How can I find the area and perimeter of [these cases], can you help me?



  • $\begingroup$ What have you tried so far? $\endgroup$ – dfnu Sep 25 '19 at 16:46
  • $\begingroup$ The perimeter of an ellipse is given by a complete elliptic integral of the second kind, so I guess there is no way to avoid integrals. On the other hand good algebraic approximations are known, for instance Ramanujan's. $\endgroup$ – Jack D'Aurizio Sep 25 '19 at 16:49
  • 2
    $\begingroup$ for area, see this and that. In particular, if you answer the figure in this answer, you will know how to calculate the perimeter. $\endgroup$ – achille hui Sep 25 '19 at 17:01
  • $\begingroup$ oops, I mean 'understand' the figure.. $\endgroup$ – achille hui Sep 25 '19 at 17:30

enter image description here

Let [.] denote areas of various shapes below.

Case 1.

One of the little grey areas is

$$I_1 = [ABCD] - [CDS] - 2[DAS]$$

where the areas of the triangle CDS and the circle sector DAS are given by

$$[CDS] = \frac{\sqrt 3}{4}L^2, \>\>\> [DAS] = \frac{\pi}{12}L^2$$


$$ I_1 = \left(1- \frac{\sqrt 3}{4} - \frac{\pi}{6}\right)L^2 $$

and its perimeter is $\left( 1+\frac{\pi}{3}\right)L$.

Case 2.

One of the four areas in the second case is

$$ I_2 = [ABCD] - [ADB] - 2I_1 = \left(\frac{\sqrt 3}{2}-1 + \frac{\pi}{12}\right)L^2 $$

and its perimeter is $\frac{\pi}{2}L$.

Case 3.

The area in the case of the middle area,

$$ I_3 = (1-4I_1-4I_2)L^2=\left(1-\sqrt 3 + \frac{\pi}{3}\right)L^2$$

with perimeter $\frac{2\pi}{3}L$.

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Call the area in the first picture A, the one in the second picture 4B, the one on the third picture 4C.

Then using equivalence of areas we get that: \ $A+4B+4C=L^{2} \\ A+3B+2C=\frac{\pi L^{2}}{4} \\ A+2B=\frac{(\pi - 2)L^{2}}{2}$ \

Solve the equations for A, B, C.

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  • 1
    $\begingroup$ Degenerate equations $\endgroup$ – Quanto Sep 25 '19 at 16:54
  • $\begingroup$ I guess we need another equation then, but I can't quite spot it. Considering geometrical shapes seems a good path though. $\endgroup$ – Stefano Sep 25 '19 at 17:05


An area of $A+2B+C$ is formed by the union of two sectors of angle $60^o$ of a circle of radius $L$. These sectors overlap in an equilateral triangle of side $L$. This gives us the extra equation we need $$A+2B+C=\frac{1}{3} \pi L^2-\frac{\sqrt(3)}{4}L^2.$$

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