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At the start of the day, a painter rested a

3 m ladder against a vertical wall so that
the foot of the ladder was 50 cm away
from the base of the wall.
During the day, the ladder slipped down
the wall, causing the foot of the ladder to
move 70 cm further away from the base
of the wall.
How far down the wall, in centimetres, did
the ladder slip?
Give your answer to the nearest 1 cm.

1 Answer

4 votes

The ladder initially slipped down approximately 295.84 cm. After moving the foot 70 cm further away, the final position resulted in a slip of approximately 8.27 cm down the wall.

The scenario involves a right-angled triangle formed by a 3 m ladder, a vertical wall, and the ground. Initially, the ladder, the hypotenuse, measures 3 meters, with its foot 50 cm away from the wall's base. Let's denote the distance down the wall as "x".

By the Pythagorean theorem, initially:

x^2 + 50^2 = 300^2

As the ladder slips, the foot moves 70 cm further away, making the new distance 120 cm:

(x + 70)^2 + 50^2 = 300^2

Solving these equations gives "x".

Firstly, solving the initial equation:

x^2 + 50^2 = 300^2

x^2 + 2500 = 90000

x^2 = 87500

x is approximately 295.84

Now, solving the final equation:

(x + 70)^2 + 50^2 = 90000

(295.84 + 70)^2 + 2500 = 90000

x^2 + 4900 + 2500 = 90000

x^2 + 7400 = 90000

x^2 = 82600

x is approximately 287.57

The ladder slipped approximately 295.84 - 287.57 ≈ 8.27 cm down the wall.

User Hiren Rathod
by
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