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Intr-un triunghi dreptunghic ABC ,m (∡A)=90°, AD⊥BC ,D∈(BC) ,se dau:
a)AD=9.6 dm si BD=7,2 dm.Calculati CD,BD si [tex] A_{ABC} [/tex].
b)BD=2 dm si BC=10 dm.Calculati [tex] A_{ABC} [/tex]
c)CD=96 cm si BC =15 dm.Calculati AD si [tex] A_{ABC} [/tex]

Va rog!!!Ofer 15 punctee!!(+8 cel mai bun raspuns)

Răspuns :

Renatemambouko
a)
teorema inaltimii
AD²=BD×CD
9,6²=7,2×CD
CD=92,16/7,2=12,8 dm
BC-7,2+12,8=20 dm
A(ABC)=BC×AD/2=20×9,6/2=96 dm²

b)
CD=10-2=8 dm
teorema inaltimii
AD=√BD×CD=√2×8=√16=4 dm
A(ABC)=BC×AD/2=10×4/2=20 dm²

c)
CD=96 cm=9,6 dm
BD=15-9,6=5,4 dm
teorema inaltimii
AD=√BD×CD=√5,4×9,6=√51,84=7,2 dm
A(ABC)=BC×AD/2=15×7,2/2=54 dm²

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