a)Conform desenului atasat
[tex]O_{1}O_{2}=T_{1}X_{1}[/tex]
[tex]T_{2}X_{1}=r_{1}-r_{2}=20cm-8cm=12cm[/tex]
Din triunghiul dreptunghic [tex]T_{1}T_{2}X_{1}[/tex]
conform teoremei lui Pithagora
[tex]T_{1}X_{1} = \sqrt{ T_{1}T_{2}^{2}-T_{2}X_{1}^{2} } = \sqrt{256+144} = \sqrt{400} =20[/tex]
b) pntru ca sa nu schimb desenul am utilizat notatii inverse caci ajung la acelasi raspuns - in mod normal e nevoie de un desen nou
[tex]MT_{2}=8 \sqrt{3} \\ MT_{1}=6 \sqrt{3} \\ r_{1}=6cm[/tex]
Avem 2 triunghiuri congruente
[tex]T_{2}MO_{2} [/tex]≡[tex]T_{1}MO_{1}[/tex]⇒
[tex]\frac{r_{1} }{M_{1}T_{1}} =\frac{r_{2} }{M_{2}T_{2}}[/tex] ⇒
[tex]r_{2} =\frac{r_{1} }{M_{1}T_{1}}*M_{2}T_{2}=\frac{6}{6\sqrt{3}}*8\sqrt{3}=8cm[/tex]
[tex]T_{1}T_{2}=MT_{2}-MT_{1}=2\sqrt{3}cm[/tex]
[tex]T_{2}X_{1}=r_{2}-r_{1}=8cm-6cm=2cm[/tex]
si acum iarasi conform teoremei lui Pithagora
[tex]T_{1}X_{1} = \sqrt{ T_{1}T_{2}^{2}-T_{2}X_{1}^{2} } = \sqrt{12+4} = \sqrt{16} =4cm[/tex]
[tex]O_{1}O_{2}=T_{1}X_{1}=4cm[/tex]
c) Aici e simplu! Iarasi avem 2 triunghiuri congruente
[tex]T_{1}MO_{1}[/tex] ≡[tex]T_{2}T_{1}X_{1}[/tex]⇒
masura unghiului [tex]T_{1}MO_{1}[/tex] = [tex]T_{2}T_{1}X_{1}[/tex]
[tex]T_{2}T_{1}X_{1}=16-4=12cm[/tex]
[tex]ctg( 30^{o} )=\frac{T_{1}X_{1}}{T_{2}X_{1}}=\sqrt{3}[/tex] unde
[tex]T_{1}X_{1}=O_{1}O_{2}[/tex]⇒
[tex]O_{1}O_{2}=\sqrt{3}*T_{2}X_{1}=12\sqrt{3}cm[/tex]
