Question

Cine ma ajuta la ex 6 repede va rog

Answer 1

6)  a)  fie medianele AA', BB', CC'      AA' = BA' = CA' = BC/2 = 10cm
in ΔABB'    AB' = AC/2 = 5√3    (BB')² = 100 + 75    BB' = 5√7cm
in ΔACC'    AC' = AB/2 = 5    (CC')² = 300 + 25     CC' = 5√13cm
b) fie AD_|_ BC   inaltimi : AD = AC/2 = 5√3 cm    AB = 10cm      AC = 10√3cm
c)fie AM, BN,CP bisectoare ⇒ in Δ ABD    BD² = AB² - AD² = 100 - 75 = 25
BD = 5cm  ⇒ CD = 15 cm  ⇒ inΔADM  DM = CD-BD = 10cm  AD = 5√3
AM = √(100 + 75) = 5√7 cm      in ΔABN  AN = BN/2 ⇒  BN² - BN²/4 = 100
3BN² = 400  BN = 20√3/3 cm    inΔABC cu PC bisect. ⇒PB/PA= BC/AC=2/√3
(PB + PA)/PA = (2+√3)√3      10/PA = (2+√3)/ √3 ⇒ PA = 10√3/(2+√3)
PA = 10√3 (2-√3)/1 = 20√3 - 30  inΔAPC    PC² = 1200 - 1200√3 +900 + 300
PC² = 2400 - 1200√3 = 10√[2(12 - 6√3)] = 10 √[2(3-√3)²]
PC = 10√2(3-√3) cm
d) R = BC/2 = 10cm    A ΔABC = AB·AC/2 = (AB + BC + CA)/2 ·r
50√3 = (30+10√3)/2 ·r     r = 100√3 /(30 + 10√3) = 100√3(30 - 10√3)/600
r = 3(√3 - 1)/6 = (√3 -1)/2 cm