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.

A 24 m 32 m C B E D 40 m 40 m


Answer:

56 m

Step by Step Explanation:
  1. 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=CD2AD2=[(40)2(24)2] m2=[1600576] m2=1024 m2AC=1024=32 m.
  2. From right ΔCBE , we have CE2=CB2+BE2[ By Pythagoras' theorem ]CB2=CE2BE2=[(40)2(32)2] m2=[16001024] m2=576 m2CB=576=24 m.
  3. Therefore, width of the street = AC+CB = 32 m+24 m=56 m.

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