Question

A ladder is resting on a wall of height $10\sqrt{7}m$ such that the foot of the ladder when placed $10\sqrt{7}m$ away from the wall, half of the ladder is extending above the wall. When the tip of the ladder is placed on the tip of the wall, how far is the foot of the ladder from the wall?

• A. $100\sqrt{7}$
• B. $70\sqrt{7}$
• C. $70$
• D. $70\sqrt{7}$

Answer 1

The correct option is C

$70$

Consider the following figure:

$AD$ is the part of the ladder in the first case and $AC$ is the ladder in the second case

$AB=BD=10\sqrt{7}m$ [given]

$2AD=AC$ [since, $AD$ is half of the ladder $AC$]

Therefore, by pythagoras theorem in $△ABD$,

${AD}^2={AB}^2+{BD}^2$

$\Rightarrow {AD}^2=({10\sqrt{7})}^2+({10\sqrt{7})}^2=2({10\sqrt{7})}^2$

$\Rightarrow AD=10\sqrt{14}$

Since, $AD$ is half of the ladder, the total length of the ladder,

$AC=2AD=20\sqrt{14}$

Now, in $△ABC,{AC}^2={AB}^2+{BC}^2$

$\Rightarrow {BC}^2={AC}^2-{AB}^2={\left(20\sqrt{14}\right)}^2-({10\sqrt{7})}^2$

$=\left({10\sqrt{7})}^2\right[{\left(2\sqrt{2}\right)}^2-1]=({10\sqrt{7})}^2\times 7$

$\Rightarrow BC=10\sqrt{7}\times \sqrt{7}=70m$