Question

ajutati maaaa ex 1....​

Answer 1

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

$\boxed{BD=18cm}$

BC=BD+CD

BC=18+8

$\boxed{BC=26cm}$

TEOREMA CATETEI:

AB²=BD*BC

$AB=\sqrt{BD*BC}=\sqrt{18*26} =6\sqrt{13} \\\boxed{AB=6\sqrt{13} cm}$

TEOREMA CATETEI:

AC²=DC*BC

$AC=\sqrt{CD*BC}=\sqrt{8*26} =4\sqrt{13} \\\boxed{AC=4\sqrt{13} cm}$

ARIA TRIUNGHIULUI DREPTUNGHIC:

$A=\frac{AB*AC}{2} = \frac{AD*BC}{2}$

$A=\frac{CB*AD}{2} =\frac{26*12}{2} =156 \ cm^{2}$

-----------------------------------------rezolvare b)-----------------------------------------------

BC=BD+CD

BC=10,8+19,2

BC= 30dm

Teorema inaltimii:

AD²=BD*CD

$AD=\sqrt{BD*DC}=\sqrt{10,8*19,2} =14,4 \\\boxed{AD=14,4\ dm}$

TEOREMA CATETEI:

AB²=BD*BC

$AB=\sqrt{BD*BC}=\sqrt{10,8*30} =18 \\\boxed{AB=18\ dm}$

TEOREMA CATETEI:

AC²=DC*BC

$AC=\sqrt{CD*BC}=\sqrt{19,2*30} =24 \\\boxed{AC=24\ dm}$

ARIA TRIUNGHIULUI DREPTUNGHIC:

$A=\frac{AB*AC}{2} = \frac{AD*BC}{2}\\ A=\frac{AB*AC}{2}=\frac{24*18}{2} =216\ dm^{2}\\\boxed{A=216\ dm^{2}}$

-----------------------------------------rezolvare c)-----------------------------------------------

BC=BD+CD

CD=BC-BD

CD=15-5,4

CD=9,6 dm

Teorema inaltimii:

AD²=BD*CD

$AD=\sqrt{BD*DC}=\sqrt{5,4*9,6} =7,2 \\\boxed{AD=7,2\ dm}$

TEOREMA CATETEI:

AB²=BD*BC

$AB=\sqrt{BD*BC}=\sqrt{5,4*15} =9 \\\boxed{AB=9\ dm}$

TEOREMA CATETEI:

AC²=DC*BC

$AC=\sqrt{CD*BC}=\sqrt{9,6*15} =12 \\\boxed{AC=12\ dm}$

ARIA TRIUNGHIULUI DREPTUNGHIC:

$A=\frac{AB*AC}{2} = \frac{AD*BC}{2}$

$A=\frac{AB*AC}{2} =\frac{9*12}{2} =54 \ dm^{2}\\\boxed{A=54 \ dm^{2}}$