Question

∆ABC-dr.
m(<A)=90°
AD⊥BC, D ∈ (BC)
a) 4BD-CD=12
CD/BD= 1 întreg și 7/9

b)CD-BD=56/5
BC=40

aflați: BD=? CD=? AB=? AC=? AD=?

Rezolvată prin teorema înălțimii vă rog!​

Answer 1

Răspuns:

Explicație pas cu pas:

a) 4BD = CD+12,  CD = BD*16/9

4BD = BD*16/9 +12

36BD -16BD =108

20BD =108,  BD = 108/20 = 27/5

CD = 27/5*16/9 = 16/15

BC = BD+CD = 27/5 + 16/15 = 97/15

AD^2 = BD*CD = 27/5*16/15 = 16*9/25 (t.inaltimii)

AD = 4*3/5 = 12/5

AB^2 = BD^2 +AD^2= (27/5)^2 +(16/15)^2

AB^2 = 729/25 + 256/225 =(9*729+256)/225

AB^2 = 6817/225,   AB = √6817/15

AC^2 = CD^2 +AD^2 = (16/15)^2 +(12/5)^2

AC^2 = 256/225 +144/25 = (256+9*144)/225

AC^2 = 1552/225,  AC= √1552/15

b) CD -BD = 56/5

BC = CD+BD = 40  le adunam

2CD = 56/5 +40 = (56 +200)/5 = 256/5

CD = 256/10 = 128/5

BD = 40 -128/5 = (200-128)/5 = 72/5

AD^2 = CD*BD = 128/5*72/5 =9216/25

AD = 96/5

etc.