\(CD=16cm\)
mà CD cắt AB là đường kính \(\Rightarrow\) \(HC=HD=\dfrac{CD}{2}=\dfrac{16}{2}=8cm\)
\(AB=20cm\Rightarrow R=10cm\)
Xét \(\Delta COH\) vuông tại \(H\)có:
\(CO^2=HO^2+HC^2\) (đ/l Pytago)
\(\Rightarrow HO^2=CO^2-HC^2\)
\(\Rightarrow HO=\sqrt{10^2-8^2}\)
\(\Rightarrow HO=6cm\)
OC=OA=OB=0,5AB=0,5.20=10 (cm).
CH=DH=0,5CD=0,5.16=8 (cm).
OC2=CH2+OH2 (Pythagorean theorem).
\(\Rightarrow\) OH=\(\sqrt{OC^2-CH^2}=\sqrt{10^2-8^2}=6\left(cm\right)\).
The length of OH is 6 cm.
OC=OA=OB=0,5AB=0,5.20=10 (cm).
CH=DH=0,5CD=0,5.16=8 (cm).
OC2=CH2+OH2 (Pythagorean theorem).
⇒⇒ OH=√OC2−CH2=√102−82=6(cm)OC2−CH2=102−82=6(cm).
The length of OH is 6 cm.
