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Pythagoras probably could have helped :(
well i got that...i just have a limited about of numbers that help me at all
i have a volume that =340 and values of x1-12 that equal the base lengths
Is it a regular square based pyramid (i.e. the 4 triangles are equilateral)? If so, then that is probably enough. If not, you're f*cked
I was trying to help. No need to get all dorky on me.
lol...i am a dork to begin with...so im sorry but you are going to have to retard this out for me
not helping..im having issues trying to figure out what goes in the calculator
What is x1? just a random number?
ok so i got the perimeter which is \[p=x \times 4 \]
now i need to get the side length which is \[a^2+b^2=c^2 so what is a b and c\]
OK I will try and do this quickly - what EXACTLY do you know?
i know the volume is 340 inches cubed..and that 1 is the side length
Do you mean 1 is the length of the 4 edges of the base? Or the length of the slants?
oops yes that is what i ment
Well if that is the case, the volume = 340 = (1/3) * 1^2 * height , so height = 1020... which sounds retardedly big. But if that was right, the slant length ^2 = 1^2 + 1020^2 And surface area = 2 * slant length * base length + base length squared Weird question, but I have to bye. bye babe xx