# VOLUME OF SPHERES PRACTICE AND PROBLEM SOLVING A/B ANSWER KEY

A rectangular prism of volume mm 3 has a rectangular base of length 10 mm and width 8 mm. Suppose the kkey and bottom of a frustum are circles of radius R and rrespectively, and that the height of the frustum is hwhile the height of the original cone is H.

Find the height h of the prism.

To find the area of the curved surface of a cone, we cut and open up the curved surface to form a sector with radius las shown below. It allows us to say that the volume of any rectangular prism, right or oblique, is given by the area of the base multiplied by the height.

To find the surface area, we find the area of each face and add them together.

Hence, if the radius of the base circle of the cylinder is r and its sphefes is hthen:. We can think of the given shape as a larger rectangular prism of dimensons 8, 3 and 10 cm from which a smaller prism of dimensions x, x and 3 cm has been cut.

Of these, the answerr, obtained by slicing a cone by a plane as shown in the diagram below, is studied in some detail in junior secondary school. Conical drinking cups and storage vessels have also been found in several early civilisations, confirming the fact that the cone is also a shape of great antiquity, interest and application.

The word sphere is simply an English form of the Greek sphaira meaning a ball. Take a hemisphere of radius and look at the area of a typical cross-section at height above the base.

Conical and pyramidal shapes are volume of spheres practice and problem solving a/b answer key used, generally in a truncated form, to store grain and other commodities. Hence, using the formula for the volume of the sphere, we have. When it was built, the Great Pyramid of Cheops in Egypt had a height of Prcatice we draw the four long diagonals as shown, then we obtain six square-based pyramids, one of which is shaded in the diagram.

Here is a interesting formula that uses the idea of approximating the spherss by pyramids with a common vertex at the centre of the sphere.

Take two square pyramids of height hone with base square length 2 h and one with base square length 2 l. More references to triangles and geometry.

Also in that module, we defined the surface area of a prism to be the sum of the areas of all its faces. A History of Mathematics: In the earlier module, Area Volume and Surface Area we developed formulas and principles for finding the volume and surface areas for prisms. The later Greek Mathematician Pappus AD discovered the following remarkable method for finding the volume of a solid of revolution generated by rotating a plane region with area A about a fixed axis.

### Cones_Pyramids_and_Spheres

The volume of each pyramid is equal to Arwhere A is the area of the base. The area of one square face of a cube is equal to 64 cm 2. The volume of the solid is equal to the product of the area A of the plane region and the distance travelled by its geometric centroid as it moves through one revolution. It remains to say that a similar method can be used to progress from a pyramid with rectangular base to one whose base is a regular polygon, although the technical volume of spheres practice and problem solving a/b answer key are more complicated and will not be given here.

The height of the pyramid is 4 cm and the side length of the base is 6 cm, find the surface area of the pyramid. Hence the total volume of these pyramids is rnA.

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When developing the formula for the volume of a cylinder volume of spheres practice and problem solving a/b answer key the module Area Volume and Surface Areawe approximated the cylinder using inscribed polygonal prisms. Adding in the area of the circles at each end of the cylinder, we obtain. This method can be used very effectively to find the volume of solids which do not have uniform cross-section, and may have curved boundaries. Suppose there are n of these pyramids in the sphere, each with base area A.

By taking more and more sides in the polygon, we obtained closer and closer approximations to the volume of the cylinder. We will use the second principle to show that the formula holds for any square pyramid of height h. The portion of a right cone remaining after a smaller cone is cut off is called a frustum.