A ladder 15 m long reaches a window which is 9 m above the ground on one side of the street. Keeping its foot at the same point, the ladder is turned to the other side of the street to reach a window 12 m high. Find the width of the street.

From figure,

Let width of the street be AB = AC + BC

Let CD and CE be the ladder at different positions.

By Pythagoras theorem,

Hypotenuse² = Perpendicular² + Base²

In triangle ADC,

⇒ CD² = AD² + AC²

⇒ 15² = 9² + AC²

⇒ 225 = 81 + AC²

⇒ AC² = 225 - 81

⇒ AC² = 144

⇒ AC = $\sqrt{144}$

⇒ AC = 12 m

In triangle BCE,

⇒ CE² = BE² + BC²

⇒ 15² = 12² + BC²

⇒ 225 = 144 + BC²

⇒ BC² = 225 - 144

⇒ BC² = 81

⇒ BC = $\sqrt{81}$

⇒ BC = 9 m

AB = AC + BC = 12 + 9 = 21 m.

Hence, the width of the street is 21 m.