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 $\sqrt{7}$ m
• B. 70$\sqrt{7}$ m
• C. 70 m
• D. 70$\sqrt{11}$ m

Answer 1

The correct option is C

70 m

Consider the above 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 the half of the ladder AC]

$\therefore$ By Pythagoras Theorem in $△$ABD,

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

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

⇒$AD=10\sqrt{7}$$\sqrt{2}$

Since AD is half of the ladder, the complete length of the ladder

AC = 2AD = 20 $\sqrt{2}$$\sqrt{7}$

Now,
In ΔABC,
${AC}^2$ = ${AB}^2$ + ${BC}^2$

⇒ ${BC}^2={AC}^2-{AB}^2$
$={\left(20\sqrt{7}\sqrt{2}\right)}^2$ - $({10\sqrt{7})}^2$
$=\left({10\sqrt{7})}^2\right[{\left(2\sqrt{2}\right)}^2-1]$
= $({10\sqrt{7})}^2\times 7$

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