Q2 of 95 Page 5

A ladder 26 m long reaches a window 24 m above the ground. Find the distance of the foot of the ladder from the base of the wall.


Let AC be the position of a window from the ground and BC be the ladder, then the height of the window, AC =24m and length of the ladder, BC = 26m


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


In ∆CAB, using Pythagoras Theorm,


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


⇒ (AC)2 + (AB)2 = (BC)2


⇒ (24)2 + (AB)2 = (26)2


⇒ (AB)2 = (26)2 – (24)2


⇒ (AB)2 = (26 – 24)(26+24)


[∵ (a2 – b2)=(a+b)(a – b)]


⇒ (AB)2 = (2)(50)


⇒ (AB)2 = 100


⇒ AB = ±10


⇒ AB = 10 [taking positive square root]


Hence, the distance of the foot of the ladder from base of the wall is 10m


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