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#1 2005-12-04 06:35:03

johny
Member
Registered: 2005-11-19
Posts: 34

A mighty Confusion!!!!

I am basically doing a frustrum of a cone, what i need to do in the question is re-arrange the volume for cone to make H the subject and substitute this H in the Surface area formula for cone. If u get what i mean!!!!!!!!! Now the problem is this, i made a diagram of a truncated cone, and put some dotted lines to make it into a whole cone.........I said the dotted lines to be as 1/3 of the radius and height. SO when i did the volume expression it came something like this:

V = 1/3*pi*r^2*h    -   1/3*pi*r^2/9*h/3.  ........................(i)

When u simplify the equation it come up something like this:

h = 81V/26*pi*r^2

I had to make volume fixed to 600.

so:  h =  1869 / pi*r^2.

Now this is where the problem comes:

Now when i substitute this equation into surface area as i said i had to:

S.A. = Pi*R*S   where S= square root of H^2 + R^2

S.A. = Pi*r  Square root of 1869/pi*r^2  + r^2

Well here it is, if i have already subtracted 1/3 of height and radius in the volume formula in equation (i)
, then do i need to do, divide remaining r^2 with 3??? Or have i already subtracted the r and h in equation (i)
.
S.A. = Pi*r/3   square root of 1869/Pi*r^2  +  r^2/3          if u get what i mean???

Please try to quickly explain this, as i need to hand this in by tommarrow!!!!!!

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#2 2005-12-05 06:38:11

irspow
Member
Registered: 2005-11-24
Posts: 457

Re: A mighty Confusion!!!!

Following your reasoning the height and radius are related to the original height and radius so no further manipulation is needed.  However I came up with this:

A = square root of [(6561 V^2 / 676 pi r^3) + r^2]

Don't worry too much about the values I used, I kept it exact instead of rounding, but I think that you failed to square the term for h in the area formula.

Sadly, I doubt that this is what your teacher is looking for anyway.

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