# Surface Area

The surface of a solid is the two-dimensional boundary that separates the space inside of the three-dimensional figure from the space outside it. You might think of the surface as the "skin" of the solid. Some curricula compare it to a "wrapper" or "jacket" of the solid. (For more, see surface.)

The surface area of a solid is the area of its surface. It is the area you would need to paint in order to cover the solid. If the object has a surface that can be "unfolded" flat without stretching or distorting it (not possible for all solids), then the area of that flattened surface is the surface area of the solid.

When students first encounter surface area and volume, they may have a hard time distinguishing them because in their earlier study of 2 dimensions, they may well have thought of area as what's "inside" the figure, with perimeter being the measure of the "outside." In three dimensions, the volume is the "inside" and the surface area is a measure of the "outside."

## Discovering formulas for surface area

### Prisms and cylinders

The reasoning in the sections below applies to *right* prisms and cylinders, ones that "stand straight up" on their bases (i.e., have axes that are perpendicular to the bases). These are the figures students most commonly encounter in school. The reasoning is easily adaptable for prisms and cylinders that "slant."

#### Triangular prism

Here are two views ofa triangular prism: . We can "unfold" the prism's surface (you might think of it as the "skin" or "jacket" or "wrapper" of the solid) to see it entirely in a single plane. Here is what it looks like as we begin the unfolding . When the surface is entirely unfolded, it is called the net of the polyhedron, and looks like this: . The net shows the pair of congruent triangles that were the two bases of the prism, and three rectangles that were the lateral faces. The three rectangles can also be viewed as one large rectangle: . The surface area of the triangular prism then is the sum of the areas of these two triangles and the rectangle. Because the triangles are congruent, they have the same area, so the surface area can also be expressed as

2 x Area_{triangle} + Area_{rectangle}.

Notice that one of the dimensions of the rectangle is the height of the prism and the other dimension is the perimeter of the base triangle. Therefore, the formula can be written as:

_{triangle}) + (height x Perimeter

_{triangle})

.

#### Rectangular prisms

We can do the same thing with a rectangular prism . Choosing the green faces as bases,^{[1]} we may unfold to see the net , and may think of that net as a pair of congruent bases and a large rectangle that represents the remaining faces . As before, one of the dimensions of this large rectangle is the height of the prism and the other dimension is the perimeter of the (green) base.

Again, the total area of the net -- which is the same as the total surface area of the prism -- is twice the area of the base (that is, the area of the two green faces) plus the area of the large rectangle (purple, in this figure) that folds around those bases to form the remaining lateral faces.

_{base-rectangle}) + (height x Perimeter

_{base-rectangle}).

#### Other prisms

The surface of a regular right pentagonal or hexagonal prism can be unfolded to form a net with the same characteristic form: two congruent bases and a long rectangle whose dimensions are the height of the prism and the perimeter of the prism's base.

Will this work for prisms whose bases are not regular polygons?

The net of any of these solids consists of the two bases and a rectangle one of whose dimensions matches the perimeter of the base. Thus, the formula applies to all right polygonal prisms.

_{base}) + (height x Perimeter

_{base})

#### Cylinder

What about cylinders?

Picture removing the circular top and bottom of a cylindrical oatmeal box and then, using one, straight, vertical cut, slitting open the side and flattening it out. The result is three pieces: the circular bases, and the rectangular wall of the box. One dimension of that wall is the height of the box; the other dimension matches the circumference of the circular base.

Although, in the case of the circle, we normally refer to the measurement around its interior as circumference rather than perimeter, the principle is the same! The surface area of any of these solids -- right prisms or cylinders -- is:

(2 x Area_{base}) + (height x Perimeter_{base})

### Pyramid and Cone

Again, the reasoning here applies directly only to the *right* pyramid and cone.

The net of a pyramid with a square base consists of a square and four congruent triangles.

The triangles can be rearranged to form a parallelogram. The base of this parallelogram is half the perimeter (semiperimeter) of the square and its height is the slant height of the pyramid. Thus, the formula for the surface area of this pyramid is

Area of square + 1/2 perimeter of square * height of triangle