Question

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

Answer 1

Let $AB=A^{\prime }{B}^{\prime }$ be the length of the ladder. Let the window be at point $A$ such that $AC=8m$ and $BC=8m$

Since, wall be perpendicular to the ground, therefore,

In right $\Delta ABC$, By Pythagoras Theorem,

$\Rightarrow AB=\sqrt{A{C}^2+B{C}^2}$

$\Rightarrow AB=\sqrt{8^2+6^2}$

$\Rightarrow AB=\sqrt{100}=10$

$\therefore AB=10m$

Now, $A^{\prime }{B}^{\prime }=10m$ and $B^{\prime }C=8m$

Again, in $\Delta A^{\prime }{B}^{\prime }C$, By Pythagoras Theorem,

$\Rightarrow A^{\prime }C=\sqrt{A^{\prime }{B}^{\prime 2}-B^{\prime }{C}^2}$

$\Rightarrow A^{\prime }C=\sqrt{{10}^2-8^2}$

$\Rightarrow A^{\prime }C=\sqrt{36}=6$

$\therefore A^{\prime }C=6m$