The heptagon has to be as close to a regular heptagon as possible, with the constraint that 2 diagonals have to be perpendicular. Thus it has 3 equal longer sides and 4 equal shorter sides. The simplest idea is to put together half of a regular hexagon and half of a regular octagon with the same circumcircle. Unfortunately, such heptagon does not have any perpendicular diagonals. Perhaps the order of the longer and shorter sides can be changed to make 2 diagonals perpendicular, but that's guessing. Assume the diagonals $BE \perp GD$ of a cyclic heptagon $ABCDEFG$ are perpendicular, $x = AB = GA,$ and $BC = CD = EF = FG.$ We are looking for such $x$ that $DE = x$ as well. For any $DE,$ the heptagon is symmetrical with respect to the line $OA,$ where $O$ is the circumcenter. Thus $DE \perp AO,\ BG \perp AO$ and the perpendicular diagonals $BE \perp GD$ meet at $P \in AO.$ Let $R$ be the heptagon circumradius. Let $BG$ meet $AO$ at $Q.$ Using Pythagorean theorem, we get $BG = 2x \sqrt{1 - \frac{x^2}{4R^2}},\ \ \ QA = \frac{x^2}{2R}$ Since the triangle $\triangle BPG$ is isosceles right, $PQ = \frac{BG}{2} = x \sqrt{1 - \frac{x^2}{4R^2}}$ and $PA = PQ + QA = \frac{x^2}{2R} + x \sqrt{1 - \frac{x^2}{4R^2}}.$ Denote $\Sigma = \sqrt{1 - \frac{x^2}{4R^2}}.$ Power of the point $P$ to the circumcircle $(O)$ is $PA \cdot (2R - PA) = 2R \cdot PA - PA^2 =$ $= x^2 + 2Rx \cdot \Sigma - \frac{x^4}{4R^2} - \frac{x^3}{R} \cdot \Sigma - x^2 \left(1 - \frac{x^2}{4R^2}\right) = 2xR \left(1 - \frac{x^2}{2R^2}\right) \cdot \Sigma$ On the other hand, power of the point $P$ to the circumcircle $(O)$ is $PB \cdot PE,$ where $PB = PQ \sqrt 2 = x \sqrt 2 \cdot \Sigma$ is known. Assuming $DE = x,$ then from the isosceles right triangle $\triangle DPE,$ we get $PE = DE \frac{\sqrt 2}{2} = \frac{x \sqrt 2}{2}.$ Hence $PB \cdot PE = x^2 \cdot \Sigma.$ Thus from $PB \cdot PE = PA \cdot (2R - PA),$ we get an equation for $x$: $2xR \left(1 - \frac{x^2}{2R^2}\right) \cdot \Sigma = x^2 \cdot \Sigma$ Reducing by $x \neq 0,\ \Sigma \neq 0$ and solving the resulting quadratic equation for $\frac x R,$ $2R \left(1 - \frac{x^2}{2R^2}\right) = x,\ \ \ \frac{x^2}{R^2} + \frac{x}{R} - 2 = 0,\ \ \frac x R = -2$ or $+1$ The only acceptable root is $x = R > 0,$ hence the heptagon $ABCDEFG$ is indeed obtained by rearranging the sides of half a regular hexagon and half a regular octagon. Therefore, its area is $S_7 = \frac{S_6 + S_8}{2} = R^2 \left(\frac{3 \sqrt 3}{4} + \sqrt 2\right).$

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sorry, I can't understand the first two sentences in your post.is it just a supposition or proof? however, I have a simple solution: the heptagon is $A{}_1A{}_2A{}_3A{}_4A{}_5A{}_6{}_7$ let $\alpha_i=A{}_iOA{}_{i+1}$ then suppose that$A{}_1A{}_4\perp A{}_2A{}_5$ then $\alpha_1+\alpha_3+\alpha_4=\pi ,\alpha_2+\alpha_5+\alpha_6+\alpha_7=\pi$ and the area og the heptagon is $\sum_{i=1}^7 {\frac12 R^2 \sin{\alpha_i}}\leq 3\sin(\frac13(\alpha_1+\alpha_3+\alpha_4)+4\sin\frac14(\alpha_2+\alpha_5+\alpha_6+\alpha_7)=\frac12R^2(\frac{3\sqrt{3}}{2}+2\sqrt{2})$ there is another two case ,use the similar method we got the maxium of the area(there is another max number for a cae but it is smaller than this one,you can check it) is$\frac12R^2(\frac{3\sqrt{3}}{2}+2\sqrt{2})$