# Is it possible to find the area of a shape from its perimeter?

Is it possible to find the area of a free form shape knowing the perimeter? An example would be a clover leaf shape. If the perimeter is 96 how would I know what the area would be?

• No, it's not possible to determine area from perimeter.
– user147263
Jan 29 '15 at 4:19
• Yves brings up a good point. By "clover leaf shape" do you mean you know the shape precisely up to similarity (i.e., the shape but not its size)? If so, the answer is yes, and you can easily write down a formula given the perimeter and area for any particular size of that object. If you simply have a perimeter, then the answer (as indicated by several answers) is no. Jan 29 '15 at 11:01

## 7 Answers

Contrary to all other answers, I say yes you can find the area $a$ of a known shape (clover leaf) from the length of its perimeter $p$.

Taking a similar model of the clover leaf, measure its area $A$ and perimeter $P$, using a curvimeter and a planimeter. You can also do that from a digital image (photoscan), but I don't know of ready-made tools for that.

Then, for any clover leaf (of the same shape), this proportionality rule holds:

$$\frac aA=\left(\frac pP\right)^2,$$ so that $$a=A\left(\frac pP\right)^2=F_{clover}p^2.$$

For any shape there is a corresponding conversion factor that you can compute once for all.

For instance, with the picture below, you can estimate an area of $19852$ pixels and a perimeter of $750$ pixels (this is an inaccurate measurement).

Then $F_{clover3hearts}\approx0.0353$, and your leaf has an area of $325$ square units.

This is impossible. We can prove this by constructing two shapes with the same perimeter but different areas.

Consider, for instance, the unit equilateral triangle, with perimeter $3$ and area $\sqrt{3}/4$, and the square with side lengths $3/4$ and area $9/16$. Since these two shapes have the same perimeter, but different areas, one cannot uniquely determine area from perimeter.

Perimeter is basically arc length. You could just as easily have a line segment of length 96 as a square of perimeter 96. Then the side length is 24, and so the area is $24^2.$ On the other hand, if the circumference of a circle is 96, then $96=\pi d,$ so that means that $r=96/ 2\pi$. This means that the area is $(96/2)^2,$ which is different from $24^2.$

In general, no, though the Isoperimetric Inequality gives and upper limit on area: For a bounded region in the plane with perimeter $L$, the area $A$ must satisfy $$A \leq \frac{L^2}{4 \pi},$$ and equality holds iff the region is a circle. By constrast, if the region is a square, we have $A = \left(\frac{L}{4}\right)^2 = \frac{L^2}{16}$, which (since $\pi < 4$) is smaller.

If you know the shape of the boundary, you can compute the area of the region by computing an integral just over the boundary, for example, with Green's Theorem. This is the principle exploited by the planimeter, a machine which does exactly this.

No, this is not possible, because you can easily increase the perimeter of any polygon (while retaining its area) by modifying its edges like this:

You can even use this technique to create fractals with an inifinite perimeter, such as the "Minkowski Sausage".

PS: As Yves Daoust and Travis have noted, this answer only applies if the shape is unknown.

If "perimeter" is the lenght of..., is impossible, as already answered. But if the form is know is possible calculate the area using Green's theorems. The Planimeter is an application of this fact.

The answer strictly speaking is no. take a US quarter for instance has a perimeter (circumference) of 3 inches exactly. A Circle has the maximum area for a given perimeter for a shape. A US Quarter has a width of 0.955 inches so its total area (for one of its circular faces) has an area of 0.7163 inches^2

As an example, a shape known as a reuleaux triangle can have the exact same width at any given axis (0.955 inches), the exact same perimeter (3 inches), and have an area of 0.643 inches^2 (rounded). There's a whole infinite group of "reuleaux" shapes that could be created from the shared area of intersecting circles of radius 0.955 inches centered on the vertices of equal-sided polygons that would all also have the exact same width on any given axis and the exact same perimeter. The more vertices of the corresponding polygon, the closer the area becomes to that of a circle.

On the other extreme, a 1.5inch line can also be said to have a perimeter of 3 inches but an area of 0 (zero) inches^2 if you think of it as a rectangle with a width of 1.5inches and a height of 0 inches.

Interesting yes?