Check the picture below.
so the volume will simply be the area of the hexagonal face times the height.
![\textit{area of a regular polygon}\\\\ A=\cfrac{1}{4}ns^2\stackrel{\qquad degrees}{\cot\left( \frac{180}{n} \right)}~~ \begin{cases} n=\stackrel{number~of}{sides}\\ s=\stackrel{length~of}{side}\\[-0.5em] \hrulefill\\ n=6\\ s=12 \end{cases}\implies A=\cfrac{1}{4}(6)(12)^2\cot\left( \frac{180}{6} \right) \\\\\\ A=216\cot(30^o)\implies A=216\sqrt{3} \\\\[-0.35em] ~\dotfill\\\\ \stackrel{\textit{area of the hexagon}}{(216\sqrt{3})}~~\stackrel{height}{(10)}\implies 2160\sqrt{3}~~\approx ~~3741.2~cm^3](https://tex.z-dn.net/?f=%5Ctextit%7Barea%20of%20a%20regular%20polygon%7D%5C%5C%5C%5C%20A%3D%5Ccfrac%7B1%7D%7B4%7Dns%5E2%5Cstackrel%7B%5Cqquad%20degrees%7D%7B%5Ccot%5Cleft%28%20%5Cfrac%7B180%7D%7Bn%7D%20%5Cright%29%7D~~%20%5Cbegin%7Bcases%7D%20n%3D%5Cstackrel%7Bnumber~of%7D%7Bsides%7D%5C%5C%20s%3D%5Cstackrel%7Blength~of%7D%7Bside%7D%5C%5C%5B-0.5em%5D%20%5Chrulefill%5C%5C%20n%3D6%5C%5C%20s%3D12%20%5Cend%7Bcases%7D%5Cimplies%20A%3D%5Ccfrac%7B1%7D%7B4%7D%286%29%2812%29%5E2%5Ccot%5Cleft%28%20%5Cfrac%7B180%7D%7B6%7D%20%5Cright%29%20%5C%5C%5C%5C%5C%5C%20A%3D216%5Ccot%2830%5Eo%29%5Cimplies%20A%3D216%5Csqrt%7B3%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20%5Cstackrel%7B%5Ctextit%7Barea%20of%20the%20hexagon%7D%7D%7B%28216%5Csqrt%7B3%7D%29%7D~~%5Cstackrel%7Bheight%7D%7B%2810%29%7D%5Cimplies%202160%5Csqrt%7B3%7D~~%5Capprox%20~~3741.2~cm%5E3)
550×10 I think how do you want it to be explained
The square is 130 in squared and the other shape is 54 in squared. If you're looking for the sum of the whole thing, its 185 in squared.
square: 13*10=130
triangle: 7*4= 28/2=14
rectangle: 10*4=40
Answer:
x = -2
Step-by-step explanation: