Question

a ladder is placed so as to reach a Window 63 dm high the ladder is then turned over to the opposite side of the street and is found to reach a point 56 dm high if the ladder is 65 dm long find the width of the street

Answer 1

Given:

The length of the ladder = 65 dm

The ladder is first placed to reach a window at a height of 63 dm and then to another window at a height of 56 dm which is to the opposite of the street.

To find:

The width of the street

Solution:

Referring to the figure let's make some considerations,

AB = height of the first window = 63 dm

AE = DE = length of the ladder = 65 dm

CD = height of the second window = 56 dm

BC = width of the street

Now,

Considering ΔABE, we have

Base = BE

Hypotenuse = AE

Perpendicular = AB

Applying Pythagoras theorem, we get

Hypotenuse² = Perpendicular² + Base²

⇒ AE² = AB² + BE²

⇒ 65² = 63² + BE²

⇒ BE² = 65² - 63²

⇒ BE² = 4225 - 3969

⇒ BE = $\sqrt{256}$

⇒ BE = 16 dm

Considering ΔDCE, we have

Base = CE

Hypotenuse = DE

Perpendicular = CD

Applying Pythagoras theorem, we get

DE² = CD² + CE²

⇒ 65² = 56² + CE²

⇒ CE² = 65² - 56²

⇒ CE² = 4225 - 3136

⇒ CE = $\sqrt{1089}$

⇒ CE = 33 dm

∴ The width of the street, BC = BE + CE = 16 + 33 = 49 dm

Thus, the width of the street is 49 dm.

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