A 5 m ladder reaches a wall at height 4 m. If the foot is moved \(1.6\,\text{m}\) towards the wall, by how much does the top slide up?
\(0.8\,\text{m}\)
Step 1: The ladder, the wall, and the ground form a right-angled triangle. - Hypotenuse (ladder) = \(5\,\text{m}\) - Vertical height = \(4\,\text{m}\) - Horizontal base distance = ?
Step 2: Use Pythagoras theorem: \(\text{Base}^2 = (\text{Hypotenuse})^2 - (\text{Height})^2\) \(\text{Base}^2 = 5^2 - 4^2 = 25 - 16 = 9\) \(\text{Base} = 3\,\text{m}\).
Step 3: Now the foot of the ladder is moved closer to the wall by \(1.6\,\text{m}\). New base = \(3 - 1.6 = 1.4\,\text{m}\).
Step 4: Again apply Pythagoras theorem to find the new height: \(\text{Height}^2 = (\text{Hypotenuse})^2 - (\text{Base})^2\) \(\text{Height}^2 = 25 - (1.4)^2 = 25 - 1.96 = 23.04\) \(\text{Height} = \sqrt{23.04} = 4.8\,\text{m}\).
Step 5: The ladder was initially reaching 4 m high. Now it reaches 4.8 m. So, the rise of the ladder’s top = \(4.8 - 4 = 0.8\,\text{m}\).
Final Answer: The top of the ladder slides up by \(0.8\,\text{m}\).