Yet Another Area Problem (Try 2 - I did not square something that needed squaring)

Let the radius of the circle be $r$ , and the internal angle opposite side with length $k$ be $\theta_k$ , where $k=1,2,3, \cdots 7$ .

Then we have $r = \dfrac k{2\sin \left(\frac {\theta_k}2\right)}$ and $\displaystyle \sum_{k=1}^7 \theta_k = 2\pi$ , $\displaystyle \implies \sum_{k=1}^7 \frac {\theta_k}2 = \sum_{k=1}^7 \sin^{-1} \left(\frac k{2r}\right) = \pi$ , Solving for $r$ numerically (I solved it with a simple Excel spreadsheet; see below), we get $r \approx 4.747537257$ .

Let the area of the isosceles triangle with base length $k$ be $A_k$ . Then area of the heptagon is:

$\begin{aligned} A_{\text{heptagon}} & = \sum_{k=1}^7 A_k & \small \color{#3D99F6} \text{Using Heron's formula} \\ & = \sum_{k=1}^7 \sqrt{s_k(s_k-r)^2(s_k-k)} & \small \color{#3D99F6} \text{where }s_k = \frac {2r+k}2 \\ & = \frac 14 \sum_{k=1}^7 k\sqrt{4r^2 -k^2} \\ & \approx \boxed{54.725} \end{aligned}$

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Reference: Heron's formula

The radius is approximately 4.74753725744023.

The area is approximately 54.7249466221096, which was computed to the same answer by two different methods.

Also, I checked the Euclidean distance between vertices.

The red and blue lines show the sides of isosceles triangles. The green lines show the altitudes of the isosceles triangles.

The apex angles:

$\begin{array}{cl} 1 & 12.0909{}^{\circ} \\ 2 & 24.3192{}^{\circ} \\ 6 & 78.3819{}^{\circ} \\ 3 & 36.8367{}^{\circ} \\ 5 & 63.5505{}^{\circ} \\ 4 & 49.8297{}^{\circ} \\ 7 & 94.991{}^{\circ} \\ \end{array}$

The methodology: compute the apex angles with the radius being an variable $r$ , solving the resultant trigonometric equation to determine the radius and then summing the areas of the isosceles triangles using the now known radius to compute the green altitudes of the isosceles triangles. A little more work was done to check the result and to draw the answer illustration.

For those who can follow Wolfram Mathematica:

$\text{sides}=\{2,6,3,5,4,7,1\}$

$\text{angles}=\text{Table}\left[2 \cos ^{-1}\left(\frac{\sqrt{r^2-\left(\frac{\text{side}}{2}\right)^2}}{r}\right),\{\text{side},\text{sides}\}\right]$

$r=r\text{/.}\, \text{FindRoot}[\text{Evaluate}[\text{Plus}\text{@@}\text{angles}]=2 \pi ,\{r,5.\}]$

r is now known.

$\text{area}=\text{Plus}\text{@@}\text{Table}\left[\frac{1}{2} \text{side} \sqrt{r^2-\left(\frac{\text{side}}{2}\right)^2},\{\text{side},\text{sides}\}\right]$

the area is now known