangle is $height = opposite = adjacent*\frac {opposite}{adjacent} = (\frac 12 s)*\tan(\frac{\theta} 2)$.

So the area of the triangle is $\frac 12*s* (\frac 12 s)*\tan(\frac{\theta} 2)=\frac 14 s^2 \tan(\frac{\theta} 2)$.

So the area of a regular $n$-gon is $n*\frac 14 s^2 \tan(\frac{\theta} 2)=\frac 1{4n}*P^2\tan(\frac{n-2}n*90)$.

For $n=3$ we get $\frac 1{12}*P^2 \tan 30 = \frac 1{12}*P^2*\frac {\sqrt 3}3=(\frac P3)^2\frac{\sqrt{3}}4 = s^2\frac{\sqrt{n}}4$ which is what we had for equilateral triangle above.

Likewise if $n=4$ we get $\frac 1{16}P^2 \tan 45 = (\frac {P}4)^2=s^2$ as we had for squares.

Believe it or not:

Area of circle = $\lim_{n\rightarrow \infty} \frac 1{4n}*P^2\tan(\frac{n-2}n*90) =P^2 \lim_{n\rightarrow \infty} \frac 1{4n}\tan(\frac{n-2}n*90) =P^2 *\frac 1{4\pi}= (\frac {P}{2\pi}) ^2 \pi = \pi r^2$


All needed is to calculate the area of a triangle with perimeter $ P $.

One could user Heron Formula.
Yet this would require the knoledge of the length of each lateral.
What can be done it put a boundary on the area of the Triangle.

For instance see:

Maximize area of a triangle with fixed perimeter

Simple proof that equilateral triangles have maximum area

Now, knowing it is bounded by a trianlge with each lateral begin $ P / 3 $ you can solve the question.


Considering perimeter is $2p$ and looking the circle we have that the area is $A_c=p^2/\pi$ and then $A_c$ is constant. Considering the square, the area is: $$A_s=p^2/4=(\pi/4)A_c \lt A_c$$ For the retangle, we have the relation for lengths $a+b=p$. So using $AM\ge GM$ and $A_r=ab$ we get: $$p/2 \ge \sqrt{ab} \Rightarrow p^2/4 \ge A_r \Rightarrow A_r \le (\pi/4)A_c \lt A_c$$ Using Heron's formula for the triangle we get: $$A_t=\sqrt{p(p-a)(p-b)(p-c)}$$ and using $AM\ge GM$ we get $$p \ge \sqrt{A_t} \Rightarrow A_t \le \pi A_c$$ but we still can't conclude anything about the relation between $A_t$ and $A_c$, except a right boundary, so let's look for a left boundary for $A_t$. But if we can choose $a+b=c=p$ we have the limit situation for the triangle exist and then the area is zero. Let's keep $c$ constant and make $a+b=c-2\epsilon \Rightarrow c=p+\epsilon$, $\epsilon >0$, and we can smoothly properly change $\epsilon$ and get new triangle with area slightly bigger than zero. So we can conclude that even if we keep the perimeter we can get triangles with area bigger then the circle and others triangle with area smaller than the circle. Another way to see that is using cosseno law. $$c^2=a^2+b^2-2ab\cos\alpha \Rightarrow c^2=(2p-c)^2-2ab(1+\cos\alpha)$$ and using that $A_t=(ab\sin\alpha)/2$ we get $$A_t=p(p-c)\left(\frac{\sin\alpha}{1+\cos\alpha}\right)$$ and keeping $c$ constant and analyse the function $f(\alpha)=\frac{\sin\alpha}{1+\cos\alpha}$ when $\alpha$ vary from $0$ to $\pi$.


To generalize: It is easily shown that any regular $n$-gon with side length $a$ the radius $r_n$ of its inscribed circle is $$r_n=\frac{a/2}{\tan(\pi/n)},$$ hence the ratio of its perimeter $P$ and the diameter of its inscribed circle is $$\pi_n:=n\cdot\tan(\pi/n).$$ Let $r_n$ be the radius of its inscribed circle. Now verify that the polygon's area is $\pi_n\cdot r_n^2$ and its circumference equals $2\pi_n\cdot r_n$.

(For example take $n=3$, a triangle with side $a$. Then $r_3=\frac{a/2}{\sqrt{3}}$ and $\pi_3=3\cdot\tan(\pi/3)=3\sqrt{3}$. We calculate its area to $\pi_3\cdot r_3^2=\frac{\sqrt{3}}{4}\cdot a^2$.)

Given $P$, the radius $r_n=\frac{P}{2\pi_n}$, hence the area is $$\frac{P^2}{4}\cdot\frac{1}{n\tan(\pi/n)}.$$ Now $\displaystyle\frac{1}{n\tan(\pi/n)}$ is strictly increasing with limit $\pi$.


Let $C$, $S$, and $E$ denote the areas of the Circle, Square, and Equilateral triangle of a given perimeter. It is straightforward to show that $C\gt S\gt E$.

Now let $R$ and $T$ denote the areas of an arbitrary Rectangle and Triangle of the same given perimeter. A modicum of algebra and/or calculus shows that $S\ge R$ and $E\ge T$.

Combining these, we can certainly say $C\gt S\ge R$ and $C\gt S\gt E\ge T$. But we cannot combine these into a string of inequalities that compare $R$ and $T$, because $R$ and $T$ can each be made as close to $0$ as one desires.


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