Solve Triangle Pyramid Volume: V=72cm3, h=12cm

[chawki]

Homework Statement

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.

Homework 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

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

 

[LCKurtz]

Homework Statement

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.

Homework 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

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]

Well, I can see your picture and it isn't even of a triangle!

Did you simply misread the problem?

 

[chawki]

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

 

[chawki]

some please give me a hint

 

[LCKurtz]

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

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]

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]

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]

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

 

[LCKurtz]

Homework Statement

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.

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]

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]

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]

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

 

[HallsofIvy]

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

 

[chawki]

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