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In the simplest form possible please.

2007-03-08 07:12:12 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

height x width x depth = volume

using the formula in the above answer ( for an irregular object ) would require this formula to calculate the volume of the water ! ( lol )

2007-03-08 07:16:48 · answer #1 · answered by Anonymous · 0 3

(1) If the object is a prism or a cylinder:
V = Area of the base x height

(2) If the object is a pyramid or a cone:
V = 1/3 x Area of the base x height

(3) If the object is a sphere:
V = 4/3 x Pi x R^3 (where R is the radius)

A simple case of (1) is when you have a rectangular prism (i.e., when the base is a rectangle). A shoe box is a good example of a rectngular prism. Then the area of the base is A = width x length (of the rectangle) , hence the volume is obtained as
V = width x length x height

An even simpler case is a cube, where width = length = height. So if the edge of a cube has length a, then
V = a x a x a = a^3

If you have a very oddly shaped object, "break" it into smaller pieces shaped as prisms, pyramids, cones, half-spheres.... then the Volume of the big object is the sum of the volumes of all the small pieces.

Hope this helped.

2007-03-08 15:20:44 · answer #2 · answered by M 6 · 1 0

Put the object in a bucket marked with volume marks. Add water to completely submerge the object, take the object out. Measure the water and subtract from the volume of your bucket. The object must be water proof!

2007-03-08 15:17:58 · answer #3 · answered by irf 4 · 0 2

Place it in pool of water and measure the change of water level. You will have the volume in waters' known weight to Volume ratio

2007-03-08 15:16:46 · answer #4 · answered by occluderx 4 · 0 1

The easiest way is to immerse it in a liquid and measure the amount of liquid before and after.

2007-03-08 15:15:55 · answer #5 · answered by H.C.Will 3 · 0 1

firstly you times the width by the length and the depth
and there you go!

2007-03-09 04:21:01 · answer #6 · answered by Anonymous · 0 0

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