
#1
Mar1911, 09:49 PM

P: 510

1. The problem statement, all variables and given/known data
The volume of a pyramid can be evaluated by using the equation V=1/3*A*h where h is the height of the pyramid and A is the area of the base of the pyramid. The problem is to design such a triangular pyramid where the volume is V = 72 cm3 and the height is h = 12 cm. The base of the pyramid is a rectangular isosceles triangle. 2. Relevant equations a) Determine the area of the base of the pyramid. Pay attention to the unit of your answer. b) Draw a picture of the base triangle of the pyramid in a proper scale 3. The attempt at a solution a) V=1/3*A*h A=72/(12/3)=18cm^{2} b) A = 18 = b*y b=y so: 18=2*b => b = 9cm 



#2
Mar1911, 11:21 PM

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#3
Mar2011, 07:56 AM

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Well, I can see your picture and it isn't even of a triangle!
Did you simply misread the problem? 



#4
Mar2011, 08:59 AM

P: 510

Volume of pyramidbut then we have the area of a triangle that equal 18m^{2} 18=(b*y)/2 To be able to draw that triangle we need to find the length of just one side (assuming the sides are equal) and the problem is..we don't have that 



#5
Mar2211, 12:00 PM

P: 510

some please give me a hint




#6
Mar2211, 12:23 PM

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#7
Mar2211, 02:31 PM

P: 510

i'm not good in drawing but i have an image of it in my mind..and i can't see how i will find the length of the side of that triangle




#8
Mar2211, 02:48 PM

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Earlier you had that V = (1/3)Ah and you are given that V = 72 and h = 12.
So A = 3V/h = 3*72/12 = 18. You had that a long time ago. So the area of your isosceles "rectangular" triangle is 18. I presume you know what isosceles means and I presume that "rectangular" triangle means what is usually called a right triangle. And as I pointed out in post #2, the area of a triangle is (1/2)*base* height. What exactly are you stuck on, given you have all this information? 



#9
Mar2211, 06:02 PM

P: 510

how you know that the triangle is isosceles ? isn't suppose to be equilateral ?




#10
Mar2211, 06:54 PM

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#11
Mar2211, 07:37 PM

P: 510

ok i must be blind :/
but still we can't find the dimension of that triangle..it's about the triangle and we need the height of the triangle, not the pyramid 



#12
Mar2211, 11:05 PM

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#13
Mar2311, 06:42 AM

P: 510

here it is...
i guess that the base of this triangle = it's height. 



#14
Mar2311, 06:45 AM

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Yes, and taking "s" to be the length of one of the legs, the area of the base is
[tex]\frac{1}{2}s^2= 18[/tex]. 



#15
Mar2311, 07:03 AM

P: 510

s^{2} = 36 s = 6 cm and then we find the hypotenuse = approximately 8.5 cm 


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