A 13 m-long ladder reaches a window of a building 12 m above the ground. Determine the distance of the foot of the ladder from the building.

Let AC be the position of a window from the ground and BC be the ladder, then the height of the window, AC =12m and length of the ladder, BC = 13m

Let AB = x m be the distance of the foot of the ladder from the base of the wall.

In ∆CAB, using Pythagoras Theorem,

(Perpendicular)² + (Base)² = (Hypotenuse)²

⇒ (AC)² + (AB)² = (BC)²

⇒ (12)² + (AB)² = (13)²

⇒ (AB)² = (13)² – (12)²

⇒ (AB)² = (13 – 12)(13+12)

[∵ (a² – b²)=(a+b)(a – b)]

⇒ (AB)² = (1)(25)

⇒ (AB)² = 25

⇒ AB = ±5

⇒ AB = 5 [taking positive square root]

Hence, the distance of the foot of the ladder from base of the wall is 5m