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Tue 28 Jun, 2005 02:14 am
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.
Thanks!!!
The volume of a cone is (h*pi*r^2)/3. In this case you have two cones with a common base. The radius of that base is 12/5 (area of right triangle is 6, therefore altitude from hypotenuse must be 6*2/5).
The volume of the cones is:
(x*pi*(12/5)^2)/3 + (y*pi*(12/5)^2)/3
where:
x+y = h (which is 5)
so you get:
(5*pi*(12/5)^2)/3 = 9.6*pi
Seems to me, you got it right.
I get the same answer
I get the same answer
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
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Muki!
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