Question 1027385
This would work:
{{{drawing(300,300,-3,17,-1,19,
line(-3,0,17,0),
line(0,0,0,16),
green(rectangle(-0.3,16,0.3,20)),
locate(0.4,17.5,green(window)),
line(3,0,3,11),
red(line(14,0,0,14)),
locate(14,3,red(ladder)),red(arrow(14,2.5,11.5,2.5)),
rectangle(0,0,0.7,0.7),rectangle(3,0,3.7,0.7),
locate(-2.8,10,bldng),locate(-2.1,9,wall),
locate(3.1,6,11ft),locate(8,1,11ft),
locate(1,1,3ft),locate(7.2,3,fence),arrow(7,2.5,3,2.5)
)}}} There are three isosceles right triangles {{{drawing(300,300,-3,17,-1,19,
line(-3,0,17,0),
line(0,0,0,16),
triangle(0,14,3,11,0,11),rectangle(0,11,0.5,11.5),
locate(1,11,3ft),locate(0.1,12.5,3ft),locate(0.1,6,11ft),
locate(0.4,17.5,green(window)),
line(3,0,3,11),
red(line(14,0,0,14)),
locate(14,3,red(ladder)),red(arrow(14,2.5,11.5,2.5)),
rectangle(0,0,0.7,0.7),rectangle(3,0,3.7,0.7),
locate(-2.8,10,bldng),locate(-2.1,9,wall),
locate(3.1,6,11ft),locate(8,1,11ft),
locate(1,1,3ft),locate(7.2,3,fence),arrow(7,2.5,3,2.5)
)}}}
The largest of those isosceles right triangles has legs measuring {{{3ft+11ft=14ft}}} . Its hypotenuse is the length of the ladder
{{{sqrt((14ft)^2+(14ft)^2)=sqrt(2)*14ft="about 19.8 ft"}}} (rounded to the nearest tenth of a foot).
 
A slightly longer ladder would do too:
{{{drawing(300,300,-3,17,-1,19,
line(-3,0,17,0),
line(0,0,0,16),
green(rectangle(-0.3,16,0.3,20)),
locate(0.4,17.5,green(window)),
line(3,0,3,11),
red(line(14,0,0,16)),locate(0.1,8,16ft),
locate(14,3,red(ladder)),red(arrow(14,2.5,11.82,2.5)),
rectangle(0,0,0.7,0.7),rectangle(3,0,3.7,0.7),
locate(-2.8,10,bldng),locate(-2.1,9,wall),
locate(3.1,6,11ft),locate(8,1,11ft),
locate(1,1,3ft),locate(7.2,3,fence),arrow(7,2.5,3,2.5)
)}}} The length of that ladder is {{{sqrt((14ft)^2+(16ft)^2)=sqrt(196ft^2+256ft^2)=sqrt(452ft^2)="about 21.3 ft"}}} (rounded to the nearest tenth of a foot).
 
The question is what your teacher's idea of the right answer is.