Question

URGENT!!!!!!!!!!!!!!!!!​

Answer 1

∆ABC

AM ; BN mediane G centru de greutate

aria∆ ABC=120cm²

a) ∆MNG≈∆ABG {MN linie mijlocie ;deci MN ll AB}

sau mai explicit

<MAB=<AMN alterne interne AM secantă

<MGN=<AGB opuse la vârf

b)aria∆ABG

aria ∆ABC=120cm²=AB×h/2

aria∆ABG=AB×h/3×2=

AB ×h/2×1/3=120×1/3=40cm²