Realizeaza figura. AB e baza mare, de jos, CD e baza mica de sus.
BD bisectoare ABC de 60 grade => ABD=DBC=30 grade.
In triunghiul DAB, A are 90 grade, B are 30 grade = > AD=BD/2. => BD=2AD.
Aplici Pitagora > BD² = AD²+AB² => (2AD)²= AD² + 15² => 4AD² =AD² +225 => 3AD² = 225 => AD² = 75 => AD = 5√3.
Duci CE perpendiculara pe AB. Astfel CE=AD = 5√3. si AE=CD
In triunghiul CEB, E are 90 grade, B are 60 grade => C are 30 de grade deci BE = BC/2 => BC=2BE
Aplici Pitagora > BC²=CE²+BE² => (2BE)²=(5√3)²+BE² => 4 BE² = 25*3 + BE² => 3BE² = 25*3 => BE²=25 => BE = 5. si BC=2BE=10.
CD=AE.
AE = AB - BE = 15 - 5 = 10 => CD = 10.
Perimetrul = AB+BC+CD+DA = 15 + 10 + 10 + 5√3 = 35+5√3
