Q. 65.0( 1 Vote )

# 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

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