Thanks, but easy. I dragged a line from B and make a ractangle and marked it as "b". So AB=6-b . and two equations: (6-b)^2+X^2=57.49 and b^2+X^2=81 X=7.51
@santiagoarosam4303 ай бұрын
EC=√(6²-5²)=√11→Si F es la proyección ortogonal de B sobre DC→ Razón de semejanza entre CFB y CED =s=9/6=3/2→ X=BF=5*s=5*3/2=15/2. Gracias y saludos.
@petrusneacsu98583 ай бұрын
Why did you calculate EC? It is of no use.
@santiagoarosam4303 ай бұрын
@@petrusneacsu9858 Es un detritus mental no eliminado. La falta de higiene es imperdonable. Dejo ese error sin corregir para ejemplo y aviso de las generaciones futuras. Gracias por la observación. Un saludo cordial.
@santiagoarosam4303 ай бұрын
@@petrusneacsu9858 Es un detritus mental no eliminado. La falta de higiene es imperdonable. Dejo ese error sin corregir para ejemplo y aviso de las generaciones futuras. Gracias por la observación. Un saludo cordial.
@ناصريناصر-س4ب3 ай бұрын
Let h be the perpendicular projection of B on CD. Triangles Bhc and CDE are similar, hence Bh=AD=x=15/2
@petrusneacsu98583 ай бұрын
What you say is method 1. See at 03:30
@ناصريناصر-س4ب3 ай бұрын
@@petrusneacsu9858Yes, but I found the solution on my own.
@paatabidzinashvili61562 ай бұрын
6*SinC=5, SinC=5/6, x=9*SinC, x=9*5/6=15/2
@michaeldoerr58103 ай бұрын
The answer is 15/2. I have notcied that in the second methos, the square us equal to itself because two triangle areas are equal while having one side that is part of [BCD] while having a different side that is part of a subtended right angle. I hope that that is the correct explanation. And I finally learned about what makes the first method ao powerful. This is similar to yesterday's problem!!!
@bene4090Ай бұрын
Gostei do segundo método, pois não é muito comum de ser utilizado, "Parabéns" I am from Brazil.