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gaurip2012
Average Member

USA
10 Posts

 Posted - 03/28/2013 :  17:46:34 I need to find the volume of a pentagon.So far in the picture they give us a cube (as the base) measuring 11.5in. and a triangle (on top) with the height of 7.4inthe answer choices are a)2,499.525 in.b)2,012.2 in.c)978.7 in.d)489.325 in.

the_hill1962

USA
1468 Posts

 Posted - 04/01/2013 :  10:20:21 Since this is not a "regular" shape, you can't use a formula.You need to break your shape up into parts in which you can calculate the volume of each. This type of problem is called finding the "composite volume".For your shape, you have a CUBE and a pyramid.Find the volume of each and add the answers together.If you need help in how to find the volume of a cube and/or the volume of a pyramid, let us know. Edited by - the_hill1962 on 04/02/2013 11:50:07

gaurip2012
Average Member

USA
10 Posts

 Posted - 04/01/2013 :  13:52:47 I tried that, the answer I get is not there, that is why I tried posting it here. I got answer A when i followed the formula, but my math teacher told me it was B, she didn't know how it was B, and the book she got it from didn't explain it either. But any advice on finding it, is welcome

TchrWill

USA
80 Posts

 Posted - 04/01/2013 :  18:54:11 quote:Originally posted by gaurip2012I need to find the volume of a pentagon.So far in the picture they give us a cube (as the base) measuring 11.5in. and a triangle (on top) with the height of 7.4inthe answer choices are a)2,499.525 in.b)2,012.2 in.c)978.7 in.d)489.325 in.I am not at all certain as to what figure or shape you are describing.An 11.5 in. square topped by a triangle 7.4 in. high is a planar figure of 5 sides, or pentagonal. You can calculate the area of this figure by adding the area of the square to the area of the triangle or A = 11.5^2 + 1`1.5(7.4)/3 = 174.8 sq. in., clearly no where near the choices. The figure, or shape, must be three dimensional.You cannot calculate a volume of a planar or two dimensional figure. This is why I feel the shape you describe might be a cube, 11.5^3 cub.in. topped by a pyramid with a 11.5 in. square base and 7.4 in. height, the total volume of which would be V = 11.5^3 + 11.5(7.4)/3 = 1549.2 cub.in. This shape is not pentagonal however, having 5 cubic surfaces and 4 triangular surfaces.Since neither of these answers matches the choices, you will have to be sure of the figure, or shape, you are trying to explain. Perhaps your teacher should do the calculations to prove that the answer is not one of the choices. I had a similar experience in college where the teacher gave us a problem on a test and we were all marked wrong. I searched through my library of books only to find the same problem with the answer incorrect. I think your best bet would be to politely ask your teacher to go through the calculations with you to show where the error is.Another thought might be that you are seeking the surface area of the three dimensional shape, the 11.5 in. cube plus the pyramid atop the cube. This is As = 11.5^2(5) + 9.731(11.5)4/2 = 885 sq.in. The 9.731 in. is the slant height of the triangular pyramid or h(s)= sqrt(7.4^2 + 5.75^2).Again, none of the choices.Make sure you are giving us the correct shape description. Edited by - TchrWill on 04/01/2013 22:44:00

the_hill1962

USA
1468 Posts

 Posted - 04/02/2013 :  12:41:42 The cube has a volume of 11.5 = 1520.875 and the pyramid has a volume of (1/3)(11.5*11.5)(7.4) = 326 13/60.Adding them together gives 1847 11/120That is not in the choices. Edited by - the_hill1962 on 04/02/2013 12:45:44

gaurip2012
Average Member

USA
10 Posts

 Posted - 04/02/2013 :  14:11:45

gaurip2012
Average Member

USA
10 Posts

 Posted - 04/02/2013 :  14:26:08

gaurip2012
Average Member

USA
10 Posts

 Posted - 04/02/2013 :  14:28:05 does that help?sorry about the misunderstandings [:1]

TchrWill

USA
80 Posts

 Posted - 04/02/2013 :  15:42:13 Yeah!!!! We have an object.The volume of this shape isV = 11.5^3 + (1/2)11.5(7.4)11.5 = 2010.2 cub.in.The nearest choice is 1012.2 or (b).Close enough?Another proof of that old saying, "A picture is worth a thousand words."

gaurip2012
Average Member

USA
10 Posts

 Posted - 04/02/2013 :  15:49:24 so basically i just had to find the volume of the cube with s and then find 1/2 base + base height?And that is the way to find the volume of a regular hexagon (I don't know the name, sorry)...

TchrWill

USA
80 Posts

 Posted - 04/02/2013 :  19:42:27 Volume of the cube is 11.5 cubed.Volume of the triangular cap is ((1/2)CUBE SIDE x H) x CUBE SIDE orV = [(1/2)11.5 x 7.4]11.5.This shape is not a regular hexagon however.

gaurip2012
Average Member

USA
10 Posts

 Posted - 04/04/2013 :  14:59:15 by why the extra 11.5? [(1/2)11.5 x 7.4]11.5

TchrWill

USA
80 Posts

 Posted - 04/04/2013 :  15:31:44 quote:Originally posted by gaurip2012by why the extra 11.5? [(1/2)11.5 x 7.4]11.5The volume of the triangular shaped top is the area of the triangle, At = 11.5(7.4)/2, or 42.55 sq.in., times the length of the triangular cross-section, 11.5 in., for a total of 489.325 cub.in..The cube is clearly Vc = 11.5^3 = 1520.875 cub.in..The total is therefore 2010.2 cub.in.

gaurip2012
Average Member

USA
10 Posts

 Posted - 04/04/2013 :  17:38:55 ohhhh thanks a bunch
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