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ma ajuta cineva? exercițiul 3 si 4

Ma Ajuta Cineva Exercițiul 3 Si 4 class=

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Nico234679ghd
a)
BD=0,6 cm
DC= 4/15 cm
BC= 0,6+4/15=13/15 cm
In ∆ABC dr => T. In. AD² = CD×BD
AD²= 0,6×4/15
AD ²= 4/25
AD = √4/25
AD= 2/5 cm

b)
BD= 8/3 cm
BC=25/6 cm
BC = BD+CD =>CD = 25/6-8/3=3/2 cm
In ∆ABC =>T. In. AD² = CD×BD
AD² = 8/3×3/2
AD² =4
AD=√4
AD=2
Ovdumi
3.
a) BD=3/5 cm, DC=4/15 cm
BC=BD+DC=3/5 + 4/15=13/15 cm
teorema inaltimii AD
AD^2=BD x DC=12/75 
AD=2/5 cm
b) BD=8/3 cm, BC=25/6 cm
DC=BC-BD=25/6 - 8/3
DC=3/2 cm
AD=√(BD x DC)=√(8/3 x 3/2)=2 cm
c) BD=4/5 cm, AD=12/5 cm
AD^2=BD x DC
DC=(144/25)/(4/5)=36/5 cm
BC=BD+DC=4/5+36/5=8 cm

4.
a) BD=2√3 cm, BC=8√3 cm
DC=BC-BD=6√3 cm
AD=√(BD x DC)=√(2√3 x 6√3)=6 cm
b) AD=6 cm, BD=4 cm
AD^2=BD x DC
DC=AD^2/BD=36/4=9 cm
c) AD=3√3 cm, DC=3 cm
AD^2=BD x DC
BD=AD^2/DC=27/3=9 cm


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