Question 4 Equilateral triange AB.Co is given. For every n, natural number, An, Bn, Cn are midpoints of sides Bn iCn 1, Cn 1 An 1, An 1Bn 1, respectively. Part of triangle A, B.Co that is outside of triangle A B C shade in red color. On the same way shade part of triangle A2B2C2 outside of triangle A3B3C3 and so on. Continue the process and find percentage of triangle A, B.Co that is shaded.
Answers
Given : equilateral triangle A₀, B₀, C₀ is given
for every n, natural numbers Aₙ , Bₙ , Cₙ are mid points of Bₙ₊₁ Cₙ₋₁ , Cₙ₋₁ Aₙ₋₁ and Aₙ₋₁ Bₙ₋₁
Part of triangle A₀B₀C₀ that is outside of triangle A₁B₁C₁ shade in red color
on the same way A₂B₂C₂ outside of triangle A₃B₃C₃ and so on.
To Find : percentage of triangle A₀B₀C₀ that is shaded.
Solution:
Area of ΔA₀B₀C₀ = A
Then Area of ΔA₁B₁C₁ = A/4 as sides are half
Hence Part of triangle A₀B₀C₀ that is outside of triangle A₁B₁C₁ shade in red color = A - A/4 = 3A/4
Area of ΔA₂B₂C₂ = (A/4)/4 = A/16
Then Area of Δ A₃B₃C₃ = (A/16)/4 = A/64
Hence Part of triangle ΔA₂B₂C₂ that is outside of triangle Δ A₃B₃C₃ shade in red color = A/16 - A/64 = 3A/64
3A/4 then 3A/64 and so on
This is an GP
with a = 3A/4
r = 1/16
Sₙ = a/(1 - r) for infinite series r < 1
Hence Sum = (3A/4)/( 1 - 1/16)
= 12A/15
= 4A/5
= 80 %
80 % of triangle A₀B₀C₀ is shaded.
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