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A ladder 17m long reaches a window which is 8 m above the ground on one side of street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window at a height of 15 m. Find the width of street.

Answer:

Let AB is the street and C be the foot of the ladder. Let D and E be windows at heights of 8 m and 15 m respectively from the ground. Then CD and CE are the positions of ladder. From the right angle ΔDAC, By Pythagoras Theorem, \[C{{D}^{2}}=A{{C}^{2}}+A{{D}^{2}}\] \[A{{C}^{2}}=C{{D}^{2}}A{{D}^{2}}\] \[=\text{ }{{17}^{2}}{{8}^{2}}\] = 289 ? 64 = 225 \[AC=\sqrt{225}=15m\]                                                                                                             Again from right ∆EBC, by Pythagoras Theorem, \[C{{E}^{2}}=B{{C}^{2}}+B{{E}^{2}}\] \[B{{C}^{2}}=C{{E}^{2}}B{{E}^{2}}\] \[={{17}^{2}}{{15}^{2}}\] = 289 ? 225 = 64 \[BC\text{ }=\text{ }\sqrt{64}\text{ }m\] = 8 m ∴ Width of street, AB = AC + BC                                                                                                                                     = 15 + 8 = 23 m