fie h1 si h2 inaltimile treiunghiurilor de arii 15 si respectiv, 10
atunci
40 *h1/2+40*h2/2=15+10
40 * latime/2=15+10=25
40*latime=50
latime =50:40=1,25
arie dreptunghi =40*1,25=50
50=15+5+10+x=30+x
x=20
prima problema
3/8=AB/7=MN/9
⇒AB=21/8 si MN=27/8
Perim AMN=3+3/8+27/8=3+30/8=3+15/4=12/4+15/4=27/4