Răspuns:
a) BD= 18cm; BC=26cm; AB=6√13 cm; AC=4√13 cm; A=156 cm²
b) BD= 30dm; AD= 14,4dm; AB= 18dm; AC= 24dm; A= 216dm²
C) DC= 9,6dm; AD= 7,2dm; AB= 9dm; AC= 12dm; A= 54dm²
Explicație pas cu pas:
-----------------------------------------rezolvare a)-----------------------------------------------
Teorema inaltimii:
AD²=BD*CD
144=BD*8
BD=144:8
[tex]\boxed{BD=18cm}[/tex]
BC=BD+CD
BC=18+8
[tex]\boxed{BC=26cm}[/tex]
TEOREMA CATETEI:
AB²=BD*BC
[tex]AB=\sqrt{BD*BC}=\sqrt{18*26} =6\sqrt{13} \\\boxed{AB=6\sqrt{13} cm}[/tex]
TEOREMA CATETEI:
AC²=DC*BC
[tex]AC=\sqrt{CD*BC}=\sqrt{8*26} =4\sqrt{13} \\\boxed{AC=4\sqrt{13} cm}[/tex]
ARIA TRIUNGHIULUI DREPTUNGHIC:
[tex]A=\frac{AB*AC}{2} = \frac{AD*BC}{2}[/tex]
[tex]A=\frac{CB*AD}{2} =\frac{26*12}{2} =156 \ cm^{2}[/tex]
-----------------------------------------rezolvare b)-----------------------------------------------
BC=BD+CD
BC=10,8+19,2
BC= 30dm
Teorema inaltimii:
AD²=BD*CD
[tex]AD=\sqrt{BD*DC}=\sqrt{10,8*19,2} =14,4 \\\boxed{AD=14,4\ dm}[/tex]
TEOREMA CATETEI:
AB²=BD*BC
[tex]AB=\sqrt{BD*BC}=\sqrt{10,8*30} =18 \\\boxed{AB=18\ dm}[/tex]
TEOREMA CATETEI:
AC²=DC*BC
[tex]AC=\sqrt{CD*BC}=\sqrt{19,2*30} =24 \\\boxed{AC=24\ dm}[/tex]
ARIA TRIUNGHIULUI DREPTUNGHIC:
[tex]A=\frac{AB*AC}{2} = \frac{AD*BC}{2}\\ A=\frac{AB*AC}{2}=\frac{24*18}{2} =216\ dm^{2}\\\boxed{A=216\ dm^{2}}[/tex]
-----------------------------------------rezolvare c)-----------------------------------------------
BC=BD+CD
CD=BC-BD
CD=15-5,4
CD=9,6 dm
Teorema inaltimii:
AD²=BD*CD
[tex]AD=\sqrt{BD*DC}=\sqrt{5,4*9,6} =7,2 \\\boxed{AD=7,2\ dm}[/tex]
TEOREMA CATETEI:
AB²=BD*BC
[tex]AB=\sqrt{BD*BC}=\sqrt{5,4*15} =9 \\\boxed{AB=9\ dm}[/tex]
TEOREMA CATETEI:
AC²=DC*BC
[tex]AC=\sqrt{CD*BC}=\sqrt{9,6*15} =12 \\\boxed{AC=12\ dm}[/tex]
ARIA TRIUNGHIULUI DREPTUNGHIC:
[tex]A=\frac{AB*AC}{2} = \frac{AD*BC}{2}[/tex]
[tex]A=\frac{AB*AC}{2} =\frac{9*12}{2} =54 \ dm^{2}\\\boxed{A=54 \ dm^{2}}[/tex]
