ROTATE FIGURE-GEOMETRY

Hello!

My problem is :If you have a right triangle ABC the one side is a=3cm and b=4 cm rotate (round) over hypotenuse . Volumen of this new figure is 19,2 Pi cm^3.?(2 Cone) How???

I´did this:
c^2=a^2+b^2=5
V=V1+V2= r^2*Pi*x/3+r^2*Pi*y/3= r^2*Pi/3*(x+y)...
(x+y)=c=5
Heron´s formula:
P(area)=c*vc/2=sqr[s*(s-a)*(s-b)*(s-c)]
s=(a+b+c)/2=6
5*vc/2=sqr[6*(6-3)*(6-4)*(6-c)]=vc=2,4
vc=r
2,4^2*Pi/3*5=5,76*Pi/3*5=9,6*Pi

So my resault is 9,6*Pi cm^3 ,where did I´go wronge.Please help me!!
I´know that 9,6+9,6 is 19,2 but this is probably not corect. If you be so kinde to give me some steps of your work.


Thanks!!!

Hello, hope!

I don't agree with either answer . . .

You have a right triangle ABC; sides are a=3cm and b=4 cm.
Rotate about its hypotenuse.
Volume of this new figure is 19.2(pi) cm³. How?
.

Click to expand...

We have a 3-4-5 right triangle.
From the similiar triangles, we can determine all the above lengths.

The left triangle forms a cone with: . r = 2.4, h = 3.2
. . Its volume is: . V₁ .= .(π/3)(2.4)²(3.2) .≈ .19.30 cm³

The right triangle forms a cone with: . r = 2.4, h = 1.8
. . Its volume is: . V₂ .= .(π/3)(2.4)²(1.8) .≈ .10.86 cm³

The total volume is: . 19.30 + 10.86 . = . 30.16 cm³

Let's see if I can guess what it is you are doing.

You have a Right triangle.
The legs are known to be 3 cm and 4 cm, making the hypotenuse 5 cm.
Spin this triangle using the hypotenuse as the axis of rotation.
Find the volume of the object so defined.

First: "c^2=a^2+b^2=5" What does that mean? I think it's you finding the length of the third side of the right triangle, but one of those "="s is not telling the truth.

Second: Your 3D object is not a cone. Maybe that is what you meant by "2 Cone". It is 2 Right Circular cones, connected at the base, pointing opposite directions.

Third: The volume of a right circular cone is (1/3)*pi*r²*h. I think you have that, but you introduced 'x' and 'y' without telling us what they are. Write definitions clearly. It will help organize your thoughts.

x = the height of one cone.
y = the height of the other cone = 5-x

It looks like you are splitting up the hypotenuse to find the heights of the cones. That is good. Unfortunately, I don't see where you explicitly solved for 'x' or 'y'. On the other hand, maybe you didn't need to do that, since the JOINT volume formula can be factored, giving

(1/3)*pi*r²*x + (1/3)*pi*r²*y =
(1/3)*pi*r²*(x+y) =
(1/3)*pi*r²*5 =
(5/3)*pi*r².

Fourth: What you don't yet know is the RADIUS of the cones. Your method to find the height/radius by equating areas is ingenious. I did it a harder way. Draw an altitude to the hypotenuse on your original right triangle. Label the length of the altitude 'h'. If x > y, we have

4² = x² + h²
3² = y² + h²

or

4² = x² + h²
3² = (5-x)² + h²

One can solve for h = 12/5, which is what you have.

This gives the exact answer:

(5/3)*pi*(12/5)² =
(12²/(3*5))*pi =
((12*4)/5)*pi =
(48/5)*pi =
9.6*pi = <== Your answer and ½ the book's answer
30.159 <== Soroban's answer

It seems like you managed the correct answer, but I found your work exceptionally difficult to follow. Three times, I found myself thinking that you were wondering off, but on later examination, you did what I expected you to do, I just couldn't tell at first.

Write more clearly and if you are missing something, I missed it, too. Don't be afraid to argue with the book, if need be, but it will take a clear demonstration to provide a different answer with confidence.

Edit: Hey, Soroban, you beat me to it, this time.

Hello!

The big triangle is a 3,4,5 triangle with a total area of (1/2*4*3) 6cm^2
(I recognize a right triangle with legs 2 and 4 as a 3,4,5 triangle w/o needing Pythagoras---too many years with a carpenters square)

if it is rotated around the hypotenuse it makes 2 cones with a common base (I'll call this base b). This base is also the base of two right triangles with heights x and y. Moreover x+y is equal to the hypotenuse of the 3,4,5 triangle. So x+y=5, and the sum of the area of the triangles is equal to the area of the 3,4,5 triangle. or

At=Ax+Ay
6=1/2bx+1/2by=1/2b(x+y)=5/2b
& rearranging I get
b=2*6/5=12/5cm

The formula for the volume of a cone is Vc=1/3pi*r^2*h

In this case this volume is the sum of two cones with b=12/5cm and combined height of 5cm

Vt=Vx+Vy=1/3pi*(12/5)^2*x+1/3pi*(12/5)^2*y
Vt=1/3pi*(12/5)^2*(x+y)=1/3pi*(12/5)^2*5=144/15*pi=9.6*pi cm^3


I´hope this will help, my english is not very good so please forgive me.

Well, there's the problem. We don't know who you are. One has only to guess, sometimes.

Your result looks fine.

I was impressed that there were three different ways to calculate the altitude. Mine was by far the hardest.