anonymous
  • anonymous
The shown net forms a regular square pyramid with all lateral and base edges the same length.find the following: @terenzreignz
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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anonymous
  • anonymous
http://www.chegg.com/homework-help/questions-and-answers/9-lateral-area-right-regular-triangular-prism-18-cm-2-total-surface-area-answer-decimal-fo-q2070861
anonymous
  • anonymous
welcome to os
anonymous
  • anonymous
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anonymous
  • anonymous
got a slant height of\[\sqrt{3}\]
anonymous
  • anonymous
Hi @peoplesay1123. A Warm Welcome to OS.
anonymous
  • anonymous
im sorry thats way hard for me :(
anonymous
  • anonymous
np
terenzreignz
  • terenzreignz
Okay... show me what you've done so far...
anonymous
  • anonymous
well i got the slant height as \[\sqrt{3}in.\] and i got the perimeter of the base as 8 in. and the base area as 4 in^2 and idk if any of those are correct but thats all i could came up with. Then i had the Lat area as about 6.93 in^2 and the surface area as 10.93 in^2. I think some of it is wrong though idk.
anonymous
  • anonymous
I posted the attachment for the diagram of this problem in these comments, just so u know.
terenzreignz
  • terenzreignz
Remember, those triangles at the side are equilateral triangles...
anonymous
  • anonymous
yea
terenzreignz
  • terenzreignz
Well, I vividly remember telling you the formula for the area of an equilateral triangle yesterday... we even derived it. Did you pay attention? :P
anonymous
  • anonymous
yeah i remember that but how do i use that formula to find thae slant height and base perimeter and base area and lat area.
terenzreignz
  • terenzreignz
Don't worry, your slant height, base perimeter and base area are correct.
anonymous
  • anonymous
isnt the formula for a the lat area of a regular pyramid 1/2(perimeter of base)(slant height)
anonymous
  • anonymous
?
terenzreignz
  • terenzreignz
I suppose. Yes, that's right. Work that out, you get...?
anonymous
  • anonymous
about 6.93 in.^2
terenzreignz
  • terenzreignz
Nope... try again... or you could show me how you got that...
anonymous
  • anonymous
1/2(8)(sqr root of 3)= 4(sqr root of 3)= 6.928= 6.93
terenzreignz
  • terenzreignz
ahh... the base length is 2, not 8.
anonymous
  • anonymous
but isnt the formula 1/2Pl ?
anonymous
  • anonymous
not 1/2Bl
terenzreignz
  • terenzreignz
Oh... sorry, right. I thought it was just one face. Then it's right, what you did ^_^
anonymous
  • anonymous
oh ok and the surface area is about 10.93 .
anonymous
  • anonymous
?
terenzreignz
  • terenzreignz
Right. did you actually need help with this? ^_^
anonymous
  • anonymous
oh i guess i wanted to make sure. Because if you look at the diagram it show Lateral Area= and i thought i was not getting the right answer becuase my lateral area didnt EQUAL 6.93 it was ABOUT 6.93. so wouldnt i use
anonymous
  • anonymous
\[\approx \]
anonymous
  • anonymous
and not>>> =
terenzreignz
  • terenzreignz
Whatever floats your (or your instructor's) boat, I assume. And lol, that sign looks fancier XD
anonymous
  • anonymous
oh alright i guess it dosent matter then its all the same i guess.. thanks for the reassuring help lol

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