The Third Dimension Examples

Sketch the orthogonal drawings from the front, right, left, and top.

Looking from the front (the orange faces) and going left to right, the figure is 3 cubes high and 4 cubes wide, with a middle step of 2 cubes. The view from the front should look like this:

From the right, the figure's height increases going left to right.

The view from the top would look like this:

The view from the left is like the view from the front, only there is no change in depth.

Identify the solid and name each base, face, edge, and vertex.

Vertices are the points where edges intersect. In the figure, the vertices are all the named points: A, B, C, D, E, F, G, H, I, and J.

Edges are line segments where faces meet. The edges are: AB, BC, CD, DE, AE, AJ, BF, CG, DH, EI, FG, GH, HI, IJ, and FJ.

The faces are the different polygons that make up the polyhedron's surface. The faces consist of five rectangles and two pentagons. The rectangles are BCGF, CDHG, DEIH, EIJA, and ABFJ.

Unlike Washington, D.C., which has only one pentagon as its base, this solid has two. The pentagons are ABCDE and FGHIJ.

These polygons are regular and they are both parallel and congruent to each other. That means this solid is a regular pentagonal prism.

Which cross-sections will result if this solid is split by a perfectly vertical and perfectly horizontal planes through the center?

The solid we're given is a cone and basically, it's asking which shapes we get if we cut it up different ways. Cutting it with a vertical plane in the center, the pointed top and the flat bottom gives us a triangle. Cutting it with a horizontal plane will give us a circle. Cutting it with a laser beam would give us a great deal of fun followed by a trip to the eye doctor.

Draw a net for the solid and calculate its surface area.

If we crack the pyramid open at its vertex and let the triangular faces fall flat, we end up with a net that looks like this:

To find the surface area, we just have to find the area of the square base and the four triangles. The area of a square is equal to s² and the area of a triangle is ½bh. Since we have one square and four triangles, our surface area equation is:

SA = s² + 4(½bh) = s² + 2bh

Looking at the net we drew, side s is the same as b. They both equal 10. The height of the triangle is 15. Let's plug in those values and solve.

SA = 10² + 2 × 10 × 15
SA = 100 + 300
SA = 400 units²