Hi Jessica,

I am not sure what formulas you are using. I don't rremember very many formulas but I can show you how I find surface areas of the figures you describe.

Cylinder

Imagine that your cylinder is a soup can with the top and bottom removed. Suppose the radius of the end is r and the height is h.

Cut along the line indicated and roll out the can. You now have a rectangle of dimensions 2 r by h and hence an area of

2 r h.

If you want to include the ends of the can in the surface area you need to add

2 r²

Regular Pyramid

Suppose the the sqare base of the pyramid has side length x, then the area of the base is x² The remainder of the surface is three, congruent triangles.

Let P be the apex of the pyramid and A be the midpoint of one of the sides of the base. |PA| is the height of the triangular face and hence you can find the area of this triangle if you can find |PA|.

If you know the height of the pyramid (h in the diagram) then PCA is a right triangle wit sides of length h and ^x/₂. |PA| can then be found using Pythagoras' Theorem.

PBA is also a right triangle so if you know |PB| you can use Pythagoras' Theorem again to find |PA|.

Right Cone

For the surface area of I right cone look at Walter's answer to an earlier question.

Prism

The diagram I drew is for a right, triangular prism but the idea is the same regardless of the shape of the polygon that forms the base.

The surface area is

2 (the area of the base) + (the sum of the areas of the sides)

In a right prism the sides are rectangles so each side has area

(length of a side of the base) (height)

If the prism is not a right prism then the sides are parallelograms and you then need to use whatever facts you know about the prism to find the dimensions of the parallelograms.

Penny

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