Question

Va rog frumos exercitiul 2!

Answer 1

a)25+75=BC^2 (T.PITAGORA)
BC=10 cm
b)t.catetei : AB^2=BD×BC
25=BD×10 =>BD=2,5 cm
t.catetei : AC^2=DC×BC
75=DC×10 =>DC=7,5 cm
t.inaltimii : AD^2=BD×DC
AD^2=2,5×7,5=18,75 =>AD=4,33 cm
c)A=(AB×AC)/2=25rad3/2 cm^2

Answer 2

a) AB=5 cm
AC=5√3 cm
In ∆ABC dr=>T. P
AB2+AC²=BC²
5²+(5√3)²=BC²
BC² =25+75
BC² =100
BC=√100
BC=10 cm

b) in ∆ABC dr =>T. Cat
AB²= BD×BC
5²= BD×10
25= BD×10
BD=25:10
BD=2,5 cm
BC=CD+BD=>CD= BC-BD=10-2,5= 7,5 cm
In ∆ABC dr =>T in
AD²= BD×CD
AD² = 2,5×7,5
AD² =18,75
AD=√18,75
AD=5√3/2

C)
A=b×h/2 = 5×5√3/2= 25√3/2 cm