Concept explainers
Consider the construction of a Koch snowflake starting with a seed triangle having sides of length 81 cm. Let M denote the number of sides, L the length of each side, and P the perimeter of the "snowflake" obtained at the indicated step of the construction. Complete the missing entries in Table 12-1 Q.
Table 12-1
M | L | P | |
Start | 3 | 81 cm | 243 cm |
Step 1 | 12 | 27 cm | 324 cm |
Step 2 | |||
Step 3 | |||
Step 4 | |||
Step 5 |
To complete:
The given chart.
Answer to Problem 1E
Solution:
The given chart is,
M | L | P | |
Start | 3 | ||
Step 1 | 12 | ||
Step 2 | 48 | ||
Step 3 | 192 | ||
Step 4 | 768 | ||
Step 5 | 3072 |
Explanation of Solution
Given:
The given chart is,
M | L | P | |
Start | 3 | 81 cm | 243 cm |
Step 1 | 12 | 27 cm | 324 cm |
Step 2 | |||
Step 3 | |||
Step 4 | |||
Step 5 |
Calculation:
Draw an equilateral triangle with side length of
Figure-1
To the middle third of each of the sides of this seed, add an equilateral triangle with sides of length 27 cm. The result is a 12 sided snowflake with perimeter,
Figure-2
To the middle third of each of the 12 sides of snowflake add an equilateral triangle with sides of length one third of the length of the side. The result is a snowflake with 48 sides each side length of 9 cm and the perimeter is,
Figure-3
M | L | P | |
Start | 3 | ||
Step 1 | 12 | ||
Step 2 | 48 | ||
Step 3 | 192 | ||
Step 4 | 768 | ||
Step 5 | 3072 |
Therefore, the given chart calculations are stated above.
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Chapter 12 Solutions
EXCURSIONS IN MODERN MATH. >ANNOT.<