# Infinite sum of areas

Above the height of the equilateral triangle ABC is constructed an equilateral triangle A1, B1, C1, of the height of the equilateral triangle built A2, B2, C2, and so on. The procedure is repeated continuously. What is the total sum of the areas of all triangles if the ABC triangle has a length? And?

Correct result:

s =  1.732 a2

#### Solution:

$S=\sqrt{ 3 }/4 \ a^2 \ \\ a_{1}=\sqrt{ 3 }/2 \ a \ \\ \ \\ S_{1}=\sqrt{ 3 }/4 \ a_{1}^2=\sqrt{ 3 }/4 \cdot \ ( \sqrt{ 3 }/2 \ a)^2 \ \\ S_{1}=3/4 \ S \ \\ \ \\ S_{2}=3/4 \ S_{1} \ \\ ... \ \\ \ \\ q=S_{2}/S_{1}=S_{1}/S=.. \ \\ q=\dfrac{ 3 }{ 4 }=0.75 \ \\ q<1 \ \\ \ \\ s=\dfrac{ \sqrt{ 3 }/4 }{ 1-q }=\dfrac{ \sqrt{ 3 }/4 }{ 1-0.75 }=\sqrt{ 3 }=1.732 \ \text{a}^2$ Our examples were largely sent or created by pupils and students themselves. Therefore, we would be pleased if you could send us any errors you found, spelling mistakes, or rephasing the example. Thank you!

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