Solving by Pythagoras theorem, qe get BE² = (1.25)² – 1² Hence, BE = 0.5m

The various forces acting are

• The weight mg of the ladder acting vertically downward at C
• The horizontal force F due to the wall at B
• Reaction of the floor f acting along AB. This is the force exerted by the ground on the ladder. It is the resultant of f_x and f_y

 Resolving the forces, to find solution to the problem

• Resolving the horizontal components of force, we get F = f_x which is the force exerted by the wall on the ladder
• Resolving the vertical components of force, we get mg = f_y which is 20 * 9.8 = f_y . Hence, f_y = 196N
• We know that the algebraic sum of moment of forces about A = 0

Hence, F * BE = mg *AD

          F * (0.5) = 196 * (0.5)

          F = 196 N which is also f_x which is the force exerted by the wall on the ladder

So the resultant is f = (f_x² + f_y²)^1/2 =  [ (196)² + (196)²]² = 196 √2 N which is the force exerted by the ground on the ladder