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Mathematics 28 Online
OpenStudy (anonymous):

The shown net forms a regular square pyramid with all lateral and base edges the same length.find the following: @terenzreignz

OpenStudy (anonymous):

welcome to os

OpenStudy (anonymous):

OpenStudy (anonymous):

got a slant height of\[\sqrt{3}\]

OpenStudy (anonymous):

Hi @peoplesay1123. A Warm Welcome to OS.

OpenStudy (anonymous):

im sorry thats way hard for me :(

OpenStudy (anonymous):

np

terenzreignz (terenzreignz):

Okay... show me what you've done so far...

OpenStudy (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.

OpenStudy (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...

OpenStudy (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

OpenStudy (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.

OpenStudy (anonymous):

isnt the formula for a the lat area of a regular pyramid 1/2(perimeter of base)(slant height)

OpenStudy (anonymous):

?

terenzreignz (terenzreignz):

I suppose. Yes, that's right. Work that out, you get...?

OpenStudy (anonymous):

about 6.93 in.^2

terenzreignz (terenzreignz):

Nope... try again... or you could show me how you got that...

OpenStudy (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.

OpenStudy (anonymous):

but isnt the formula 1/2Pl ?

OpenStudy (anonymous):

not 1/2Bl

terenzreignz (terenzreignz):

Oh... sorry, right. I thought it was just one face. Then it's right, what you did ^_^

OpenStudy (anonymous):

oh ok and the surface area is about 10.93 .

OpenStudy (anonymous):

?

terenzreignz (terenzreignz):

Right. did you actually need help with this? ^_^

OpenStudy (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

OpenStudy (anonymous):

\[\approx \]

OpenStudy (anonymous):

and not>>> =

terenzreignz (terenzreignz):

Whatever floats your (or your instructor's) boat, I assume. And lol, that sign looks fancier XD

OpenStudy (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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