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A pole 23 m long reaches a window which is $$3\sqrt 5 \,{\text{m}}$$ above the ground on one side of a street. Keeping its foot at the same point, the pole is turned to the other side of the street to reach a window $${\text{4}}\sqrt {15} \,{\text{m}}$$ high. What is the width (in m) of the street?
Answer & Solution
Correct Answer:
Option
C
In ΔABC,
By pythagoras theorem-
(BC)2 = (23)2 - ($$3\sqrt 5 $$ )2
(BC)2 = 529 - 45
(BC)2 = 484
BC = 22
Again ΔCDE,
(CD)2 = (23)2 - ($${\text{4}}\sqrt {15} $$ )2
(CD)2 = 529 - 240
(CD)2 = 289
CD = 17
Then width of the street BD = BC + CD
= 22 + 17
= 39
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