Question

The foot of a ladder is $$6$$m away from a wall and its top reaches a window $$8$$m above the ground. If the ladder is shifted in such a way that its foot is $$8$$m away from the wall, then to what height does its top reach?

Answer 1

Given: Let say Wall height $$= AC=8\space\mathrm{m}$$ and Distance of ladder's foot from the wall $$= BC=6\space\mathrm{m}$$

Using the Pythagoras theorem, we get

$$(AB)^2=(AC)^2+(BC)^2$$

$$\Rightarrow (AB)^2=(8)^2+(6)^2$$

$$\Rightarrow (AB)^2=64+36$$

$$\Rightarrow (AB)^2=100$$

$$\Rightarrow (AB)=\sqrt {100}$$

$$\Rightarrow AB=10$$

$$\therefore$$ Length of the ladder $$=10\space\mathrm{m}$$

When the ladder is shifted:

Let the height of the ladder after it is shifted be $$H\space\mathrm{m}$$ w.r.t. ground.

Then by using the Pythagoras theorem, we can find the height of the ladder after it is shifted.

$$8^2+H^2=10^2$$

$$\Rightarrow H^2=10^2-8^2$$

$$\Rightarrow H^2=100-64$$

$$\Rightarrow H^2=36$$

$$\Rightarrow H=\sqrt{36}$$

$$\Rightarrow H=6$$

Thus, the height of the ladder is $$6\space\mathrm{m}$$.