A ladder 40 m long reaches a window which is 24 m above the ground on one side of a street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 32 m high. Find the width of the street.
Answer:
56 m
- Let AB be the street and let C be the foot of the ladder. Let D and E be the given windows such that AD=24 m and BE=32 m.
Then, CD and CE are the two positions of the ladder.
Clearly, ∠CAD=90∘, ∠CBE=90∘ and CD=CE=40 m.
From right ΔCAD, we have CD2=AC2+AD2[ By Pythagoras' theorem ]⟹AC2=CD2−AD2=[(40)2−(24)2] m2=[1600−576] m2=1024 m2⟹AC=√1024=32 m. - From right ΔCBE , we have CE2=CB2+BE2[ By Pythagoras' theorem ]⟹CB2=CE2−BE2=[(40)2−(32)2] m2=[1600−1024] m2=576 m2⟹CB=√576=24 m.
- Therefore, width of the street = AC+CB = 32 m+24 m=56 m.