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?

A. 17

B. 35

C. 39

D. 22

Answer: Option C

Solution (By Examveda Team)

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