Heptagon's fine, but!

Well, though looks geometry, this is Moment of inertia! Attach unit masses at all the 7 vertices.

We know that moment of inertia $\textbf{I}$ of a system of mass $\textbf{m}$ , having moment of inertia $I_{com}$ about center of mass, about a point at distance $l$ from center of mass is given by $\text{I =I}_{com}+ ml^2$

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From the given side length, we conclude that circumradius, i.e. JA is $2$ units, moment of inertia of each vertex being $1\times 2^2 =4$ .

$I_{cm} = 7\times 4 = 28$ .

The center of mass will be $J$ by simple observation, so $l = JH$

$JH = \sqrt{(16-8)^2+(12-8)^2} ...\quad\quad \text{Distance formula}\\ \quad = \sqrt{80}$

$I_{com}=28 , m= 7 , l= \sqrt{80}$

Hence, $I_H = 28+7\times 80= \boxed{588}$ .

I will use a vector approach here. First of all, it is clear that the radius of the circumcircle of the heptagon is 2 units. This is because the length of side of a regular n-gon is: $s=2r\sin { \frac { \pi }{ n } }$ Which can be proved by simple trigonometry. So, for the given case: $4\sin { \frac { \pi }{ 7 } } =2r\sin { \frac { \pi }{ 7 } }$ $\therefore r=2$ Now, consider: ${ HA }^{ 2 }={ |\vec { HA } | }^{ 2 }=(\overrightarrow { HJ } +\overrightarrow { JA } ).(\overrightarrow { HJ } +\overrightarrow { JA } )={ HJ }^{ 2 }+2\overrightarrow { HJ } .\overrightarrow { JA } +{ JA }^{ 2 }$ Let d be the distance between H and J : $\therefore { HA }^{ 2 }={ d }^{ 2 }+2\overrightarrow { HJ } .\overrightarrow { JA } +{ r }^{ 2 }$ Thus, we can rewrite the original expression as follows: $7({ d }^{ 2 }+{ r }^{ 2 })+2\overrightarrow { HJ } .(\overrightarrow { JA } +\overrightarrow { JB } +\overrightarrow { JC } +\overrightarrow { JD } +\overrightarrow { JE } +\overrightarrow { JF } +\overrightarrow { JG } )$ $=7({ d }^{ 2 }+{ r }^{ 2 })+2\overrightarrow { HJ } .(\overrightarrow { 0 } )$ Which can be explained easily by symmetry. Thus, the required answer is: $=7({ d }^{ 2 }+{ r }^{ 2 })=7(80+4)=\boxed{588}$

Note: This technique is essentially the same as the moment of inertia approach . This basically works as a proof for the parallel axis theorem, where J is the centre of mass of the system.