anonymous
  • anonymous
Find the surface area of the composite solid if the base is a regular hexagon. Round your answer to the nearest hundredth. A.471.53 m2 B.615.53 m2 C.709.06 m2 D.802.59 m2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
did you first break up the 3D figure into 2 different 3D figures?
anonymous
  • anonymous
No. I didnt

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anonymous
  • anonymous
Am I suppose to multiply the numbers?
jim_thompson5910
  • jim_thompson5910
let's focus on the hexagonal prism
jim_thompson5910
  • jim_thompson5910
|dw:1436843876436:dw|
jim_thompson5910
  • jim_thompson5910
what is the area of each side panel?
jim_thompson5910
  • jim_thompson5910
|dw:1436844004000:dw|
jim_thompson5910
  • jim_thompson5910
each side panel looks like this |dw:1436844037861:dw|
anonymous
  • anonymous
okay
jim_thompson5910
  • jim_thompson5910
what is the area of that panel?
anonymous
  • anonymous
well is the area of the whole hexagon 93?
jim_thompson5910
  • jim_thompson5910
93.53 roughly
jim_thompson5910
  • jim_thompson5910
so that is the bottom face
jim_thompson5910
  • jim_thompson5910
the top face of the hexagonal prism gets covered up, so it doesn't count
anonymous
  • anonymous
okay
jim_thompson5910
  • jim_thompson5910
each panel is 42 square meters there are 6 of them 42*6 = 252 square meters so far, the total surface area is 93.53+252 = 345.53
jim_thompson5910
  • jim_thompson5910
we haven't dealt with the pyramid on top though so we're not done yet
jim_thompson5910
  • jim_thompson5910
what is the area of this triangle here |dw:1436844618751:dw|
anonymous
  • anonymous
45
jim_thompson5910
  • jim_thompson5910
there are 6 such triangles 6*45 = 270 so 270 square meters is the surface area of the pyramid up top (ignoring the bottom face)
jim_thompson5910
  • jim_thompson5910
in total, the composite 3D figure has 93.53+252+270 = 615.53 square meters of surface area
anonymous
  • anonymous
Thank you very much. I posted another question, if you will. Thank you

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