A ladder is lying against a wall which is 5 metres high. If the ladder slips 2 metres away from the wall, the top of the ladder touches the foot of the wall. The length of the ladder is

Let us draw the diagram of the ladder against the wall as given in the question.

Let the length of the ladder be t metres.

We can find BC, i.e the distance of the ladder from the foot of the wall using Pythagorus theorem

$$\ BC^2=\ t^2-5^2$$.,

=>$$\ BC^2=\ t^2-25$$

=> BC=$$\ BC^{ }=\ \sqrt{\ t^2-25}$$

Now, when the ladder is slips 2 m away from the wall, the ladder lies on the ground completely. This can be depicted as:

Now, BC+ BC'= t 

=> $$\ \ \sqrt{\ t^2-25}+2=t$$

=> $$\ \ \sqrt{\ t^2-25}=t-2$$

Squaring both sides,

$$\ \ t^2-25=\left(t-2\right)^2$$

=> $$\ \ t^2-25=\left(t^2-4t+4\right)$$

=>$$4t=29.\ And\ so,\ t=7.25$$