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?

Result

s =  1.732 a2

Solution:

S=3/4 a2 a1=3/2 a  S1=3/4 a12=3/4 (3/2 a)2 S1=3/4 S  S2=3/4 S1 ...  q=S2/S1=S1/S=.. q=34=0.75 q<1  s=3/41q=3/410.75=31.7321=1.732 a2S = \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 } \doteq 1.7321 = 1.732 \ a^2







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