# Volume of pyramid

by chawki
Tags: pyramid, volume
 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)=18cm2 b) A = 18 = b*y b=y so: 18=2*b => b = 9cm Attached Thumbnails
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 Quote by chawki 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)=18cm2 b) A = 18 = b*y b=y so: 18=2*b => b = 9cm
Your picture didn't come through, but I'm guessing your formula A = 18 = b*y is wrong. Remember the area of a triangle is (1/2)base * height. And even if it were correct, b = y wouldn't give you 18 = 2b, it would be 18 = b2. Two things to fix.
 PF Patron Sci Advisor Thanks Emeritus P: 38,405 Well, I can see your picture and it isn't even of a triangle! Did you simply misread the problem?
P: 510

## Volume of pyramid

 Quote by HallsofIvy Well, I can see your picture and it isn't even of a triangle! Did you simply misread the problem?
Yes i did my mistake, i thought it's a pyramid with 4 faces..

but then we have the area of a triangle that equal 18m2
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
Attached Thumbnails

 P: 510 some please give me a hint
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 Quote by chawki Yes i did my mistake, i thought it's a pyramid with 4 faces.. but then we have the area of a triangle that equal 18m2 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
 Quote by chawki some please give me a hint
You still haven't drawn a picture of your pyramid. When you do I think you will find what you need in my earlier post. (reply #2).
 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
 PF Patron HW Helper Thanks P: 6,742 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?
 P: 510 how you know that the triangle is isosceles ? isn't suppose to be equilateral ?
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 Quote by chawki 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.
 Quote by chawki how you know that the triangle is isosceles ? isn't suppose to be equilateral ?
 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
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 Quote by chawki 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
I asked you before if "rectangular" triangle means right triangle. You didn't answer that, but I assume it does. You have an isosceles right triangle and you know its area is 18. Draw a picture of it. You should be able to figure out its dimensions from the picture.
 P: 510 here it is... i guess that the base of this triangle = it's height. Attached Thumbnails
 PF Patron Sci Advisor Thanks Emeritus P: 38,405 Yes, and taking "s" to be the length of one of the legs, the area of the base is $$\frac{1}{2}s^2= 18$$.
P: 510
 Quote by HallsofIvy Yes, and taking "s" to be the length of one of the legs, the area of the base is $$\frac{1}{2}s^2= 18$$.
ahh ok..now i get it..
s2 = 36
s = 6 cm

and then we find the hypotenuse = approximately 8.5 cm
Attached Thumbnails

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