Two parallel chords

In a circle 70 cm in diameter, two parallel chords are drawn so that the center of the circle lies between the chords. Calculate the distance of these chords if one of them is 42 cm long and the second 56 cm.

Result

v1 =  49 cm
v2 =  7 cm

Solution:

d=70 cm r=d/2=70/2=35 cm t1=42 cm t2=56 cm  x1=r2(t1/2)2=352(42/2)2=28 cm x2=r2(t2/2)2=352(56/2)2=21 cm  v1=x2+x1=21+28=49 cmd=70 \ \text{cm} \ \\ r=d/2=70/2=35 \ \text{cm} \ \\ t_{1}=42 \ \text{cm} \ \\ t_{2}=56 \ \text{cm} \ \\ \ \\ x_{1}=\sqrt{ r^2 - (t_{1}/2)^2 }=\sqrt{ 35^2 - (42/2)^2 }=28 \ \text{cm} \ \\ x_{2}=\sqrt{ r^2 - (t_{2}/2)^2 }=\sqrt{ 35^2 - (56/2)^2 }=21 \ \text{cm} \ \\ \ \\ v_{1}=x_{2}+x_{1}=21+28=49 \ \text{cm}
v2=x2x1=2128=7 cmv_{2}=|x_{2}-x_{1}|=|21-28|=7 \ \text{cm}



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