Question

Fie C(O,r). Aflati distanta de la centrul cercului la coarda AB, daca:
a)AB= 8cm, r = 5 cm
b)AB= 24 cm, r=13cm
c)AB=a), r= b)

Answer 1

[tex]\text{Ducem razele OA si OB. } \\ \Delta ABC ~este ~isoscel ~deoarece ~ OA = OB = r \\ \text{Distanta de la O la AB se masoara pe inaltimea triunghiului din O} \\ Ducem ~inaltimea~ OD, D \in AB, AD = BD. \\ a)OA = 5; ~AB=8 =\ \textgreater \ ~ AD = 4 \\ =\ \textgreater \ ~ OD= \sqrt{OA^2 - AD^2} = \sqrt{5^2 - 4^2}=\sqrt{25 - 16}=\sqrt{9} = \boxed{3\;cm} \\ \\ b)OA = 13; ~AB=24 =\ \textgreater \ ~ AD = 12 \\ =\ \textgreater \ ~ OD= \sqrt{OA^2 - AD^2} = \sqrt{13^2 - 12^2}=\sqrt{169 - 144}=\sqrt{25} = \boxed{5\;cm} [/tex]


[tex]\displaystyle b)OA = b; ~AB=a =\ \textgreater \ ~ AD = \frac{a}{2} \\ \\ =\ \textgreater \ ~ OD= \sqrt{OA^2 - AD^2} = \sqrt{b^2 - (\frac{a}{2})^2}= \\ \\ =\sqrt{b^2 - \frac{a^2}{2^2}}= \sqrt{b^2 - \frac{a^2}{4}}= \sqrt{ \frac{4b^2}{4} - \frac{a^2}{4}}= \boxed{\frac{1}{2} \sqrt{ 4b^2 - a^2}~cm} [/tex]