Formula of the area of heptagon?

Let the areas and the side lengths of the three heptagons be $A_k$ and $a_k$ respectively, where $k=1,2,3$ . We know that $A_k \propto a_k^2$ . Therefore, we have $\dfrac {A_i}{A_j} = \dfrac {a_i^2}{a_j^2}$ . From the right-angle $\triangle ABC$ , by Pythagorean theorem :

$\begin{aligned} a_2^2 & = a_3^2 - a_1^2 & \small \color{#3D99F6} \text{Dividing both sides by }a_1^2 \\ \frac {a_2^2}{a_1^2} & = \frac {a_3^2}{a_1^2} - 1 \\ \frac {A_2}{A_1} & = \frac {A_3}{A_1} - 1 & \small \color{#3D99F6} \text{Multiplying both sides by }A_1 \\ \implies A_2 & = A_3 - A_1 \\ & = 7102 - 2017 \\ & = \boxed{5085} \end{aligned}$

Let $a < b < c$ be the sides of the right triangle. The areas of the heptagons are proportional to the square of the sides. Thus, for some proportionality constant $k$ we have $ka^2 = 2017;\ \ \ kb^2 = A;\ \ \ kc^2 = 7102.$ From the Pythagorean Theorem, $a^2 + b^2 = c^2.$ Multiplying by the proportionality constant $k$ and substitution results in $ka^2 + kb^2 = kc^2, \\ 2017 + A = 7102,$ showing that $A = 7102 - 2017 = \boxed{5085}$ .

Me being relatively clueless, I went the long way round (compared to the rest of the answers here), working backward from the areas to get the lengths and then using Pythagoras to get the third length and finally the area.

Firstly, considering the 7 congruent isosceles triangles that each hexagon can be split into, the inner angle is $\frac{360}{7} ^{\circ}$ and the base length is L. Splitting these again, into two right-angled triangles, the height ( $adj = \frac {opp}{\tan\theta}$ ) is shown to be $\frac{L}{2} \div (\tan \frac {180}{7}) = \frac{L}{2 \tan \frac {180}{7}}$

Considering the larger isosceles, the area is $\frac{L}{2} \times \frac{L}{2 \tan \frac {180}{7}} = \frac {L^2}{4 \tan \frac {180}{7}}$ . This is $\frac{1}{7}$ of the total area ( $A$ ) of the heptagon, i.e. $A = \frac {7L^2}{4 \tan \frac {180}{7}}$

I rearranged this to get $L = \sqrt {\frac {4A \tan\frac{180}{7}}{7}}$ , and for the two heptagons the two lengths $\sqrt {\frac {4 \times 2017 \tan\frac{180}{7}}{7}} \approx 23.559 cm$ and $\sqrt {\frac {4 \times 7102 \tan\frac{180}{7}}{7}} \approx 44.208 cm$

Finally, the third side of the central right-angled triangle, and the side length of the third heptagon, is $L = \sqrt {44.208^2 - 23.559^2} \approx 37.407 cm$ , and the heptagon has $A = \frac {7L^2}{4 \tan \frac {180}{7}} cm^2$ as above, $= \underline {5085 cm^2}$ exactly

Since it's a regular figure we could divide it into 7 equivalent triangles each with a vertex at origin and one side being the side of the heptagon. The area of each triangle would be proportional to $a^2$ .

Thus area of heptagon = $G\times a^2$ where G, is some "geometrical factor" [eg. $\frac{\sqrt{3}}{4}$ for an equilateral triangle] Thus dividing by 'G' and applying Pythagoras for the sides 'a', yields required area as difference of the five areas..

I observed that the area of each heptagon was proportional to the square of the side. Then, from the Pythagorean theorem, the area of the unknown heptagon is the difference between the areas of the known heptagons.

The area of any regular polygon is proportional to the square of its side, so by Pythagorean theorem it's a simple subtraction.

I knew the Pythagorean theorem about 3 rectangles surrounding a right-angled triangle but nothing about heptagons.

The following formula applies to every polygon, hence, the area of a heptagon is :

$A_7 = 7_{\text {isoceles} {\triangle}} \times a_\text {apothem length} \times \frac {s_\text {side length}}{2}$

Then we have :

$A_7 =7 \times a \times \frac {s}{2}$

So the area of a $\frac {1}{7}$ of the heptagon is an isoceles triangle :

$A_1 = a \times \frac {s}{2}$

The half part of the isoceles triangle is a right-angled triangle (the heptagon has 14 right-angled triangles).

$A_\frac{1}{2} = a \times \frac {s}{4}$

We know that the heptagon is composed of 7 isoceles triangles whose the angle in the center of the heptagon is :

$\alpha = \frac {360^o}{7} \approx 51,43^o$

Then, half of that angle belonging to the right-angled triangle is :

$\frac {\alpha}{2} = \frac {360^o}{14} \approx 25,71^o$

Let's calculate the tangent of that half alpha angle :

$\tan{\frac {\alpha}{2}} = \frac{\frac{s}{2}}{a} = \frac{s}{2a}$

$\tan{(25,71)} = \frac{s}{2a} \iff 2a \times 0.48 = s \iff a \times 0.96 = s$

$a = \frac{s}{0.96}$ Now we get the apothem value, let's use it in the area formula of the heptagon :

$A_7 =7 \times a \times \frac {s}{2} \iff A_7 =7 \times \frac{s}{0.96} \times \frac {s}{2} \iff A_7 =7 \times \frac{s^2}{1.92}$

$\color{#D61F06}{\boxed{s^2 = \frac{1.92 \times A_7}{7}}}$ We have now the squared side of a right-angled triangle.

As we have a right-angled triangle whose one side is part of each heptagon, let's use the Pythagorean theorem :

$s^2_{(7102)} = s^2_{(2017)} + s^2_{(x)}$ , $x$ is unkown.

$\color{#3D99F6}{\frac{1.92 \times 7102}{7} = \frac{1.92 \times 2017}{7} + \frac{1.92 \times x}{7}}$

Simplifying, we get : $\color{#3D99F6}{7102 = 2017 + x}$

And at last, we have : $\color{#3D99F6}{x = 5085}$

The area of the remaining heptagon is $\color{#D61F06}{\boxed{5085 cm^2}}$

blob pythagorean theorem (numberphile) :https://www.youtube.com/watch?v=ItiFO5y36kw

The area of a regular heptagon is $A = \frac {7}{4} s^2 \cot(\frac{180^\circ}{7})$ , where $s$ is the side of the heptagon.

Given the two areas, we calculate the sides of the corresponding areas using the above formula.

When $A_1=2017 \implies s_1= 23.56$ .

When $A_3= 7102 \implies s_3= 44.21$ .

By Pythagorean method, we can calculate the side of the third heptagon, since $\triangle ABC$ is a right triangle.

$(44.21)^2= (23.56)^2+x^2 \implies x= 37.21= s_2$ . ( the side of heptagon of $A_2)$

The area of heptagon $A_2 = \frac{7}{4} s_2^2 \cot (\frac{180^\circ}{7})\implies A_2= \boxed{5085}$