Question

A ladder rests against a vetical wall at a height of
42 m from the ground with its foot at a distance of
9 m from the wall on the ground. If the foot of the
ladder is shifted 3 m away from the wall, how much
lower will the ladder slide down ?​

Answer 1

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★ Point to remember:-

→ This is similar to a right angle triangle, with the length of the ladder being the hypotenuse and the point at which the top end rests on the wall being one side and the point at which it touches the ground from the wall being the other side.

→ We can find the length of the ladder using Pythagoras theorem.

★ Length of the ladder:-

→ $\sqrt({12}^{2} + {9}^{2})$

→ $\sqrt{(144 + 81)}$

→ $\sqrt{225} = 15$

Therefore, if the ladder slips by 3 m, then the base becomes 12m.

Since, $9^{2} + 12^{2} = 15^{2}$

Thus, the ladder slides down by 3m on the wall.

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Answer 2

Imagine the situation as a right angled triangled where height of wall ( perpendicular ) is 42 m and distance of ladder from wall ( base ) is 9 m.

So by Pythagoras theorem,

Hypotenuse² = Base² + Perpendicular²

☛ Hypotenuse² = 9² + 42²

☛ Hypotenuse² = 81 + 1764

☛ Hypotenuse² = 1845

or Hypotenuse = 42.95 m

Now, The foot of the ladder is shifted away from the wall 3 meters.

So,

Base = 3 + 9 = 12 m

Now, By Pythagoras Theorem:

Hypotenuse² = Base² + Perpendicular²

☛ 1845 = 12² + Perpendicular²

☛ Perpendicular² = 1845 - 144

☛ Perpendicular² = 1701

or Perpendicular = 41.24 m

Slide down of ladder = Previous Perpendicular - Current Perpendicular

☛ 42 - 41.24

☛ 0.76 m

Or 76 cm