Question

An insect $8m$ away from the foot of a lamp post which is $6m$ tall, crawls towards it. After moving through a distance, its distance from the top of the lamp post is equal to the distance it has moved. How far is the insect away from the foot of the lamp post? [Bhaskaracharya's Leelavathi]

Answer 1

Distance between the insect and the foot of the lamp post $=BD=8m$.
The height of the lamp post $=AB=6m$.
After moving a distance, let the insect be at $C$,
Let $AC=CD=xm$.
$\therefore BC=(8-x)m$.
In $△ABC$, $\angle B={90}^o$
$\therefore {AC}^2={AB}^2+{BC}^2$ ($\because$ Pythagoras theorem)
$x^2=6^2+{(8-x)}^2$
$x^2=36+64-16x+x^2$
$\therefore 16x=100$
$x=\frac{100}{16}=6.25$ Note: We can also consider $BC=x$, then $CD=AC=(8-x)$
$\therefore BC=8-x=(8-6.25)=1.75m$
The insect is $1.75m$ away from the foot of the lamp post.