Question

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. k to task 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. ​

Answer 1

21cm

Explanation:

Pythagoras Theorem for right angled triangle a² = b² + c²

hypotenuse² = opposite ² + adjacent ²

Call vertical height of wall (where ladder meets it) H

call ladder L and call ground distance G

50cm = 0.5m

3² = 0.5² + H², H² = 3² - 0.5² = 9 - 0.25 = 8.75. H = √8.75.

length of ladder is always 3m.

After ladder has slipped, G is now 0.5 + 0.7 = 1.2m.

H² = 3² - 1.2²

= 9 - 1.44

= 7.56

H = √7.56

H has changed from √8.75 to √7.56

It has slipped (√8.75 - √7.56)m

= 0.208m

= 20.8 cm

=21cm to nearest cm.