When rolling a number cube, what is the probability of rolling a number greater than 2
1 answer:
You might be interested in
If you notice, the figure has two hexagonal faces and 6 rectangles.
now the rectangles are just 4x9 each, so their area is just 6(4*9) for all 6.
the hexagons has a distance from the center perpendicular to a side of 7.8, namely the apothem is 7.8, and since each side is 9 km long, the perimeter is 9+9+9+9+9+9.
since the area of a regular polygon is (1/2)(apothem)(perimeter), we can simply get the area of those two hexagons and the area of the rectangles, sum them up, and that's the surface area of the hexagonal prism.
![\bf \stackrel{\textit{6 rectangles}}{6(4\cdot 9)}~~~~+~~~~\stackrel{\textit{2 hexagons}}{2\left[ \cfrac{1}{2}(7.8)(54) \right]}](https://tex.z-dn.net/?f=%5Cbf%20%5Cstackrel%7B%5Ctextit%7B6%20rectangles%7D%7D%7B6%284%5Ccdot%209%29%7D~~~~%2B~~~~%5Cstackrel%7B%5Ctextit%7B2%20hexagons%7D%7D%7B2%5Cleft%5B%20%5Ccfrac%7B1%7D%7B2%7D%287.8%29%2854%29%20%5Cright%5D%7D)
now the rectangles are just 4x9 each, so their area is just 6(4*9) for all 6.
the hexagons has a distance from the center perpendicular to a side of 7.8, namely the apothem is 7.8, and since each side is 9 km long, the perimeter is 9+9+9+9+9+9.
since the area of a regular polygon is (1/2)(apothem)(perimeter), we can simply get the area of those two hexagons and the area of the rectangles, sum them up, and that's the surface area of the hexagonal prism.