Recursion squares

In the square ABCD is inscribed a square so that its vertices lie at the centers of the sides of the square ABCD.The procedure of inscribing square is repeated this way. Side length of square ABCD is a = 22 cm.

Calculate:
a) the sum of perimeters of all squares
b) the sum of area of all squares

Result

Σ p =  300.45 cm
Σ S =  968 cm2

Solution:

p1=4⋅22 qp=12 Σp=4⋅22⋅11−12=300.45  cm p_1 = 4\cdot 22 \ \\ q_p = \dfrac{ 1}{ \sqrt2} \ \\ \Sigma p = 4\cdot 22 \cdot \dfrac{ 1} { 1- \dfrac{ 1}{ \sqrt2}} = 300.45 \ \text { cm }
 S1=222 qS=12 ΣS=222⋅11−12=968 cm2 \ \\ S_1 = 22^2 \ \\ q_S = \dfrac{ 1}{ 2} \ \\ \Sigma S = 22^2 \cdot \dfrac{ 1} { 1- \dfrac{ 1}{ 2}} = 968 \ cm^2

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Following knowledge from mathematics are needed to solve this word math problem:

Pythagorean theorem is the base for the right triangle calculator. See also our trigonometric triangle calculator.

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