ABCD square

In the ABCD square, the X point lies on the diagonal AC. The length of the XC is three times the length of the AX segment. Point S is the center of the AB side. The length of the AB side is 1 cm. What is the length of the XS segment?

Correct result:

x =  0.354 cm

Solution:

a=1 cm u=2⋅ a=2⋅ 1≐2 cm≐1.4142 cm  u1=3/4⋅ u=3/4⋅ 1.4142≐1.0607 cm u2=1/4⋅ u=1/4⋅ 1.4142≐0.3536 cm  u3=3/4⋅ a=3/4⋅ 1=34=0.75 cm u4=1/4⋅ a=1/4⋅ 1=14=0.25 cm  s=a/2=1/2=12=0.5 cm  s2=s−u4=0.5−0.25=14=0.25 cm  x=s22+u42=0.252+0.252=0.354 cma=1 \ \text{cm} \ \\ u=\sqrt{ 2 } \cdot \ a=\sqrt{ 2 } \cdot \ 1 \doteq \sqrt{ 2 } \ \text{cm} \doteq 1.4142 \ \text{cm} \ \\ \ \\ u_{1}=3/4 \cdot \ u=3/4 \cdot \ 1.4142 \doteq 1.0607 \ \text{cm} \ \\ u_{2}=1/4 \cdot \ u=1/4 \cdot \ 1.4142 \doteq 0.3536 \ \text{cm} \ \\ \ \\ u_{3}=3/4 \cdot \ a=3/4 \cdot \ 1=\dfrac{ 3 }{ 4 }=0.75 \ \text{cm} \ \\ u_{4}=1/4 \cdot \ a=1/4 \cdot \ 1=\dfrac{ 1 }{ 4 }=0.25 \ \text{cm} \ \\ \ \\ s=a/2=1/2=\dfrac{ 1 }{ 2 }=0.5 \ \text{cm} \ \\ \ \\ s_{2}=s - u_{4}=0.5 - 0.25=\dfrac{ 1 }{ 4 }=0.25 \ \text{cm} \ \\ \ \\ x=\sqrt{ s_{2}^2 + u_{4}^2 }=\sqrt{ 0.25^2 + 0.25^2 }=0.354 \ \text{cm}

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