Question

A ladder $25m$ long is leaning against a wall that is perpendicular to the level ground. The bottom of the ladder is $7m$ away from the base of the wall. If the top of the ladder slips down $4m$, how much will the bottom of the ladder slip?

• A. $7m$
• B. $8m$
• C. $10m$
• D. $15m$

Answer 1

The correct option is B $8m$

R.E.F first image

In $\Delta ABC,$

$A{C}^2=A{B}^2-B{C}^2$ {Using Pythagorus theorem}

$\Rightarrow AC=\sqrt{{25}^2-7^2}$

$\Rightarrow AC=24m$

As after sliping length of ladder remains same $\Rightarrow DE=25cmandAD=4m$ {REF second Image}

Let say $x$ be the length of how much the ladder slips.

New Height $=DC=AC-AD=20m;EC=(x+7)m$

In $\Delta DEC$,

$D{E}^2=D{C}^2+E{C}^2$

$\Rightarrow {25}^2={20}^2+(x+7{)}^2\Rightarrow (x+7{)}^2=225$

$\Rightarrow x+7=15(\because (x+7)>0$ , We can't take negative value $)\Rightarrow x=8m$