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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.

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

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