Two chords

Calculate the length of chord AB and perpendicular chord BC to circle if AB is 4 cm from the center of the circle and BC 8 cm from the center of the circle.

Result

a =  16 cm
b =  8 cm

Solution:

h1=4 cm h2=8 cm   r2=h12+h22  r=h12+h22=42+824 5 cm8.9443 cm  D=2 r=2 8.94438 5 cm17.8885 cm  a<D,b<D  r2=h12+(a/2)2   a=2 r2h12=2 8.9443242=16 cmh_{1}=4 \ \text{cm} \ \\ h_{2}=8 \ \text{cm} \ \\ \ \\ \ \\ r^2=h_{1}^2 + h_{2}^2 \ \\ \ \\ r=\sqrt{ h_{1}^2 + h_{2}^2 }=\sqrt{ 4^2 + 8^2 } \doteq 4 \ \sqrt{ 5 } \ \text{cm} \doteq 8.9443 \ \text{cm} \ \\ \ \\ D=2 \cdot \ r=2 \cdot \ 8.9443 \doteq 8 \ \sqrt{ 5 } \ \text{cm} \doteq 17.8885 \ \text{cm} \ \\ \ \\ a<D, b<D \ \\ \ \\ r^2=h_{1}^2 + (a/2)^2 \ \\ \ \\ \ \\ a=2 \cdot \ \sqrt{ r^2 - h_{1}^2 }=2 \cdot \ \sqrt{ 8.9443^2 - 4^2 }=16 \ \text{cm}
r2=h22+(b/2)2  b=2 r2h22=2 8.9443282=8 cmr^2=h_{2}^2 + (b/2)^2 \ \\ \ \\ b=2 \cdot \ \sqrt{ r^2 - h_{2}^2 }=2 \cdot \ \sqrt{ 8.9443^2 - 8^2 }=8 \ \text{cm}



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