Q4 of 95 Page 5

A ladder 10 m long just reaches the top of a building 8 m high from the ground. Find the distance of the foot of the ladder from the building.


Let AC be the top of the building from the ground and BC be the ladder, then the height of the building, AC = 8m and length of the ladder, BC = 10m


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


In ∆CAB, using Pythagoras Theorm,


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


⇒ (AC)2 + (AB)2 = (BC)2


⇒ (8)2 + (AB)2 = (10)2


⇒ (AB)2 = (10)2 – (8)2


⇒ (AB)2 = (10 – 8)(10+8)


[∵ (a2 – b2)=(a+b)(a – b)]


⇒ (AB)2 = (2)(18)


⇒ (AB)2 = 36


⇒ AB = ±6


⇒ AB = 6 [taking positive square root]


Hence, the distance of the foot of the ladder from building is 6m


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