A 2.6 metres long rod leans against a wall, its foot 1 metre from the wall. When the foot is moved a little away from the wall, its upper end slides the same length down. How much farther is the foot moved?

Length of rod = 2.6 m

Floor Base = 1 m

Height of wall = ?

Now using Pythagoras Theorem

(Floor Base)² + (Height of wall)² = (Length of rod)²

(Height of wall)² = (Length of rod)² – (Floor Base)²

⇒ Height of the wall = √((Length of rod)² – (Floor Base)² )

⇒ Height of the wall = √((2.6)² + 1²)

using the formula

⇒ Height of the wall = √((2.6 + 1) (2.6 – 1)

= √(3.6 × 1.6)

⇒ √5.76

= 2.4

now we got the height of the wall but in the question it is given that it is further slided down, therefore we consider that as x.

Length of rod remains same.

Floor base becomes (1 + x)

Height of wall becomes (2.4 – x)

Now using Pythagoras theorem again

⇒ (floor base)² + (height of wall)² = ( lengthof rod)²

⇒ (2.4 – x)² + (1 + x)² = 2.6²

⇒ 5.76 + x² – 4.8 + 1 + x² + 2x = 6.76

⇒ 6.76 + 2x² – 2.8x = 6.76

⇒ 2x² – 2.8x = 6.76 – 6.76

⇒ 2x² – 2.8x = 0

⇒ 2x(x – 1.4) = 0

⇒ (x – 1.4) = 0

∴ x = 1.4