Math Problem Statement

The midpoints of the three sides of an equilateral triangle in Figure 1 are used to create the vertices of a new triangle, which is then removed, leaving three triangles shown in Figure 2. This is repeated for each of the remaining triangles to create Figure 3. This process continues for the remaining triangles. The shaded area of each Figure in the pattern follows a geometric sequence. The area of the triangle in Figure 1 is 100 cm². (a) Write down the common ratio, r. (b) Find the shaded area in Figure 10. (c) Find the total shaded areas of Figures 1 to 10.

Solution

Let's break down the problem based on the provided image.

Problem Summary:

  1. Figure 1 is an equilateral triangle with a total area of 100 cm².
  2. In Figure 2, a smaller equilateral triangle (created by joining the midpoints of the original triangle) is removed, leaving three smaller triangles.
  3. This process is repeated for Figure 3, Figure 4, and so on, creating a fractal-like pattern.
  4. The shaded area of each figure forms a geometric sequence.

We need to address the following:

Part (a): Write down the common ratio, rr.

  • In Figure 1, the area is 100 cm².
  • In Figure 2, a smaller triangle (1/4 of the original area) is removed, leaving 3/4 of the area shaded.
    • So, the shaded area in Figure 2 is 100×34=75cm2100 \times \frac{3}{4} = 75 \, \text{cm}^2.

Thus, the common ratio rr is 34\frac{3}{4}.

Part (b): Find the shaded area in Figure 10.

  • The shaded area follows a geometric sequence with:
    • Initial area a=100cm2a = 100 \, \text{cm}^2
    • Common ratio r=34r = \frac{3}{4}

The formula for the nn-th term in a geometric sequence is: An=A1×rn1A_n = A_1 \times r^{n-1} For Figure 10 (n=10n = 10): A10=100×(34)9A_{10} = 100 \times \left(\frac{3}{4}\right)^{9}

Let's calculate this value.

Part (c): Find the total shaded areas of Figures 1 to 10.

  • The sum of the first nn terms of a geometric sequence is given by: Sn=a(1rn)1rS_n = \frac{a(1 - r^n)}{1 - r} For Figures 1 to 10 (n=10n = 10, a=100a = 100, and r=34r = \frac{3}{4}): S10=100(1(34)10)134S_{10} = \frac{100(1 - \left(\frac{3}{4}\right)^{10})}{1 - \frac{3}{4}}

I'll calculate both (b) and (c) now.### Results:

  • (b) The shaded area in Figure 10 is approximately 7.51 cm².
  • (c) The total shaded area of Figures 1 to 10 is approximately 377.47 cm².

Would you like further details or clarifications on any of these calculations?

Here are 5 related questions to expand on this topic:

  1. How does the area reduction pattern change if a different fraction of the triangle is removed?
  2. What would happen to the total shaded area if the process continued indefinitely?
  3. Can the total area ever become zero? Why or why not?
  4. How does this fractal pattern relate to other famous fractals like the Sierpinski triangle?
  5. How would the solution change if the initial triangle's area were different?

Tip: When working with geometric sequences, always ensure you carefully identify the initial term and common ratio before applying formulas.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Geometry
Geometric Sequences

Formulas

A_n = A_1 * r^(n-1)
S_n = (a(1 - r^n))/(1 - r)

Theorems

Geometric Sequence Sum Formula

Suitable Grade Level

Grades 9-12