Question

Calculati raza cercului circumscris unui trapez isoscel ABCD ( AB paralel cu CD ) , daca :
a) AD = DC = a si AB = 2a ;
b) AB = 25 cm , CD = 7 cm si AD = 15 cm .

Answer 1

   
Desenele sunt in fisierul atasat.


[tex]\displaystyle\\ a)\\ \text{Se da:}\\ \text{Trapezul isoscel ABCD cu }(AB || CD)\\ AB=2a\\ BC=CD=AD=a\\ \text{Se cere:}\\ \text{Raza cercului circumscris }R=?\\ \text{Desenam diagonala AC}\\ \text{Din C coboram perpendiculara }CE~cu~E\in AB\\\\ \text{Rezolvare:}\\\\ EB=\frac{AB-CD}{2}=\frac{2a-a}{2}=\frac{a}{2}\\\\ \Longrightarrow\text{ Cateta EB = cu jumatate din ipotenuza BC in }~\Delta EBC\\ \Longrightarrow~~\ \textless \ ECB=30^o\\ \Longrightarrow~~\ \textless \ EBC=60^o\\[/tex]

[tex]\displaystyle \\ \\ \Longrightarrow~~ \ \textless \ BCD=180 - \ \textless \ ABC = 180-60= 120^o \\ \text{fiind unghiuri suplimentare.}\\ \ \textless \ ADC=\ \textless \ BCD = 120^o ~~\text{deoarece trapezul este isoscel} \\ \\ \text{In triunghiul isoscel ADC avem:} \\ \ \textless \ ADC=120^o \\ \Longrightarrow~~\ \textless \ DAC = \ \textless \ ACD = \frac{180 - ADC}{2}=\frac{180 - 120}{2}=\frac{60}{2}=30^o \\ \Longrightarrow~~\ \textless \ ACB = \ \textless \ BCD - \ \textless \ ACD = 120 - 30 = \boxed{90^o} \\ \Longrightarrow~~\Delta ABC \text{este triunghi dreptunghic in C} [/tex]

[tex]\displaystyle \text{Daca trapezul ABCD este inscris in cerc, } \\ \text{atunci si triunghiul dreptunghic ABC este inscris in cerc} \\ \\ \text{Rezulta ca diametrul cercului este ipotenuza AB a triunghiului} \\ \\ \Longrightarrow ~~ D_{cerc} = AB = \boxed{2a }\\ \\ \Longrightarrow ~~R = \frac{D}{2} = \frac{2a}{2} = \boxed{a} [/tex]



[tex]\displaystyle b)\\ \text{Se da:}\\\text{Trapezul isoscel ABCD cu }(AB || CD)\\ AB=25~cm\\ BC=AD=15~cm\\ CD =7~cm\\ \text{Se cere:}\\ \text{Raza cercului circumscris }R=?\\ \text{Desenam diagonala AC}\\ \text{Din C coboram perpendiculara }CE~cu~E\in AB\\\\\text{Rezolvare:}\\\\ EB=\frac{AB-CD}{2}=\frac{25-7}{2}=\frac{18}{2}=9~cm\\\\ In~\Delta BCE~avem:\\ Ipotenuza~BC=15~cm\\ Cateta~EB=9~cm\\ CE=\sqrt{BC^2-EB^2}=\sqrt{15^2-9^2}=\sqrt{225-81}=\sqrt{144}=12~cm[/tex]

[tex]\displaystyle \\ In~\Delta ACE~avem:\\ Cateta~CE=12~cm\\ Cateta~AE=AB - EB = 25 - 9 = 16~cm\\ AC=\sqrt{CE^2+AE^2}=\sqrt{12^2-16^2}=\sqrt{144+256}=\sqrt{400}=20~cm \\ \\ In~\Delta ABC~avem:\\ AB=25~cm\\ BC= 15~cm\\ AC=20~cm \\ \text{Verificam daca triunghiul ABC este dreptunghic: } \\ \\ AB^2 =?= BC^2+AC^2 \\ 25^2 =?= 15^2+20^2 \\ 625=?= 225+400 \\ 625= 625 \\ \Longrightarrow~~\Delta ABC \text{ este triunghi dreptunghic.} [/tex]

[tex]\displaystyle \\ \\ \text{Daca trapezul ABCD este inscris in cerc, } \\ \text{atunci si triunghiul dreptunghic ABC este inscris in cerc} \\ \\ \text{Rezulta ca diametrul cercului este ipotenuza AB a triunghiului} \\ \\ \Longrightarrow ~~ D_{cerc} = AB = \boxed{25~m }\\ \\ \Longrightarrow ~~R = \frac{D}{2} = \frac{25}{2} = \boxed{12,5~cm}[/tex]