marlinsfan Posted March 13, 2005 Share Posted March 13, 2005 Im a total idiot and i can't find a pyramid or a triangular prism that was a surface area that is less than its volume by a 40 cm margin.. If you could find a triangular prism/ pryamid that has no more than 320 cm squared and at the same time has a volume of over 360 cm cubed....i would really apprieciate it thank you soo much Link to comment Share on other sites More sharing options...
jordan Posted March 13, 2005 Share Posted March 13, 2005 Use SA-40=V and substitute in the formulas for SA and V. Since it doesn't seem to give any other paramiters, you can make up something for all the variables but one and then solve for that last one. That should yeild the correct answer. Link to comment Share on other sites More sharing options...
marlinsfan Posted March 13, 2005 Author Share Posted March 13, 2005 Use SA-40=V and substitute in the formulas for SA and V. Since it doesn't seem to give any other paramiters, you can make up something for all the variables but one and then solve for that last one. That should yeild the correct answer. well thats my problem....i add in random numbers and it doesn't help...i mean i get really close but not close enough...then i try to alter it minorly and then it just goes no where...thanks though Link to comment Share on other sites More sharing options...
jordan Posted March 13, 2005 Share Posted March 13, 2005 First, what equations are you using for SA and V? Let's see, SA should be...1/2lw+3[(1/2)sqrt(h2w2)(l)] And V is...1/3(lwh) Actualy, those are probably wrong, but it was my quick guess at them. Anyway, what are you using for you formulas? Link to comment Share on other sites More sharing options...
Callipygous Posted March 13, 2005 Share Posted March 13, 2005 Im a total idiot and i can't find a pyramid or a triangular prism that was a surface area that is less than its volume by a 40 cm margin..If you could find a triangular prism/ pryamid that has no more than 320 cm squared and at the same time has a volume of over 360 cm cubed....i would really apprieciate it thank you soo much plugging in random numbers will probably give you a hard time. does it have to be 320 and 360? aim for equal side lenghts, the closer to a sphere the better. if you dont have to have 320 and 360, and it just needs to be a difference of 40, then make it bigger. Link to comment Share on other sites More sharing options...
marlinsfan Posted March 13, 2005 Author Share Posted March 13, 2005 First' date=' what equations are you using for SA and V? Let's see, SA should be...1/2lw+3[(1/2)sqrt(h[sup']2[/sup]w2)(l)] And V is...1/3(lwh) Actualy, those are probably wrong, but it was my quick guess at them. Anyway, what are you using for you formulas? SA= (2*area of base) + (Perimeter * Hieght) V= (Area of Base * Hieght * 1/3) Thats for a PRISM the thing about SA-40=V...i've tried it before, it gives an amazing margin of around 87 cm, but it doesn't meet the requirement of the volume being greater or equal to 360 cm cubed...otherwise it would be fine...if that doesn't make sense just tell me and ill further explain it...thank you for your help Link to comment Share on other sites More sharing options...
marlinsfan Posted March 13, 2005 Author Share Posted March 13, 2005 woops, i take that back...even when i do the SA-40=V, it still didn't work...i seriously need help...please Link to comment Share on other sites More sharing options...
marlinsfan Posted March 13, 2005 Author Share Posted March 13, 2005 plugging in random numbers will probably give you a hard time. does it have to be 320 and 360? aim for equal side lenghts' date=' the closer to a sphere the better. if you dont have to have 320 and 360, and it just needs to be a difference of 40, then make it bigger.[/quote'] I should have been more clear on the requirements: The Surface Area HAS to be less than 321cm and the Volume HAS to be 360 or greater...equal side lengths...i tried but it doesn't help too much Link to comment Share on other sites More sharing options...
jordan Posted March 13, 2005 Share Posted March 13, 2005 Ok, one more question for you. Can you sepcify exactly what shape we're working with here? Draw a picture or link to a picture of it because your formulas and descprition "triangular prism/ pryamid" seem to be leading in two different directions. Link to comment Share on other sites More sharing options...
marlinsfan Posted March 13, 2005 Author Share Posted March 13, 2005 Ok, one more question for you. Can you sepcify exactly what shape we're working with here? Draw a picture or link to a picture of it because your formulas and descprition "triangular prism/ pryamid" seem to be leading in two different directions. true...i really don't mind which shape i can find...if its either a rectangular-based pryamid or a triangular pyramid...or even a triangular prism...as long as it reaches those requirements...thats all that matters...it can even be a cone The Formulas for Pyramids : V= 1/3 * Area of Base * Hieght I BELIEVE THIS IS THE FORMULA FOR SA OF A PYRAMID= Area of Base = (1/2*Perimeter*Slant Hieght) Link to comment Share on other sites More sharing options...
marlinsfan Posted March 13, 2005 Author Share Posted March 13, 2005 OK...ive been using the Formula SA of Pyramid = Area of Base + (1/2 * Perimeter * Slant Hieght but i just read this:Surface Area: Add the area of the base to the sum of the areas of all of the triangular faces. The areas of the triangular faces will have different formulas for different shaped bases it doesn't seem to be the same thing...which one is correct? Link to comment Share on other sites More sharing options...
jordan Posted March 13, 2005 Share Posted March 13, 2005 I'd say just stick with a square-based pyramid and don't worry about the others, in which case you'd have the right SA formula except for one thing. I would add a 4 in front of the 1/2PH because that will give you the area of all of the triangular faces. Link to comment Share on other sites More sharing options...
marlinsfan Posted March 13, 2005 Author Share Posted March 13, 2005 I'd say just stick with a square-based pyramid and don't worry about the others, in which case you'd have the right SA formula except for one thing. I would add a 4 in front of the 1/2PH because that will give you the area of all of the triangular faces. yeah, i just noticed that....i have a horrible teacher and i copied it down off the board...so what i found of the internet is that Surface Area is just the Area of the Triangular Faces summed up is true...that helps alot Link to comment Share on other sites More sharing options...
Recommended Posts
Create an account or sign in to comment
You need to be a member in order to leave a comment
Create an account
Sign up for a new account in our community. It's easy!
Register a new accountSign in
Already have an account? Sign in here.
Sign In Now