Question

Foot of a 10 m long ladder leaning against a vertical wall is 6 m away from the base of the wall. Find the height of the point on the wall where the top of the ladder reaches.

Answer 1

Let AB be a vertical wall and AC = 10 m is a ladder.

The top of the ladder reaches to A and distance of ladder from the base of the wall BC is 6 m.

In right-angled ΔABC,

`AC^2 = AB^2 + BC^2`   ......[By pythagoras theorem]

⇒ `(10)^2 = AB^2 + (6)^2`

⇒ `100 = AB^2 + 36`

⇒ `AB^2 = 100 - 36 = 64`

∴ `AB = sqrt(64)` = 8 cm

Hence, the height of the point on the wall where the top of the ladder reaches is 8 cm.