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Volume of pyramid

by chawki
Tags: pyramid, volume
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chawki
#1
Mar19-11, 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)=18cm2

b) A = 18 = b*y
b=y
so: 18=2*b => b = 9cm
Attached Thumbnails
Volume of pyramid.JPG  
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LCKurtz
#2
Mar19-11, 11:21 PM
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Quote Quote by chawki View Post
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.
HallsofIvy
#3
Mar20-11, 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?

chawki
#4
Mar20-11, 08:59 AM
P: 510
Volume of pyramid

Quote Quote by HallsofIvy View Post
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
Volume of pyramid.JPG  
chawki
#5
Mar22-11, 12:00 PM
P: 510
some please give me a hint
LCKurtz
#6
Mar22-11, 12:23 PM
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Quote Quote by chawki View Post
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 Quote by chawki View Post
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).
chawki
#7
Mar22-11, 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
LCKurtz
#8
Mar22-11, 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?
chawki
#9
Mar22-11, 06:02 PM
P: 510
how you know that the triangle is isosceles ? isn't suppose to be equilateral ?
LCKurtz
#10
Mar22-11, 06:54 PM
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Quote Quote by chawki View Post
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 Quote by chawki View Post
how you know that the triangle is isosceles ? isn't suppose to be equilateral ?
Didn't you read your own statement of the problem?
chawki
#11
Mar22-11, 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
LCKurtz
#12
Mar22-11, 11:05 PM
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Quote Quote by chawki View Post
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.
chawki
#13
Mar23-11, 06:42 AM
P: 510
here it is...
i guess that the base of this triangle = it's height.
Attached Thumbnails
Volume of pyramid.JPG  
HallsofIvy
#14
Mar23-11, 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].
chawki
#15
Mar23-11, 07:03 AM
P: 510
Quote Quote by HallsofIvy View Post
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].
ahh ok..now i get it..
s2 = 36
s = 6 cm

and then we find the hypotenuse = approximately 8.5 cm
Attached Thumbnails
Volume of pyramid.JPG  


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