From an equilateral triangle with side 2014 we cut off another equilateral triangle with side 214, such that both triangles have one common vertex and two of the side of the smaller triangles lie on two of the side of the bigger triangle. Is it possible to cover this figure with figures in the picture without overlapping (rotation is allowed) if all figures are made of equilateral triangles with side 1? Explain the answer! [asy] import olympiad; unitsize(20); pair A,B,C,D,E,F,G,H; A=(0,0); B=(1,0); C=rotate(60)*B; D=rotate(60)*C; E=rotate(60)*D; F=rotate(60)*E; G=rotate(60)*F; draw(A--B); draw(A--C); draw(A--D); draw(A--E); draw(A--F); draw(A--G); draw(B--C--D--E--F--G--B); A=(2,0); B=A+(1,0); C=A+rotate(60)*(B-A); D=A+rotate(60)*(C-A); E=A+rotate(120)*(D-A); F=A+rotate(60)*(E-A); G=2*F-E; H=2*C-D; draw(A--D--C--A--B--C--H--B--G--F--E--A--F--B); A=(4,0); B=A+(1,0); C=A+rotate(-60)*(B-A); D=B+rotate(60)*(C-B); E=B+rotate(60)*(D-B); F=B+rotate(60)*(E-B); G=E+rotate(60)*(D-E); H=E+rotate(60)*(G-E); draw(A--B--C--A); draw(C--D--B); draw(D--E--B); draw(B--F--E); draw(E--G--D); draw(E--H--G); A=(8.5,0.5); B=A+(1,0); C=A+rotate(60)*(B-A); D=A+rotate(60)*(C-A); E=A+rotate(60)*(D-A); F=A+rotate(60)*(E-A); G=A+rotate(60)*(F-A); H=G+rotate(60)*(F-G); draw(A--B); draw(A--C); draw(A--D); draw(A--E); draw(A--F); draw(A--G); draw(B--C); draw(D--E--F--G--B); draw(G--H--F);[/asy]