The information on this page is for teachers who want to guide students through some of the fractals quiz problems. They're tricky and kids might need some clarification and hints to get them going.

A "stage 0" Koch Curve is just a line segment. No matter how long it is, consider its length to be 1 unit -so that the lengths of the later stages can be easily compared to the first one.

When the segment in "stage 1" is split in three equal parts, each of the smaller segments has a length of 1/3 of a unit. Once the equilateral triangle is made and the middle segment is removed, there are 4 segments each having a length of 1/3 of a unit. This makes the total length of the "stage 1" Koch Curve 4/3 of a unit.

In "stage 2", each of the four smaller segments is split into 3 equal parts. Since the length of each segment was 1/3 of a unit, the length of each of the smaller segments is now 1/3 of 1/3 - or 1/9 of a unit. After the middle segment in each part is erased and the equilateral triangles are formed, the students can see that there are 16 of these "1/9 unit" segments. To find the total length of the stage 2 Koch curve, add all of these together to get a result of 16/9 or 1 and 7/9 units.

At this point (or possibly earlier) the students may notice a pattern. At each stage, the length of the previous stage is multipled by 4/3 to get the length of the new stage. So the pattern will continue: 1, 4/3, 16/9, 64/27, 256/81, etc. This length keeps getting larger and larger. If you keep going long enough (through enough stages) you can make the length as large as you want. Because of this, the actual length of the Koch curve is considered to be infinite!

A "stage 0" Sierpinski gasket is just a triangle. No matter what size it is, consider its area to be 1 square unit - so that the area of the later stages can be easily compared to the first one.

When the midpoints of each side are connected in "stage 1", the first triangle is separated into four smaller triangles each of which are exactly the same and shape. This means that the area of each of the smaller triangles is 1/4 of a square unit. When the middle triangle is removed to form the first stage of the Sierpinski gasket, the remaining area is therefore 1 - 1/4 or 3/4 of a square unit.

In stage 2, each of the remaining 3 black triangles is split into four smaller triangles (again by connecting the midpoints of the sides). Since the area of each of the 3 triangles was originally 1/4 of a square unit, the area of each of the smaller triangles is now 1/4 of 1/4 - or 1/16 of a square unit. When the middle triangle is removed from each of the three original black triangles, the students will see that there a total of 9 of the smaller triangles left, each having an area of 1/16 of a square unit. If the areas of these are all added together, it gives a total area of 9/16 of a square unit for stage 2 of the gasket.

In stage 3, the process is continued. The new small triangles now formed will each have an area of 1/4 of 1/16 or 1/64 of a square unit. Since there will be 27 of them, the area of stage 3 of the Sierpinski Gasket will be 27/64 of a square unit.

At this point (or possibly earlier) the students may notice a pattern. At each stage, the area of the previous stage is multipled by 3/4 to obtain the area of the new stage. So the pattern will continue: 1, 3/4, 9/16, 27/64, 81/256, etc. This area keeps getting smaller and smaller. If you keep going long enough (through enough stages) you can make the area as small as you want. Because of this, the actual area of the Sierpinski gasket is considered to be zero.